There were many excellent comments on yesterday’s Bayesian Riddle. Here’s what I believe is the simplest and most natural analysis.
First, let’s recall the problem:
A murder has been committed. The suspects are:
- Bob, a male smoker.
- Carol, a female smoker.
- Ted, another male smoker.
- Alice, a female non-smoker.
You are quite sure that one (and only one) of these suspects is the culprit. Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker.
Now your crack investigative team, in which you have total confidence, reports that, on the basis of new evidence, they’ve determined that the culprit is definitely female.
Who’s the most likely culprit, and with what probability?
Notice that if you considered all the suspects equally likely, your estimate would have been three to one for a smoker. Since you estimated only 2-to-1, you must have believed that the individual smokers were less likely than average to be guilty. So when you find out the culprit is female, it’s the female non-smoker — that is, Alice — who is now your prime suspect.
To quantify this, 2/3 of the original probability weight is on the three smokers, which means they each have a 2/9 chance of being guilty. 1/3 of the weight is on Alice, the non-smoker. A ratio of 1/3 to 2/9 is the same as a ratio of 3 to 2. So when you narrow things down to one smoker and one non-smoker, the odds are 3 to 2 for the non-smoker. That means there’s a 60% chance your culprit is Alice.
It is, of course, in the nature of these puzzles that one can get pretty much any answer one wants by introducing unnatural assumptions. If you’d already excluded Bob before you came up with your 2-to-1 estimate, then the others all have equal probability weight, so after Ted is excluded, the odds for Alice are 50/50. For that matter, maybe you’d already excluded Carol, in which case, after the investigators rule out the boys, the odds for Alice are 100%. Or maybe the investigators concluded the killer was a woman because they knew the killer was Carol. (Or, as one commenter cleverly suggested, maybe they concluded the killer was a woman because they found a cigarette with red lipstick, which causes you to update the odds that the killer was a smoker.) Et cetera. But all of these stories assume that I withheld critical information about how you and your investigators reached your conclusions. So unless you suspect me of playing unfair, the answer is Alice with a probability of 60%.
Your solution is not correct, or if it is then the riddle is not very Bayesian. Pro tip: If the order in which you receive evidence changes your answer, then something has gone very wrong.
It depends what you mean by ” the odds are 2-to-1 that the culprit is a smoker”. If you mean the chances for these 3 people against that 1 person is 2 to 1 your argument is correct. but if it means that the evidence for example a sample from the scene shows that there is a 2-to-1 chance that the culprit is smoker then I think your answer doesn’t seem correct.
Max, the order of the evidence doesn’t matter.
What does matter is that one of the pieces of evidence is “Of the pool of these 4 suspects, there’s a 2/3 probability of the culprit being a smoker”. Even if you give that piece of evidence last, it’s still evidence that pertains to the pool of 4 suspects; you can’t suddenly take that 2/3 probability and use it on a different pool of suspects, just because you’ve already eliminated some of them for unrelated reasons.
You could eliminate Bob and Ted first, then discover that of the original 4 there was a 2/3 probability of the culprit being a smoker, do the same sums that Steve has, and still get to the same answer.
“After carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker, even though you currently have no suspects that smoke.”
You can make this statement in the real world, but not in a Bayesian problem. Right? The Bayesian odds would have to be 0% chance of the culprit being a smoker if I have no suspects that smoke. Otherwise, the original problem would be ambiguous.
In other words those original 2-1 odds are tied to the original suspect population and that population only.
The confusion arises with the phrase “the odds are”. You have already established that there are only 4 suspects. Then studying evidence of an unkown nature, you decide the odds are 2 to 1 that it is a smoker. One way you may have arrived at this conclusion is that you discover the probability of it being a non-smoker is greater than it being a smoker. When you apply this in a Baysian way, you arrive at the odds of 0.22 for each smoker and 0.33 for the non-smoker, leading to your conclusion that the odds are 2 to 1 it is a smoker.
This is quite different from knowing the probability of it being a smoker is 0.666. The probability of it being one of the four is 1. The probability of it being a woman is 1.
Looking carefully, I believe your interpretation is the most correct from the way the problem is phrased.
@Max #1: I’d imagine Steve’s rejoiner to the order mattering is that if you got your team’s information about Bob and Ted’s innocence then your analysis of the other evidence would not lead you to a 2:1 prediction.
Imagine for instance you know it was one of the 4 people and they used one of 3 weapons. Gun, rope or candlestick. Each are equally likely.
But maybe you concluded that it couldn’t have been a smoker with a candlestick. Leaving you with 9 of your original 12 options. Then your team gets rid of another 4 of these leaving you with only 5 options. Alice with gun rope or candlestick or Carrol with gun or rope.
@Steve: I’m not 100% sure that the natural assumption is that all smokers are equally effected by pre-team evidence, although I realize this has alot to do with my/your aesthetics. Of course once you specify that you consider all smokers equally likely (2/9 chance) the puzzle kinda dies.
Max (#1):
If the order in which you receive evidence changes your answer, then something has gone very wrong.
Well then—-because, in this solution, the order in which you receive the evidence does not change your answer, we can, on Bayesian grounds, upgrade our confidence that something has gone very right.
Smylers (#3): Thanks for saving me the trouble of typing this, and for saying it better than I would have.
“You are quite sure that one (and only one) of these suspects is the
culprit. Moreover, after carefully examining the evidence, you’ve
concluded that the odds are 2-to-1 that the culprit is a smoker.”
So let’s break this down. You’ve examined the crime scene. You
know the victim isn’t a smoker. You found ashes at the scene.
You’re not sure that it’s tobacco, but you feel it’s likely.
Therefore, it’s more likely than not that the murder was a smoker.
Now, you find some more evidence that definitely eliminates two
suspects. Suddenly your physical evidence changes and it’s *not*
more likely than not that the murder was a smoker?
Ron: I start out with four suspects, of whom three are smokers. With no other information, then, I start out with the prior expectation that the odds are 3-to-1 the culprit is a smoker. After carefully examining the evidence, I update those odds to 2-to-1.
It must, then, have been the case that the evidence made the smokers seem *less* likely than before, not *more* likely. (Perhaps I found evidence that the culprit had especially clean breath.)
Had I discovered that same evidence *after* my investigators narrowed things down to Carol and Alice, it still would have led me to go with Alice as the likely culprit.
Just a small point on order. If you are using Bayes’s Theorem to update your probabilities, and there is good reason why you should in general, then the order cannot matter. This is because Bayes’s Theorme prescribes multpilying and dividing verious probabilities, and multiplication commutes. SO if you find the order matters you have either erred or decided to eschew Bayes’s Theorem.
Here is a scenario where Steve would be wrong (I think)…
If I determined before I knew ANY of my suspects — and surely not the actual number of suspects — that the odds were 2-1 that the murderer was a smoker. Let’s say I saw that, at the scene of the crime, the murderer had logged into the “exactly 2/3 of us are smokers” online community.
Then, when you finally tell me that we really have just two suspects, the chance that Alice did is just 33%.
Scott H: I agree with your analysis, but I disagree that you and I are addressing the same problem. The problem on the table specifies that you’re estimating the odds at 2-to-1 *in the presence* of your knowledge about who the suspects are.
So, do I understand correctly that the statement:
1) “…after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker”
is a red herring? That is, it’s sole purpose is to mislead.
The honest, fair thing to have stated is:
2) “…after carefully examining the evidence, you’ve concluded that the odds are 2/9 that Bob did it, 2/9 that Carol did it, 2/9 that Ted did it, and 3/9 that Alice did it”
But statement (2), along with subsequent information, makes getting the answer too easy.
How to camouflage (2)? Well, it turns out that, although smoking apparently has nothing to do with the crime, Bob, Carol, and Ted are smokers, but Alice is not. Therefore, while it’s completely irrelevant, it’s still a fact that the cumulative odds of the culprit being a smoker are 6/9, and 3/9 of not being a smoker, or 2:1 a smoker, thereby giving us red herring statement (1).
Statement (1) is meant to mislead us into thinking that smoking has something to do with the solving of the crime when in reality it doesn’t.
Bob Lince: No, there’s more to this.
You are asserting (I think) that the evidence that led to 2-to-1 odds must have had nothing to do with smoking. On the contrary, it might have been quite directly related to smoking — it might be, for example, evidence that eyewitnesses (or nosewitnesses) noticed no tobacco smell.
The Bayesian lesson is that if evidence directly related to smoking leads you to a 2-to-1 confidence level that the culprit smoked, it must have been evidence that any *individual* smoker is *less* likely to be the culprit.
This has important practical real-world consequences. I’m on trial. I happen to have naturally purple hair. A DNA expert testifies that the odds are a million-to-one that the culprit had naturally purple hair. Should the jury view that as evidence for or against me? Juries are misled by this kind of thing all the time.
In other words: Yes, the sole lesson of the puzzle is that it’s easy to be misled. But that’s a lesson that has significance far beyond this particular puzzle.
A similar problem, to illustrate:
An HIV test has a 5% false positive rate for subjects which are HIV negative.
Imagine you apply this test to some random person. It comes back positive. What is the likelihood that the person is actually HIV positive?
Answer: you don’t know.
If that person is a lesbian nun then the likelihood is high that this is a false positive, if the person is an uncircumsized male homosexual IV drug user from Sub-Saharan Africa with other STDs then the likelihood is very high that the positive result is correct.
Steve,
I think you need to reverse the order of the information as it comes in, and tell us the solution again using the numbers you provided.
And I will start it off with how I see it:
You have 4 suspects as you described them, and your detectives tell you that it is 100% probability that Alice or Carol is the killer. Then the detectives tell you that it is 66% sure that the killer is a smoker. Therefore, it is 66% sure that the killer is Carol.
The ball is in your court.
Yancey Ward (#18): When the detectives tell me it’s 66% sure that the killer is a smoker, do they already know that it’s been narrowed down to Alice and Carol?
What if they don’t? Your puzzle is ambiguous, at best.
@18:
If that was the order of the evidence, Steve would not say the odds are 2-1 that the killer was a smoker at that point. He would say the odds are 2-3 that the killer was a smoker, and that would be the end. He’s saying that the odds are based on the number of mutually exclusive, collectively exhaustive suspects remaining. “2-1” represented the posterior probabilities for each suspect after taking smoking status into consideration.
Yesterday I argued that it perhaps mattered WHY the odds were given as 2-1 that the culprit was a smoker. But now I see that the odds were definitely given taking into account the number of initial suspects. Intuitively, if a non-smoker had the highest odds among the four before the men were eliminated, it wouldn’t make any sense for the non-smoker to then have the lowest odds after men were eliminated just because they were men. In other words, if smoking or non-smoking made a suspect more likely, that would persist in the odds before and after the evidence eliminating suspects by gender, which is what happens in Steve’s example where Alice is the favorite before and afterwards. It helped me to think of it as a betting market.
The order of evidence does matter critically, but not in the way people think, and yet the problem is still very properly Bayesian. Here’s how:
Bayes’ Theorem is about the probability A is true, given the conditions x. But x is not the same when the “order” of evidence is reversed. The puzzle indeed yields 60% probability for Alice as constructed, but it is confusing to think that is also yields 66% probability (per Yancey above) when using first the data of female certainty, and then the smoker evidence data. That is because the smoker evidence data, as presented, pertains to a population of all smokers and non-smokers, and all females and males. There is no indication of what the smoker evidence might be when considering solely a population of females, but it is manifestly incorrect to use the 2:1 odds, which are for a different population. By the way, by inference it would be 3:2 odds that the culprit is a smoker, when the population under consideration is solely female. This yields a 60% probability for Alice, consistent with the other evidence presented.
#20 – I don’t think it is ambiguous, but it is very easy to misinterpret. The evidence that made you think it was 2 to 1 that it was a smoker may have had something to do with smoking, but it may have had nothing to do with smoking. We do not know. As long as it resulted in the odds mentioned, the statement “you decide that the odds of it being a smoker are 2 to 1” remains valid. The phrasing is very careful, but can easily be mis-read as meaning that the probability of it being a smoker was 0.666.
@Steve 15:
OK, you’ve confused me here. I’m not sure what point you are trying to make. The purple hair fact is incriminating in this instance. The odds of having naturally purple hair are about 7 billion to 1 against.
Dave,
I am not sure you can actually make the assumption you are making.
Let’s suppose the evidence of a smoker is of the nature that you found the murder weapon, and the trigger was covered in skin oil that contains a substance that is found in smokers two thirds of the time, and in non-smokers only one third of the time. With this information, you then do what Lieutenant Landsburg does with the 4 suspects- you assign probabilities to each suspect based on their smoking habits. You then find out that the skin oil also contains a substance that is found only in women. Steve’s answer is that the highest probability is that it is Alice.
Now, turn the problem around. You first learn of the women-only skin oil substance, and you eliminate Bill and Ted. You then learn of the smoker-linked skin oil substance. You must then say Carol is the highest probability suspect.
The issue comes down to what the nature of the evidence for it being a smoker is, and how that 2/3 probability was determined, and that is not specified in the problem.
I didn’t expect this answer. Yesterday I did it like this:
P(Alice) = 1/4
P(smoker) = 2/3
P(Alice|smoker) = 1/7
However, this is double-counting. I already know the suspect pool when I come up with my P(smoker) estimate, so by the time I report it I’ve already done the Bayesian update and P(smoker) = P(~Alice) = 2/3.
Then my investigators arrive with independent data that the murderer is truly a murderess with infinite probability, so the following Bayesian update doesn’t deserve the name because all the factors are just 1s and 0s.
So I agree that P(Alice) = 60% is the correct solution. However, since solving the problem didn’t actually involve using Bayes’ theorem I was greatly deceived by the title! Not that that’s a good excuse.
Reading through the comments on the previous thread, Dave, I see that you were constructing hypotheticals similar to mine.
Yancey Ward (#20): Whaddya mean my problem is ambiguous? My problem was perfectly well specified. Your problem introduced the ambiguity: What did the detectives know when they disxovered the smoking evidence?
Steve, see my #25.
I definitely see how, when the odds statement is interpreted as pertaining to the suspect pool, this answer is correct. But if I were trying to describe that situation, I would write, “Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is one of these smokers.”
Otherwise, I imagine that the evidence about smoking is probably forensic in nature, and [I feel like] the results of forensic analyses are usually reported without any knowledge of the suspect pool.
Exactly, Ben.
Yancey Ward (#25):
Let’s suppose the evidence of a smoker is of the nature that you found the murder weapon, and the trigger was covered in skin oil that contains a substance that is found in smokers two thirds of the time, and in non-smokers only one third of the time.
Okay, so what you’re telling me (writing E for the evidence and S for the event that the culprit is a smoker) is that P(E|S)=2/3 and P(E|not-S)=1/3.
Given my suspect pool, then, I have Prob(E)=(2/3+2/3+2/3+1/3)/4 = 7/12.
Now as a good Bayesian, I compute
P(S|E) = P(E|S)P(S)/P(E) = (2/3)(3/4)/(7/12) = 6/7
But it was given in the problem that P(S|E)=2/3.
Conclusion: Your hypothesis is not consistent with what’s given in the problem. In other words, you’ve changed the problem.
If the smoker evidence is truly independent of the suspect pool (like with Steve’s oil example), then the correct answer to the puzzle is Alice with 67% probability. That is, the existence of Bob and Ted is irrelevant, once there is evidence to definitively rule them out as suspects. Said differently, the correct suspect pool is the set of Alice and Carol, when taking all evidence under consideration. This makes both intuitive, as well as Bayesian sense.
Me: “You said this was a Bayesian problem but the answer didn’t involve Bayes’ theorem! Treachery!”
Steve: “The Way of Bayes is not a theorem; it’s a way of life, punk” (dons cloak and disappears into the night)
What if one does Not assume that all of the suspects are equally likely?
If for example, Carol and Alice are related to the victim and both would gain by the victim’s death and Bob and Ted are unrelated, but were otherwise possible suspects, then the probabilities for Bob and Carol and Ted may add up to 6/9 but not necessarily be equal to one another.
If Carol had a 5/9 probability and Bob and Ted each had a 1/18 probability, wouldn’t that affect the solution?
Please be gentle in your response, I figure the pre-response probability of me making a relevant comment on this issue is slim and the post response probability will be none.
Steve,
You have four suspects and one killer at the start, and the probability of the killers being for each of them is:
Ted = 1/4
Bill = 1/4
Carol= 1/4
Alice= 1/4
You then learn information X (undefined in your problem) that tells you:
Ted+Bill+Carol = 2/3
Alice = 1/3
Correct? I mean, this is the clear English interpretation of the problem you wrote out. Ted, Bill, and Carol are the smokers, and Alice is the non-smoker, and you have determined that the killer is going to come from the Ted+Bill+Carol group 2/3 of the time, and will be Alice 1/3 of the time. Indeed, Alice considered as an individual is more likely to be the killer than each of the other three considered alone. We should agree up to this point, right? Then your detectives come and tell you that the killer must be female, then the probabilities properly collapse:
Ted+Bill+Carol = 2/3 then becomes Carol = 2/3 and Alice = 1/3. Now we disagree, don’t we?
Ken B: It is only incriminating if we know the perpetrator had natural purple hair. Steve did not say this.
I always like to relate these problems back to the original Monty Hall formulation. Given 3 boxes, and one prize, I make the initial selection knowing that I will be right 1/3 of the time and wrong 2/3 of the time. You then show me one of the boxes that I did not select is empty, and then I properly change my selection of boxes knowing the other unopened box carries the 2/3 probability of having my prize. This is precisely (I am arguing) analogous to the last piece of information given in Steve’s problem- with the elimination of Ted and Bill, the entire probability that the killer comes from the Bill+Ted+Carol grouping collapses onto Carol.
@Yancey,
This isn’t Monty Hall.
Yancey (36): “then the probabilities properly collapse:
Ted+Bill+Carol = 2/3 then becomes Carol = 2/3”
Those probability didn’t collapse. If Ted+Bill+Carol = 2/3 then Carol = 2/9. Then the final odds are Carols 2/9 to Alice’s 3/9.
Martin-1 40
You state:”If Ted+Bill+Carol = 2/3 then Carol = 2/9.”
Why assume Eqaul probabilities? Are all suspects equally likely?
Bill (41): In our state of partial knowledge yes, they are equally likely.
Steve, here’s a Bayesian puzzle from LessWrong that might make you stumble. Keep in mind I’ve read all your TBQ posts.
If I meet a mathematician on the street, and she says, “I have two children, and at least one of them is a boy,” what is the probability that they are both boys?
Sorry Yancey, I just (38). Disregard my recent replies.
Yancey Ward (#36): First, since you’ve abandoned your #25 argument, can I presume you no longer stand by it?
Second, you are now assuming that when Bob and Ted disappear, all of the 2/3 probability collapses onto Carol, and none onto Alice. So you’re telling me that
Probability (Alice|Female) = Probability (Alice)
Now let’s be good Bayesians again:
Prob(Alice)=Prob(Alice|Female)=Prob(Female|Alice)Prob(Alice)/Prob(Female)=1 Prob(Alice)/Prob(Female)
Therefore Prob(Female)=1.
In other words, you’re (implicitly) assuming that I knew for certain the killer was female before my investigators brought me this info.
I was about to say that this might be technically within the letter of the problem but clearly in violation of its spirit. But on rereading, I see that it’s actually in violation of the letter as well, since the problem specifies that your crack investigative team brings you new information that the culprit is female.
Your story assumes that you knew along the culprit was female, so that info couldn’t have been new. In other words, once again, you’ve changed the problem.
@Harold 37: No. Steve told us the DNA makes it 1 million to one the killer had naturally purple hair. Ex ante it was 7 billion to one the killer had nph. The odds have diminished on the basis of the evidence; it’s incriminating.
Bill: As I said in the post, if you have additional information beyond what’s given in the problem, then you’ll of course get different answers. Interestingly, though, one answer you can’t get is the one Yancey’s pushing.
Martin-2: Of *course* I’m using Bayes’s Theorem:
With “S” meaning the culprit is a smoker, “F” meaning the culprit is female, and “E” meaning the initial evidence, I first note that Prob(F|E)= 5/9.
Now:
Prob(Carol|E+F)=Prob(S|E+F)
=Prob(F|S+E)Prob(S|E)/Prob(F|E)
=(1/3)(2/3)/(5/9)
=2/5
That’s Bayes’s Theorem in the transition to the second line.
Steve (45): The false (but compelling) equivalence to Monty Hall suggests a companion problem:
Monty has presented you with 3 doors, one of which conceals many utils. After you choose a door an errant gust of wind opens a different door and lo, behind stands a goat of 0 utils. Do you change your selection?
Seeing the difference from the original Monty Hall is an important Bayesian lesson.
Steve (40): “I first note that P(F|E)=”
Was this supposed to say P(F|E) = 1/2?
I am continually amazed at how contentious discussions of Bayes’s Theorem always get (and not just on this blog). People rant on for days about Monty Hall.
Martin-2: It was supposed to say P(F|E)=5/9. (I’ll fix this soon.) The point being that given the evidence, there’s a 2/9 chance it’s Carol and a 3/9 chance it’s Alice, hence a 5/9 chance it’s a female.
@Martin-2:
Let me take a stab at explaining how to get the numbers.
What we will do is fill an urn with balls labelled with names. We’ll do it in a way to be consistent with what we know.
We know that there is a 1/3 chance that the killer is a smoker.
So lets start with an urn with some multiple of 3 balls. Let’s say 18 for concreteness; we can adjust later.
6 of the balls are labelled Carol, since Carol is the only smoker.
Now we have 12 balls left over.
What do we know about how they are distributed? Not much. That’s the key actually: we are ignorant about how to label the rest.
That means we should assign an equal chance to each other option. There’s a solid reason for saying that.
The amount of uncertainty of a distribution of the probabilities can be defined and measured.
Shannon did it. And the in this case the distribution which maximizes our uncertainty is the uniform distribution.
So to reflect the fact we know nothing we should use it.
That means 4 balls with each other name.
Note how the balls represent waht we know: 6 Carol and 12 others.
Now we are told, on the basis of new information, the guys are innocent.
We can reflect this by removing their balls. (Snip)
So now what have we left? An urn with 10 balls: 6 Carols, 4 Alices.
And that urn captures all we know.
It gives the result of our calculations.
@Steve re 53: Oops I see I scrambeld Carol and Alice. (Well, in the movie so did Bob & Ted). I made Carol the non-smoker.
Anyway if Steve can fix this I’d appreciate it.
Ken: If you want to resubmit, I’ll then delete #53.
The fallacy in Steve’s argument is that he assumes the Principle of indifference. That critical info was withheld. The error is explained in Chapter IV of A Treatise on Probability by John Maynard Keynes.
Advo,
I think it is precisely like the Monty Hall problem and Steve is making the exact same error made by those who fail to understand how the probability collapses unto the un-chosen and unopened box. Given the initial distribution of the probability, we had:
Ted+Bill+Carol= 2/3
Alice = 1/3
This contains the totality of the information (evidence available). Now I come along and tell Steve that Ted and Bill were at a hotel at the time of the murder having a lovers’ tryst. This is the exact equivalent of Monty opening one of the unchosen boxes and showing me that it is empty. In the Monty Hall problem, my chances of winning with my initial selection is still 1/3, but is 2/3 if I change to the remaining unchosen box. The probability that the killer comes from the Ted/Bill/Carol grouping is still 2/3, even if I have removed two of the suspects from the pool.
Here is what Steve is doing in the Monty Hall problem:
I get four boxes and make my choice- box D, which Steve tells me has a likelihood of having the prize 1/3 of the time while the other three each have the probability of having the prize 2/9 of the time. I have made the rational decision, have I not? Then Steve comes to me and shows the the empty contents of boxes A and B, and then asks me if I want to stay with box D, or do I want to change. Of course, I change my selection since I still only have 1/3 of chance of winning, but I now know that Box C has 2/3 chance of winning. Steve would have you believe that Box D chances have increased because of the new information, but they have not.
@Yancey, the issue is what exactly the first evidence says. You’re interpreting it like I think it should be interpreted — that, apart from any composition of the suspects, any given smoker is twice as likely to be the culprit as any given non-smoker. Steve is saying the first evidence is specified knowing the composition of the suspects.
Let’s say I arrive at the scene and swab the victim. I send the swab off to a lab (which knows nothing about the case except the swab) and they report that particular markers they observed on the swab occur 1.5 times as frequently when a non-smoker is the killer than when a smoker is the killer. I therefore know that P(A) = 1.5*P(B) = 1.5*P(C) = 1.5*P(T). If P(B)=P(C)=P(T), I know that P(B)+P(C)+P(T)=2/3 and P(A)=1/3. Therefore, after carefully examining the evidence, I conclude that the odds are 2-to-1 that the culprit is one of these smokers.
I think as originally written, the odds statement more appropriately applies to number that is 1/1.5 in the above example. But I see how it could be argued that it applies to the conclusion in the above example as well. @Steve, can you acknowledge that it is at least somewhat reasonable to interpret the 2:1 odds statement as a swab result similar to the example above (therefore leading to the conclusion that it’s Carol with P=2/3)?
Roger (#56): I can only conclude that you didn’t read the post.
@Roger,
while I generally think this blog could use more Keynes, I really don’t agree with Keynes criticism of the principle of indifference. Or maybe I should say – his criticism may be valid for the application of the principle, but not for the principle in principle (snicker).
From his first two examples (color of the book and country of origin), I really don’t see how the book example makes any sense whatsoever, and on the second one, Keynes uses extraneous information (British Isles cover England and Ireland) to argue against the principle, which invalidates its premise, does it not?
In fact, I would go so far as to say that the latter example is total bullsh1t, which kind of surprises me from Keynes, but maybe I just don’t understand the example.
Given the information Steve supplied, I don’t see what one could do EXCEPT apply the principle of indifference. If there isn’t more information, there isn’t more information, and random guessing isn’t going to change that.
Yancey Ward, as you pointed out, I made a similar argument yesterday to what you are saying today.
“the trigger was covered in skin oil that contains a substance that is found in smokers two thirds of the time, and in non-smokers only one third of the time. With this information, you then do what Lieutenant Landsburg does with the 4 suspects- you assign probabilities to each suspect based on their smoking habits.”
The problem with your hypothetical (and mine from yesterday) is that with this information, the smokers would EACH have a higher probability than does Alice when the 2-1 odds are given. But Alice is the favorite.
In other words, if the oil is found in smokers two thirds of the time and non-smokers one third of the time*, why would Alice the non-smoker then be considered the relative favorite (highest probability in the group) to be the culprit in the initial “2-1” statement?
The problem with your reasoning here (and mine from yesterday) is that the evidence that first makes Alice the favorite based on smoking information is somehow supposed to make Carol the favorite after the men are eliminated, which makes no sense without other information being introduced. If non-smoking causes one suspect to be the initial relative favorite, and then two other suspects are cleared based on gender, it can’t logically follow that the existing evidence would then catapult a smoker into being the favorite because she is a smoker.
* and by the way, why would these add up to 1? If you thought they had to add to 1, then this is another source of confusion.
Ben:
Therefore, after carefully examining the evidence, I conclude that the odds are 2-to-1 that the culprit is one of these smokers.
Sure. Alternatively, you might as well assume that the only reason I adopted 2-to-1 odds is that I’m a madman, so there’s no information in those odds.
The presumption here is that I had a *valid* reason for putting the odds at 2-to-1. If I acted like the person in your comment, I’d be the functional equivalent of a madman.
@Yancey:
The difference between this and Monty Hall is that what subsequent information Monty Hall gives you depends on your first selection.
Monty Hall opens a door with a goat you HAVEN’T selected. If Monty Hall were just opening a door with a goat regardless of whether you had selected it or not, the odds of any of the remaining two doors being it would be 50-50 and not 2/3-1/3.
The Monty Hall-effect depends on there being a connection between your first selection and the subsequently provided information.
Steve’s subsequent information (female) has absolutely no connection with the first information (smoker).
@Advo: If you don’t have enough info to solve the problem, and you have to make stuff up, then the principle of indifference may be as good as anything else. If making an assumption helps you solve an unsolvable problem, then go ahead, but it is an extra assumption.
Smylers (#3):
“Of the pool of these 4 suspects, there’s a 2/3 probability of the culprit being a smoker”
You’ve put quotation marks around words that are not a quote. This is generally inadvisable and is particularly so here, because you have badly misrepresented the problem as originally presented. That the culprit is one of BCTA and that the odds are 2-to-1 that the culprit is a smoker were presented as independent facts with no verbal link between them whatsoever. Thus it is not correct to conclude what you have.
(1)
JonathanKariv (#6):
I’d imagine Steve’s rejoiner to the order mattering is that if you got your team’s information about Bob and Ted’s innocence then your analysis of the other evidence would not lead you to a 2:1 prediction.
Yes, that is a likely rejoinder based on my understanding of Steven’s probable brain state at the time of posing and solving this problem. But in that case, the following statement should have been qualified, and it was not: the odds are 2-to-1 that the culprit is a smoker. If this is not an absolute, then it ought not be presented as one. Otherwise you have a crummy problem.
Steve Landsburg (#7):
Well then—-because, in this solution, the order in which you receive the evidence does not change your answer, we can, on Bayesian grounds, upgrade our confidence that something has gone very right.
I understand what you think and why you think it. I simply disagree. I think you wrote the problem poorly and that my interpretation of it is the more natural one.
Ctrl-F’ing “Max” returns no more hits, so I’ll just skip the rest of the comments and hope I’m not simply repeating something someone else has already explained.
The solution to the problem hinges on whether the probability given that the culprit is a smoker is independent of the fact that there are 3 smokers and 1 nonsmoker amongst the set of suspects. If it is not, then Steven’s solution is correct. I think the plain language of the problem suggests that it was. Apparently, reasonable minds may differ.
Yancey Ward:
I’d like to understand what you’re thinking. For that, it would help a lot if you could tell me which (if any) of the following three statements you disagree with.
1) You’re on Let’s Make A Deal and choose Door 1. The host, as always, shows you a goat behind another door. Today he shows you Door 3. The probabilities are now 1/3 for Door 1 to have the prize and 2/3 for Door 2.
2) You’re on Let’s Make A Deal and choose Door 1. A random gust of wind comes along and reveals a goat behind Door 3. The probabilities are now 1/2 for Door 1 and 1/2 for Door 2.
3) The problem in this post is more closely analogous to case 2) than to case 1).
Dave (#62):
The problem with your reasoning here (and mine from yesterday) is that the evidence that first makes Alice the favorite based on smoking information is somehow supposed to make Carol the favorite after the men are eliminated, which makes no sense without other information being introduced.
Well put!
Fun puzzle. Thanks for posting this Steve. I am not even close to being in the same stratosphere mathematically as most of the people here, but your explanation was clear and easy to follow.
Jack: Thanks for these kind words. I very much appreciate them.
Max (#69):
The solution to the problem hinges on whether the probability given that the culprit is a smoker is independent of the fact that there are 3 smokers and 1 nonsmoker amongst the set of suspects….I think the plain language of the problem suggests that it was.
On your interpetation, then, I could go on believing the odds for a smoker are 2-to-1 even if there are *NO* smokers in my set of suspects, or even if every one of them is a smoker.
But that’s nuts. So when you say “Reasonable minds may differ”, all I can say in response is that if a reasonable mind took endorsed your position, it would have to be making an unreasonable mistake.
Here’s how I approached the puzzle.
1. I assumed that the evidence indicating a 2/3 chance that the culprit is a smoker is based on something unrelated to the initial set of suspects and independent of the second premise (that culprit is female) — for example, DNA or other physical evidence that make it twice as likely that the culprit is a smoker than a non-smoker among the general population.
2. The second premise tells us the culprit is one of the two women.
3. Only one of the two is a smoker.
3. We know that there’s a 2/3 chance that the culprit is a smoker.
4. That 2/3 chance falls entirely on the one woman who is a smoker.
Conclusion: The female smoker (Carol) is the most likely culprit, with a 2/3 chance that she is the culprit.
Of course, if my assumption (my #1 above) is invalid, that changes things, but in that case that part of the puzzle should have been stated explicitly.
Am I missing something in my reasoning above?
Obviously 3,3,4 should be 3,4,5.
Steve (#15):
You said,
I’m on trial. I happen to have naturally purple hair. A DNA expert testifies that the odds are a million-to-one that the culprit had naturally purple hair. Should the jury view that as evidence for or against me?
A million-to-one odds seem a near certainty. Given the rarity of other suspects with naturally purple hair, I’d say the evidence against you is damning. I hope you have a good alibi.
But what should we, the jury, think when your fast-talking defense attorney attempts to impeach the credibility of the DNA expert’s testimony? Maybe we think like this:
The statement, “The culprit had naturally purple hair” is logically equivalent to, “Everyone who does not have naturally purple hair is not the culprit.” So however convinced we are of the a priori truth of that statement, every time we see someone who does not have naturally purple hair and who we know is not the culprit, as good Bayesians we can regard the sighting as confirming evidence for the DNA expert’s claim. At least a little bit.
Is that a good Bayesian update? Or does it indicate something problematical about being a good Bayesian?
Steve Landsburg (#74):
On your interpetation, then, I could go on believing the odds for a smoker are 2-to-1 even if there are *NO* smokers in my set of suspects, or even if every one of them is a smoker.
No you couldn’t. The set of suspects must include both smokers and nonsmokers in order for any probability between 0 and 1 to make sense. This does not imply that the probability of a smoking killer is always dependent on the ratio of smokers to nonsmokers in the set of suspects. The only cases where that is necessarily true are where the set of suspects consists of all smokers or all nonsmokers. In any instance where a mixture exists, the probability may be a dependent or an independent variable. It is simply not correct to say that interpreting the variable as independent in one instance requires interpreting it as independent in all instances.
But that’s nuts. So when you say “Reasonable minds may differ”, all I can say in response is that if a reasonable mind took endorsed your position, it would have to be making an unreasonable mistake.
You do not understand my position. As a rationalist, I am open to the idea that I have made an unreasonable mistake, but I will rate the probability of this as very low until you can tell me what it is. The unreasonable mistake you believe you’ve identified is an artifact of your misinterpretation of my interpretation of the problem.
Re:#70
1) You’re on Let’s Make A Deal and choose Door 1. The host, as always, shows you a goat behind another door. Today he shows you Door 3. The probabilities are now 1/3 for Door 1 to have the prize and 2/3 for Door 2.
Agree
2) You’re on Let’s Make A Deal and choose Door 1. A random gust of wind comes along and reveals a goat behind Door 3. The probabilities are now 1/2 for Door 1 and 1/2 for Door 2.
I was about to respectfully disagree, arguing that it doesn’t matter whether door 3 was opened by a random gust of wind or a game show host, and that what matters is that a door we didn’t pick was opened, exposing a goat.
Then, I worked it out carefully, and you’re right.
As if the Monty Hall problem wasn’t hard enough to understand as it is!
Max 78:
“In any instance where a mixture exists, the probability may be a dependent or an independent variable.”
That’s not true.
It cannot be an independent variable because the probability makes a statement ABOUT THE SPECIFIC SUSPECT POOL.
The premise of the puzzle is unequivocally:
1. We have four suspects, B, C, T, A and we are certain that it couldn’t have been anyone else.
2. A is a non-smoker.
3. GIVEN THIS SUSPECT POOL, odds are 2:1 for a smoking killer.
You can’t say “odds are 2:1 for a smoking killer” without relating to the specific suspect pool.
UNLESS THE ODDS ARE 2:1 FOR THE SPECIFIC SUSPECT POOL, the odds can’t be 2:1 at all.
The confusion probably comes from the fact that the CSI-guy on TV wouldn’t make such a statement with regard to any particular suspect pool (which you rarely have that well-defined in practice), but rather with regard to the general population.
Ken B #53 (unless it gets deleted). I think your analysis is wrong, because it depends on the order. We start with no knowledge – lets put 24 balls in the bag and distribute equally between the 4 suspects – 6 each. We then find that the men are innocent. Remove their balls, snip. We have 12 balls left. We then apply the knoweledge that there is a 1/3 chance it is a non smoker and a 2/3 chance it is a smoker. We get 4 balls for Carol (the non-smoker) and 8 balls for Alice (the smoker).
The key point is that the 2/3 chance of it being a smoker is not independant – it depends on some other knowledge about these particular suspects. We cannot remove balls on the basis of this knowledge because it is a conclusion, not evidence, and only applies when these particular four are the only suspects. Some other evidence led us to the conclusion that *because* we know it was one of these four *and* we know about their smoking *then* there is a 2/3 chance it was a smoker.
Replacing 53.
@Martin-2:
Let me take a stab at explaining how to get the numbers.
What we will do is fill an urn with balls labelled with names.
We’ll do it in a way to be consistent with what we know.
This urn will be our computer, tracking our knowledge.
We know that there is a 1/3 chance that the killer is a smoker.
So lets start with an urn with some multiple of 3 balls. Let’s say 18 for concreteness; we can adjust later.
6 of the balls are labelled Alice, since Alice is the only non-smoker.
Now we have 12 balls left over.
What do we know about how they are distributed? Not much. That’s the key actually: we are ignorant about how to label the rest.
That means we should assign an equal chance to each other option. There’s a solid reason for saying that.
The amount of uncertainty of a distribution of the probabilities can be defined and measured.
Shannon did it. And the in this case the distribution which maximizes our uncertainty is the uniform distribution.
So to reflect the fact we know nothing we should use it.
That means 4 balls with each other name.
Note how the balls represent what we know: 6 Alices and 12 others.
(Note also that the 2/3 fraction is based on the assumption the men are still suspects.
If they weren’t we’d have different information expressed as different fractions.)
Now we are told, on the basis of new information, the guys are innocent.
We can reflect this by removing their balls. (Snip)
So now what have we left? An urn with 10 balls: 4 Carols, 6 Alices.
And that urn captures all we know.
It gives the result of our calculations.
I’ll try this once more…
The problem here is that the person in charge of this case is a
logician/statistician. A good detective or forensic doctor would
*never* let the suspect pool shade the odds exposed by testing and
observation.
E.g.: “The victim died of stab wounds.”
“But nobody in the suspect pool had access to a knife!”
“Then you better enlarge your suspect pool.”
In real life, “The odds are 2:1 that the culprit is a smoker”
is a statement about the evidence, not about the suspect pool, even
if the composition of the suspect pool is known.
The problem stated, “You are quite sure that one (and only one) of
these suspects is the culprit.” It’s entirely possible to be quite
sure and quite wrong. If this was supposed to be a stat/logic
problem, that should have been stated as, “One (and only one) of
the suspects is the culprit.”
@Roger: I think my post answers your concern, giving a rationale.
As you might have noticed, I agreed with you one the other thread. I wish SL would address those issues actually, which are more interesting than this one puzzle, BUT as a brain-teaser I think Steve got it right. We are given brand new information, and that implies, absent something else, independence. Perhaps that is just a convention of phrasing puzzles, but it is pretty near universal. If later Steve said, aha I forgot to tell you I saw Carol do it, then we’d all call that cheating.
Ron:
The problem here is that the person in charge of this case is a
logician/statistician. A good detective or forensic doctor would
*never* let the suspect pool shade the odds exposed by testing and
observation.
Really? Suppose I believe the odds that the culprit is a smoker are 2-to-1. Then I discover the suspect pool contains zero smokers. Does that mean that as a “good detective”, I have to go on believing the odds for a smoker are 2-to-1?
Or suppose I believe there’s a 50% chance that OJ did it and a 50% chance that Pierre Deligne did it. Then I discover that Deligne has an unshakable alibi, so he’s removed from the suspect pool. Would a “good detective” continue to believe that the probability for OJ is 50%?
@Harold: I address your error in a new comment in 82.
There is evidence about smoking. It comes to us as a fraction, and that fraction was arrived at under the assumption that the men were suspects. Lacking anything indication to the contrary we should assume it was based on the assumption each of the 4 is equally likely a priori. That’s another example of maximum uncertainty as I discusss in 82. Had the analysts who came up with the 2/3 fraction done so with the knowledge that the guys were out they would have come up with a different fraction. So your objection is wrong.
You can in fact work out what fraction they would have arrived at. Exercise for the reader!
Gordon Landwirth:
Am I missing something in my reasoning above?
Yes.
@Gordon Landwirth 75: Here is your error:
” I assumed that the evidence indicating a 2/3 chance that the culprit is a smoker is based on something unrelated to the initial set of suspects …” I discuss this in 86.
How do we know Bob & Ted are innocent? Well they have an interesting alibi: me.
Bob & Ted & I were playing an interesting game.
They each held two balls. Of the four balls they hold, two are gold and two are ping-pong balls.
At the end of the game everyone gets to keep any balls they hold.
I could not see the balls and get to demand balls from other players, twice.
I demanded a ball from Ted. Ted gave me a ball but he got to choose which.
Alas he gave me a ping pong ball.
Now I have to demand a second ball.
Whom should I ask?
It may help to clarify this puzzle if you consider what my answer would be if he gave me a gold ball.
Ken B. Your analysis is not wrong, but I interpreted it as meaning the 1/3 was a Baysian multiplier when it is not. There is other evidence that we do not know about that results in the 1/3. If it was, we could apply it in any order, and we can’t because we get a different answer. My apologies if you did not intend it to be read this way.
Putting it another way, the analysis is correct, but it can not be the way the answer was arrived at. It could have been arrived at by determining that a non-smoker was more likely to have been the perpetrator, and applying this probability to the pre-existing probabilities to arrive at 1/3 for the non-smoker. This would be consistent Baysian approach. By stating that it was a Baysian problem, it implies that this is the approach that should be taken, and caused confusion if the wording was not examined closely. out of interest, what would the probability of it being a non smoker need to be to arrive at the 1/3 odds?
I need to re read more. In 89 I meant Bob and Ted each hold 2, and there are 2 gold and 2 ping pong, but I do NOT know how they are distributed. My bad.
Steve:
You seem to be falling into “the usual suspects” fallacy.
“Suppose I believe the odds that the culprit is a smoker are 2-to-1.
Then I discover the suspect pool contains zero smokers. Does that
mean that as a “good detective”, I have to go on believing the odds
for a smoker are 2-to-1?”
Your suspect pool doesn’t change the facts. If you’ve run a test
that shows 2:1 likelihood of the perp being a smoker, then one of
two things is true:
1. You’ve hit one of those cases where the suspect defies the odds.
or
2. You need to enlarge your suspect pool.
Or suppose I believe there’s a 50% chance that OJ did it and a 50%
chance that Pierre Deligne did it. Then I discover that Deligne has
an unshakable alibi, so he’s removed from the suspect pool. Would a
“good detective” continue to believe that the probability for OJ is
50%?
No, of course not. These odds are your own estimate, not the result
of, for instance, a forensic test. However, they don’t collapse
into 100% certainty for OJ, either. There still remains the
possibility that the one-armed man did it.
#89 – impossible to say -are you perhaps excessively interested in ping-pong?
So, Steve, you are claiming the detectives who come to you and conclusively eliminate the men are a gust of wind?
Let me restate the last one within the confines of my restating of your problem- the detectives, when they reveal the contents of boxes A and B, do so with the perfect knowledge that A and B have goats behind them.
Assuming you, Bob and Ted all actually prefer the gold ball. The balls can be distributed (B/T) 1) PP/GG, 2) GG/PP or 3) GP/GP. They get to choose, so if they can they will give you a ping pong ball. After Ted gives you a ping pong ball, we can only rule out 1. If it is 2, you must pick Bob, but if it is 3 you must pick Ted to get the gold. So it makes no difference.
After this gamne are you sure they did not have time to get back to commit the murder?
Yancey: Did the detectives *know in advance* which doors had goats behind them, and use that knowledge to guide their choice of which door to open? Because that’s what’s critical to your Monty Hall story, and missing here.
So maybe rather than try to shoehorn this example into a mold where it does not fit, you could take a look at the careful Bayesian calculations scattered throughout this comment section and tell me exactly which step in the calculation contains an arithmetic mistake.
PS: I listed three statements and asked which ones you disagree with. From your most recent response, it *appears* that you agree with 1) and 2), but not 3). But it sure would help keep the discussion on track if you could explicitly confirm that.
Steve,
It doesn’t matter whether the detectives knew beforehand. It only matters that when they open doors with goats, that they know exactly what doors to open. This is where you are going wrong. I agree with 1 and 2, but not your assertion that is #3.
In the original Monty Hall problem, the host and his minions don’t need to know where the goats are when the contestant makes the initial selection, they only need to know afterwards. Your detectives, when they come to you with the new information are precisely analogous to the host peaking behind the doors before revealing non-winning ones.
Ron:
However, they don’t collapse
into 100% certainty for OJ, either.
They don’t stay 50% either, though, do they?
And, it really isn’t shoehorning. How about you tell me where you disagree:
You, Detective Landsburg have stated the following:
P(Bill+Ted+Carol)= 2/3
P(Alice)=1/3
Your crack detective team comes along, takes a closer look at Bill and Ted, and realizes that they are mannequins incapable of murder. Surely, you aren’t going to tell me that that Alice is the favorite suspect, right?
Harold, the good news is that you will always be welcome at my casino.
Before my first demand what the possibilities? Ted can hold
PP
PG
GP
GG
Note that PG and GP are distinct cases.
We should assume each of these is equally likely, for the reasons I gave in other comments (and common sense).
Once Ted gives me a ball things change. I know more.
Say he gives me a G. Then he must have started with GG. Ted, give me your ball!
Say he gives me a P. Now he can only have started with
PP
PG
GP
all equally likely. So it is now 2/3 that his remaining ball is G.
If I ask for it, I get it.
What about Bob? He must hold
GG
PG
GP
If I ask him then 2 times out of three he will surrender a P. Because he can.
Ted, give me your ball!
This little bit of fun can be connected to the post in several ways. In particular note how the conclusion that Ted has a 2/3 chance of being guilty of holdsing my gold ball depends on the presence of Bob.
I interepreted the 2 to 1 odds that the culprit was a smoker as independent of who the suspects were. Say for instance that there was a cigarette butt at the scene that some forensics expert says indicates 66% odds that the killer smoked. The number of suspects wouldn’t change those odds.
Pat,
I interpreted it the same way, but I don’t think it really matters now. Given the defined suspect list and their smoking habits, Detective Steve has determined the probabilities that the killer is
Bill, Ted, or Carol 2/3 of the time and is
Alice 1/3 of the time.
We know beyond any doubt that in the Bill/Ted/Carol group that at least two of them are innocent, but we don’t know any specific pair that is innocent until the other detectives come to Steve and reveal that Bill and Ted could not have committed the crime.
@101: That is a stronger assumption ie it assumes more information than Steve’s reading. And it nicely ties into Roger’s point.
Imagine your 2/3 conclusion is based on a report by Floogle Investigations. Floogle found out Ted had an alibi. But due to poor communication skills they didn’t tell you that, the only said, “Hey we think the odds are 2/3 a smoker dunnit.” The 2/3 works right? But now a Floogler, learning that Bob also has an alibi, will conclude it is 50-50 between Alice and Carol.
But you are reading the Floogler as saying Carol is a 2-1 favorite.
@Ken B, I like your problem better than the original one!
Yancey: I’ve asked you twice to tell me what you disagree with, and you’ve ignored the request both times. It’s quite impossible to have a conversation with you if you’re elusive about what you’re trying to say. So for the time being, I’m disallowing further comments from you (including your recent followup comments that ignored my request). Let me know if/when you’re ready to engage in productive discussion.
Pat T (101):
Say for instance that there was a cigarette butt at the scene that some forensics expert says indicates 66% odds that the killer smoked. The number of suspects wouldn’t change those odds.
Of *course* a change in the suspect pool would change those odds!
If, for example, I narrow the suspect pool down to three non-smokers, I surely cannot continue to believe that the odds are 2-to-1 that the culprit smoked.
That’s an extreme example, but it illustrates a general point. When the suspect pool changes, the odds generally change.
I answered your question from #70 in #98, Steve. I disagree with your assertion that the problem is more analogous to your item 2.
I will point out that I can’t be clearer in what I am saying than in the held-up comment #103. I think your fundamental mistake is taking the equation: P(Bill) + P(Ted) + P(Carol) = 2/3 and when learning that P(Bill) and P(Ted) both = 0 ignoring that you ever thought that the sum was 2/3.
And I will note that your holding up my comments is a really classy thing to do on your part considering I not only responded to your request for information, but have not been rude or nasty about the disagreement.
Ken B. Yes, there are two ways to hold one of each. Of course!
#102 -Yes, but that is only because the 2/3 came from knowing something about Ted individually. If say Bob and Carol were the only ones with red hair, they could equally have said it was 2:1 a red-head did it. As you say, it would require very bad communication. It is different from Floogle deciding that the odds of it being a smoker were 2:1 as an independant bit of data. The original scenario was careful not to claim that it was an independant bit of information, but given that it is posed as a Baysian problem, it is understandable that people mis-intepret it as such, since that is the way it is more usually done. It takes careful reading of the problem to get it right.
Steve (@99):
However, they don’t collapse
into 100% certainty for OJ, either.
They don’t stay 50% either, though, do they?
No, they don’t.
However, your forensic test (in the other example) doesn’t change
its value, even if your suspect pool doesn’t have any smokers in it.
It’s a result that must be disclosed to the lawyer of whoever you
decide is the culprit. You can bet they’ll bring this up in court.
Objective test results that provide odds are quite different from
your own assessment of the odds of a given set of people being
guilty.
Yancey: You did indeed answer in #98, and I overlooked it. My considerable apologies. I’ve approved all your posts.
And, on the assumption that you are actually still reading my comments, even if being held up. I would argue that you have the following that must be true:
P(Carol) + P(Bill) +P(Ted) = 2/3, but that this equation cannot be satisfied by asserting the following about the individual probabilities:
P(Carol)= 2/9, P(Bill)= 2/9, and P(Ted) = 2/9
but must, instead, be satisfied by one of the following sets of equations:
P(Carol)= 2/3, P(Bill)= 0, P(Ted)= 0; or
P(Carol)= 0 , P(Bill)=2/3, P(Ted)= 0; or
P(Carol)= 0 , P(Bill)= 0 , P(Ted)= 2/3.
And we know this because we know that at least two of the three are innocent (you stated there is one and only one killer). The last bit of info eliminates the last two members of the set, leaving you with the first one.
Steve,
Then I apologize for my comment in 110. My opinion of you has always been high, I just think you are wrong in this problem.
Or, to put it another, and more explicit way, Detective Steve believed one of the following before the new information arrived:
(A) P(Alice)= 1/3, P(Carol)= 2/3, P(Bill)= 0, and P(Ted)= 0; or
(B) P(Alice)= 1/3, P(Carol)= 0 , P(Bill)= 2/3, and P(Ted)= 0; or
(C) P(Alice)= 1/3, P(Carol)= 0 , P(Bill)= 0, and P(Ted)= 2/3.
The detectives arrive and say, “Lietenant Steve, we can now tell you to ignore beliefs B and C are wrong.
And for those constructing bags of balls with names, or hats with slips of paper, the proper thing to do is to have 3 bags or hats:
Bag 1 has 3 Alice balls, 6 Carol balls, and 0 balls for Ted and Bill.
Bag 2 has 3 Alice balls, 0 Carol balls, 6 Ted balls, and 0 for Bill.
Bag 3 has 3 Alice balls, 0 Carol balls, 0 Ted balls, and 6 for Bill.
The detectives then arrive and tell you which bag you have to draw from.
Advo (#80):
“In any instance where a mixture exists, the probability may be a dependent or an independent variable.”
That’s not true.
It cannot be an independent variable because the probability makes a statement ABOUT THE SPECIFIC SUSPECT POOL.
If the probability makes a statement about the specific suspect pool, then of course I agree that it is a dependent variable. But I disagree with your claim that the probability necessarily makes a statement about the specific suspect pool. If you can persuade me that it must, then I will concede the point. On the other other hand, if I can persuade you that it need not, then you should concede the point. Agreed?
The premise of the puzzle is unequivocally:
1. We have four suspects, B, C, T, A and we are certain that it couldn’t have been anyone else.
2. A is a non-smoker.
3. GIVEN THIS SUSPECT POOL, odds are 2:1 for a smoking killer.
You should be more careful with the word “unequivocally.” Your insertion of the phrase “GIVEN THIS SUSPECT POOL” substantially alters the nature of the problem. With its inclusion, I agree that the probability necessarily makes a statement about the specific suspect pool. However, that phrase was not included in the original problem. To persuade me that it MUST have been implied, please quote the precise language from the OP which admits of no other reasonable interpretation.
You can’t say “odds are 2:1 for a smoking killer” without relating to the specific suspect pool.
UNLESS THE ODDS ARE 2:1 FOR THE SPECIFIC SUSPECT POOL, the odds can’t be 2:1 at all.
http://rationalwiki.org/wiki/Argument_by_assertion
Suppose Multivac (or God, if you prefer) informs us that 2/3 of all murders are committed by smokers. Is your claim that one still couldn’t say “odds are 2:1 for a smoking killer” without relating to the specific suspect pool? This sounds unreasonable to me. It seems like you are denying the possibility of possessing knowledge about a system that does not include knowledge about all of its constituent parts. Have I understood you correctly, or am I missing something?
Max:
> The premise of the puzzle is unequivocally:
> 1. We have four suspects, B, C, T, A and we are certain that it couldn’t have been anyone else.
> 2. A is a non-smoker.
> 3. GIVEN THIS SUSPECT POOL, odds are 2:1 for a smoking killer.
You should be more careful with the word “unequivocally.” Your insertion of the phrase “GIVEN THIS SUSPECT POOL” substantially alters the nature of the problem. With its inclusion, I agree that the probability necessarily makes a statement about the specific suspect pool. However, that phrase was not included in the original problem. To persuade me that it MUST have been implied, please quote the precise language from the OP which admits of no other reasonable interpretation.
Gladly:
You are quite sure that one (and only one) of these suspects is the culprit. Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker.
@Al V: Well, you are and always have been a gentleman of superior taste and discernment!
@107, 101:
You two are talking past each other. Pat is saying that the 2/3 an absolute probability, that can be known and measured regardless of the suspect pool. Steve is noting that that does not make sense really. How can a statement *about the suspect pool* ignore the suspect pool?
So does the 2/3 make sesne as anything BUT a statement about the suspect pool? I think not. Pat proposes a reading that sounds more like “the chances that when you identify the murder you will find he smokes are 2/3” than “of the known suspects the chances are 2/3 you will find the culprit amonst the smokers.” Which are different probabilities.
Max,
Suppose God (or, in this case, Steve Landsberg) says:
1. The murder was committed either by B, C, T or A.
2. Odds are 2/3 that the killer is a smoker.
Then it follows that:
3. With regard to the suspect pool containing B, C, T and A, chances are 2/3 that the killer is a smoker.
1. and 2. cannot both be true at the same time unless 3. is true as well.
Given the truth of statement 1., statement 2. can relate to no other population than the suspect population under statement 1., because no one else could possibly have committed the murder.
Steve set up the puzzle very precisely, there’s no way around this.
Ken B:
Pat is saying that the 2/3 an absolute probability, that can be known and measured regardless of the suspect pool. Steve is noting that that does not make sense really. How can a statement *about the suspect pool* ignore the suspect pool?
Exactly.
I see it’s Landsburg with a “u”. Sorry about that.
Ken @101
I agree with your answer, but when I thought about the probable distributions of the balls I imagined Ted picking two balls from a bag containing the four balls (somehow he can’t tell the difference between ping-pong balls and gold balls – now there’s someone I’d like in my casino!). In this scenario doesn’t Ted 1/6 (1/2 x 1/3) probability of picking the two gold balls, 1/6 probability of picking two ping-pong balls and 4/6 of picking one of each? Is there a reason to prefer one distribution over the other?
So unless you suspect me of playing unfair, the answer is Alice with a probability of 60%.
I don’t suspect you of “playing unfair”, but I do think that it is an additional assumption that each of the smokers are equally guilty. It is a simplifying assumption, and possibly a reasonable one in the face of no more information, but an additional assumption that isn’t in your original problem set up.
Steve Landsburg (#119):
You are quite sure that one (and only one) of these suspects is the culprit. Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker.
Why is it not reasonable to conclude that these facts are independent of one another?
“Why is it not reasonable to conclude that these facts are independent of one another?”
Because of the word “moreover”. Also, #122.
By the way, how do you do formatting?
Steve Landsburg (#119):
You are quite sure that one (and only one) of these suspects is the culprit. Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker.
Why is it not reasonable to conclude that these facts are independent of one another?
Advo (#122):
Suppose God (or, in this case, Steve Landsberg) says:
1. The murder was committed either by B, C, T or A.
2. Odds are 2/3 that the killer is a smoker.
Then it follows that:
3. With regard to the suspect pool containing B, C, T and A, chances are 2/3 that the killer is a smoker.
1. and 2. cannot both be true at the same time unless 3. is true as well.
Of course. There is no dispute that the italicized words above are correct. However.
Suppose God (or, in this case, Steve Landsberg) says:
1. The murder was committed either by B, C, T or A.
2. Odds are 2/3 that the killer is a smoker.
3. The murder was committed either by C or A.
Then it follows that:
4. With regard to the suspect pool containing B, C, T and A, chances are 2/3 that the killer is a smoker.
5. With regard to the suspect pool containing C and A, chances are 2/3 that the killer is a smoker.
1. and 2. and 3. cannot all be true at the same time unless 4. and 5. are true as well.
You should agree that this is also correct. Where we differ, then, is when you say the following:
Given the truth of statement 1., statement 2. can relate to no other population than the suspect population under statement 1.
This is clearly wrong, as I have just shown. The truth of statement 1. does not imply that statement 2. can relate to no other population than the suspect population under statement 1.
Steve set up the puzzle very sloppily, there’s no way around this.
Advo (#127):
Because of the word “moreover”.
It’s funny that you would say this. The word “moreover” is precisely why I believe the independent interpretation to be the more reasonable one. Consider these examples of its usage pulled from an online dictionary:
“The cameras will deter potential criminals. Moreover, they will help police a great deal when a crime actually is committed.”
Aren’t these just two independent facts about what cameras will do?
“swimming alone is against the rules and, moreover, it’s dangerous”
Aren’t these just two independent facts about what swimming alone is?
By the way, how do you do formatting?
Do you mean italics? Put and (without the spaces) around any text you want italicized.
Erm. Apparently that tag works even if you include spaces. My bad.
Put brackets like with an i in between them in front of text you want italicized. Put those same brackets with /i in between them after the text you want italicized.
@Max,
the order the information emerges is relevant.
Once 3. becomes known (The murder was committed either by C or A.), 1. and 2. are no longer true.
The suspect pool shrinks and the odds of the murderer being a smoker are adjusted accordingly.
Oh dear.
The sort of bracket I’m referring to looks like a sideways V. To make them on my keyboard, you type shift-comma and shift-period. Hope this helps. =)
Advo (#131):
the order the information emerges is relevant.
Check comment #1.
Once 3. becomes known (The murder was committed either by C or A.), 1. and 2. are no longer true.
This is definitely wrong according to the precise wording of the problem as we have re-written it. Once 3. becomes known, 1. and 2. are still true. That the murder was committed either by C or A is not in conflict with the murder being committed either by B, C, T or A. So 1. is still true. And it is not in conflict with the probability being 2/3 that the killer is a smoker. So 2. is still true.
The suspect pool shrinks and the odds of the murderer being a smoker are adjusted accordingly.
This is true for a different problem than the one we are dealing with here. I agree that it is reasonable to interpret the problem Steven posed as being like the one you have in mind. I disagree that this is the only reasonable interpretation or even that it is the most natural one.
@Max,
Once we know that either C or A committed the murder, we cannot say anymore that B or C have committed the murder.
Therefore, saying that B, C, T or A have committed the murder would then be false. Do you agree?
Correction:
Once we know that either C or A committed the murder, we cannot say anymore that B or T have committed the murder.
Therefore, saying that B, C, T or A have committed the murder would then be false. Do you agree?
Advo (#135):
Once we know that either C or A committed the murder, we cannot say anymore that B or T have committed the murder.
Anymore? But we never said that B or T have committed the murder in the first place. We said only that “[t]he murder was committed either by B, C, T or A,” and this is still true.
Therefore, saying that B, C, T or A have committed the murder would then be false. Do you agree?
No, I do not agree, and I am very confident that you are mistaken. If I say that I am either the President of the United States or I am not, and it turns out that I am not, then it is still true that I am either the President of the United States or I am not.
Any calculation that starts with P(Alice)=1/3 and P(Carol)=2/9 is going to lead you astray here. Given that there is only one killer and four suspects, and given Steve’s analysis that the odds are 2/3 that it is a smoker, then you can only believe one of two comparisons of the probabilities with regards to Alice and Carol:
P(Alice) = 1/3 and P(Carol) = 2/3
Or
P(Alice) = 1/3 and P(Carol) = 0
The new information tells you which one to believe, but it does not change those probabilities.
No, Max.
The original post says:
“The suspects are:
Bob, a male smoker.
Carol, a female smoker.
Ted, another male smoker.
Alice, a female non-smoker.
You are quite sure that one (and only one) of these suspects is the culprit.”
After we establish that the culprit is woman, Ted and Bob are no longer suspects.
I guess the problem here is that the statement I used was an imprecise reflection of SL’s original statement.
So let me restate:
1. For each of the following persons, there is a likelihood of >0 that they committed the murder:
B,C, T or A.
Once we establish that the murder was committed by A or C, we know that the likelihood of T or B having committed the murder = 0.
So statement 1. is no longer true.
Max:
You are quite sure that one (and only one) of these suspects is the culprit. Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker.
Why is it not reasonable to conclude that these facts are independent of one another?
I’ve answereed this multiple times. I’ll answer it exactly once more. Please don’t ask me to answer it yet again.
If you are certain that your suspect pool consists entirely of non-smokers, then you cannot believe that the odds the culprit is a smoker are 2-to-1. This shows that you cannot formulate these odds independent of your knowledge of the suspect pool. That’s a full and sufficent argument right there.
Moreover, it would be insane to throw away pertinent knowledge before formulating odds. The problem tells you quite clearly that you know who the suspects are at the time when you formulate the odds. Assuming you’re not insane, you’ll have incorporated that knowledge into your calculation.
Advo (#138):
I guess the problem here is that the statement I used was an imprecise reflection of SL’s original statement.
No, the problem here is that there is no precise reflection of SL’s original statement. Steve set up the puzzle very sloppily, there’s no way around this.
So let me restate:
1. For each of the following persons, there is a likelihood of >0 that they committed the murder:
B,C, T or A.
Once we establish that the murder was committed by A or C, we know that the likelihood of T or B having committed the murder = 0.
So statement 1. is no longer true.
This is no mere restatement; you have substantially altered the substance of the problem that we were talking about.
Yancey: You keep saying that new information can’t change the probabilities of 2/3 for a smoker and 1/3 for a non-smoker.
What if you get new information that Alice is definitely the killer.
1) Do you agree that this new information changes the probabilities from 2/3 and 1/3 to 0 and 1?
2) If one bit of information (“It’s Alice) can change the probabilities, then why can’t another bit of information (like “It’s a female”)? What are your rules for which kinds of information are allowed to change the probabilities and which aren’t?
@Max,
I restated 1. to match Steve’s original statement. The reformulated statement leaves nothing out and adds nothing.
First there’s 4 suspects.
Then there’s 2 suspects.
Now that there’s 2 suspects, you can’t truthfully say that there’s 4 suspects.
Steve Landsburg (#140):
I’ve answereed this multiple times. I’ll answer it exactly once more. Please don’t ask me to answer it yet again.
I apologize. I haven’t been reading any of the comments that weren’t addressed to “Max.”
If you are certain that your suspect pool consists entirely of non-smokers, then you cannot believe that the odds the culprit is a smoker are 2-to-1. This shows that you cannot formulate these odds independent of your knowledge of the suspect pool. That’s a full and sufficent argument right there.
I have already answered this argument in comment #78, which was addressed to you. I’ll answer it exactly once more. Please don’t ask me to answer it yet again.
The set of suspects must include both smokers and nonsmokers in order for any probability between 0 and 1 to make sense. This does not imply that the probability of a smoking killer is always a function of the ratio between smokers and nonsmokers in the set of suspects. The only cases where that is necessarily true are where the set of suspects consists of all smokers or all nonsmokers. In any instance where a mixture exists, the probability may be a dependent or an independent variable. It is simply not correct to say that interpreting the variable as independent in one instance requires interpreting it as independent in all instances.
I said something else relevant in comment #118:
Suppose Multivac (or God, if you prefer) informs us that 2/3 of all murders are committed by smokers. Is your claim that one still “cannot formulate these odds independent of your knowledge of the suspect pool”? This sounds unreasonable to me. It seems like you are denying the possibility of possessing knowledge about a system that does not include knowledge about all of its constituent parts. Have I understood you correctly, or am I missing something?
Moreover, it would be insane to throw away pertinent knowledge before formulating odds. The problem tells you quite clearly that you know who the suspects are at the time when you formulate the odds. Assuming you’re not insane, you’ll have incorporated that knowledge into your calculation.
You are committing the same error here that you chastised someone else for in comment #47. “[I]f you have additional information beyond what’s given in the problem, then you’ll of course get different answers.” I agree that one would incorporate knowledge of the suspect pool into one’s calculation of the odds that the murderer is a smoker, assuming that were possible. But there is no indication given in the problem – none whatsoever – that this is the case.
Advo (#143):
I restated 1. to match Steve’s original statement. The reformulated statement leaves nothing out and adds nothing.
When you say that the reformulated statement adds nothing, you mean to say that it adds nothing to Steve’s original statement. Yet you must agree that it adds something to your original interpretation of Steve’s statement: “The murder was committed either by B, C, T or A.”
First there’s 4 suspects.
Then there’s 2 suspects.
Now that there’s 2 suspects, you can’t truthfully say that there’s 4 suspects.
Of course. I’ve never claimed otherwise.
I see where I went in the wrong direction. I read through the
details of the problem, and analyzed it as if it were a real-world
murder mystery.
What I neglected to take into account was the title of the article
itself, “A Bayesian Solution”. That indicates a different approach
and a different slant on just what the quoted odds mean.
Although I’m certainly not alone in this, it did clearly differ from
the author’s intentions.
Max:
Suppose Multivac (or God, if you prefer) informs us that 2/3 of all murders are committed by smokers. Is your claim that one still “cannot formulate these odds independent of your knowledge of the suspect pool”?
Of course that’s not my claim. My claim is that of course you will account for the fact that 2/3 of all murders are committed by smokers when you formulate the probability that this particular murder was committed by a smoker. But you cannot formulate that probability without knowledge of the suspect pool.
Ron:
I see where I went in the wrong direction. I read through the
details of the problem, and analyzed it as if it were a real-world
murder mystery.
I devoutly hope (and fully expect) that real world detectives are pretty good Bayesians.
Note, incidentally, that your story is inconsistent on its own grounds. According to you, you believe ex ante that the probability for a smoker to have committed the crime is 2/3. You also presumably believe ex ante that the probability for Carol is less than 2/3 (because it might have been Bob or Ted).
When the new report comes in (“It’s a female!”), you’re willing to update the odds for Alice (from less-than-2/3 up to 2/3) but you’re not willing to update the odds for a smoker. In other words you’re trying to be both a Bayesian and not-a-Bayesian at the same time. I can’t imagine any theory of probability that could countenance both these positions at the same time.
Max:
Since you read only comments addressed to you, let me repeat something I said in a comment addressed to Ron:
Note, incidentally, that your story is inconsistent on its own grounds. According to you, you believe ex ante that the probability for a smoker to have committed the crime is 2/3. You also presumably believe ex ante that the probability for Carol is less than 2/3 (because it might have been Bob or Ted).
When the new report comes in (“It’s a female!”), you’re willing to update the odds for Alice (from less-than-2/3 up to 2/3) but you’re not willing to update the odds for a smoker. In other words you’re trying to be both a Bayesian and not-a-Bayesian at the same time. I can’t imagine any theory of probability that could countenance both these positions at the same time.
Before the detectives come back with the new information, here is the full set of possibilities:
P(Bill) = 1, P(everyone else, grouped and ungrouped)=0
P(Ted) = 1, P(everyone else, grouped and ungrouped)=0
P(Carol)= 1, P(everyone else, grouped and ungrouped)=0
P(Alice)= 1, P(everyone else, grouped and ungrouped)=0;
P(Alice)= 1/3 P(everyone else)= 2/3 with the 2/3 distributed in three different ways.
However, the first four of these are irrelevant- in all you already know who the killer is and the suspect list is meaningless- it is only the last one, the one in which you have segregated the suspects into two pools that matters- it is the only one with unknowns. Now, lets suppose you thought Alice was more likely to be the killer based on her shifty eyes, and you know shifty-eyed people are more likely to be killers than non-shifty eyed people, and you did use that information to make the assignment of probability in that last case. Of course, if you had only one other possible suspect, you no doubt would have P(Alice)=6/10 and P(other guy)=4/10. However, you had 4 possible suspects in total and you assign as above. Your detectives come along and tell you which of the two suspects in the grouping of three that could not have committed the crime under any circumstances, and so they are eliminated. You are now claiming you can go back to your idea about shifty-eyed people minus the two suspects eliminated. I am arguing it is this process that is invalid.
Think about it this way, and I know I have already done this above, but it bears repeating. Had you shown me two boxes and told me that Box A has a prize 4/10 of the time and Box B has it 6/10 of the time and ask me to choose, I, of course, select Box B. However, if you show me 4 boxes and tell me that the prize is in Box B 1/3 of the time, and is in one of the other three boxes the rest of the time, I of course still select Box B (its weighted probability is still higher than the other boxes). You as the host then, knowing like the detectives in the original problem, how to open 2 of the unchosen boxes without a prize and show me the contents and ask me if I want to change my selection. Based on your reasoning, I should stick with Box B, right?
@Max,
alright. So now we return to post #122.
We have established that statement 3. invalidates statement 1. (in its restated version).
Once we know 3. (the culprit is a woman), statement 1. is no longer true, thus (in this specific case, since the smoker/non-smoker composition changes), statement 2. cannot be true any longer either.
@Yancey
The big difference between this problem and the Monty Hall game is that, in the Monty Hall game, the TV host tricks you : he does not open an empty door at random : he chooses one that he knows is empty.
If we try to apply it to our bayesian crime, it is a bit like if our crack team would announce that the killer is female *when they in fact know who the killer is*. And then obviously the two clues (it’s a smoker 2-to-1 and it’s a female) are not independant anymore, and it ruins all your calculations.
Steve, I’m surprised that you seem quite dismissive of my interpretation. Are you suggesting there is no possible way that forensic results can be reported without knowledge of the suspect pool, or that the way you’ve written the problem cannot possibly be interpreted as such a reporting?
Ben: I am suggesting both.
Certainly forensic science cannot tell you the odds of the killer being a smoker in the absence of some prior assumption about the suspect pool. The easiest way to see this is to imagine the case where a smoldering cigarette is found near the crime scene. Given the known (real world) population, that makes it likely the killer is a smoker. In a different locality where cigarette smoking is unheard of, but burning cigarettes are routinely carried for good luck, it means something very different. That’s an extreme case, but the same logic *always* applies.
And even if you didn’t realize that this logic always applies, certainly the problem statement makes it clear that the logic applies *in this case* because you are told quite explicitly that you *know the suspect pool* at the time when you calculate the odds. You could, of course, assume that you are the worst detective in history and therefore choose to ignore what you know about the suspect pool, but if you’re going to do that, you might as well assume that you’re insane in any of countless other ways.
Steve:
When the new report comes in (“It’s a female!”), you’re willing
to update the odds for Alice (from less-than-2/3 up to 2/3) but
you’re not willing to update the odds for a smoker. In other words
you’re trying to be both a Bayesian and not-a-Bayesian at the same
time. I can’t imagine any theory of probability that could
countenance both these positions at the same time.
Okay, let me tell you a mystery story…
A murder occurred on an Army base. A group of 1000 men are the only
suspects, since they were in a restricted area at the time of the
murder (along with the victim, of course). We pick up the story
with the forensic pathologist reporting to the detective. A
mathematician happens to be there at the time.
Pathologist: “The victim was killed by a knife. From the wounds,
it really looks like the work of a trained knife fighter. I’d
give it a 90% probability that this was the case.”
Detective: “That doesn’t help much. These men are all commandos
here for refresher training. They’re all skilled knife fighters.
All, that is, except for one, who’s the base’s driver.”
Mathematician: “Given the makeup of the suspect pool, shouldn’t
you up your probability to 99.9%?”
Pathologist: “No. Regardless of the suspect pool, there’s still a
10% chance that the wounds I see were deadly by blind luck rather
than skill. I’m not going to assign a certainty that I can’t
justify purely by the examination of the body. Don’t talk to me
about suspect pools; these wounds are such that 9 times out of 10,
they were made by a trained knife fighter.”
It’s later determined that all but two of the men were in
classes at the time of the murder. Only the driver and one
commando were unaccounted for.
The mathematician computes the probabilities and says, “Aha!
The driver did it! The Bayesian odds make him clearly the
most likey murderer.
. . .
Spoiler alert:
Via long and brilliant work, the detective was able to prove that
the victim’s wife snuck onto the base and killed her husband.
@Yancey you need to learn about a concept called “conditional probability”. It will make it much easier to reason about this puzzle, and other probability puzzles.
Here’s a simple example : I roll two dice, one red, one blue, and hide them under a cup. What’s the chance I got 12?
I hope you’ll agree that P(R+B=12) = 1/36.
Now, I peek at the red die, hoping it’s a 6. What’s the chance the red die shows a 6?
I hope you’ll agree that P(R=6) = 1/6.
Suppose I’m lucky, and sure enough, the red die shows a 6. NOW what’s the chance that my total is 12?
I hope you’ll agree that the chance is 1/6. This doesn’t mean that P(R+B=12) has changed. That’s still 1/36. P(R+B=12) refers to my a priori probability of getting 12. That was 1/36 at the start, remains 1/36 now, and will still be 1/36 when the blue die is revealed.
The chance of getting 12 in two dice is 1/36. However, the chance of getting 12 on two dice, given that the first dice is known to show a 6, is much higher – it’s 1/6.
The formal way to write this is
P(R+B=12 | R=6) = 1/6.
Bayes rule says that P(A|B) = P(A and B) / P(B).
For this example,
P(R+B=12 | R=6) = P(Total is 12 | R shows 6)
= P(Total is 12 AND R shows 6) / P(R shows 6)
= P(R and B both show 6) / P(R shows 6)
= (1/36) / (1/6)
= 1/6
For Steve’s Bayesian puzzle (that very few people are explicitly applying Bayes’ rule to), don’t confuse
* P(Alice is guilty)
* P(Alice is guilty | evidence about smokers)
* P(Alice is guilty | crack team’s evidence)
* P(Alice is guilty | evidence about smokers AND crack team’s evidence)
They’re all different numbers.
Yancey (if you’re still reading): Let me try to make an argument on *your* terms, taking your Monty Hall analogy seriously.
Initially, the probabilities for Bob, Carol, Ted and Alice are 2/9, 2/9, 2/9, 1/3.
Now we eliminate Bob and Ted.
Your argument, as I understand it is: The probability for Alice was 1/3, and nothing has happened to change that. Therefore, it’s still 1/3.
Let me now apply the *exact same argument* to Carol: The probability for Carol was 2/9, and nothing has happened to change that. Therefore, it’s still 2/9.
We’ve now proved — with *your* logic — that the probability for Alice is 1/3, and the probability for Carol is 2/9. And we know these are the only two remaining possibiilities.
1) Does it bother you that these probabilities don’t add to 1?
2) Do you see that even if we ignore that inconvenient fact, the probability for Carol is still less than it is for Alice? That is, do you agree that 2/9 is less than 1/3?
Ron:
Pathologist: “No. Regardless of the suspect pool, there’s still a
10% chance that the wounds I see were deadly by blind luck rather
than skill.
In other words, your story starts from the assumption that the pathologist is an idiot. I hope this is unrealistic.
Steve:
In other words, your story starts from the assumption
that the pathologist is an idiot. I hope this is
unrealistic.
The pathologist simply doesn’t have faith in the omniscience
of whoever delivered the suspect pool. Note also that
“Round up the usual suspects” is easy, but it’s not
necessarily the best technique.
If you read past the spoiler, you’ll have noted that the
murderer was not even contained in the original suspect
pool.
The odds the pathologist quoted were correct, based on the
extensive medical literature and his many years of
experience. They’re independent of the suspect pool.
The kind of odds you want mean something different, just
as “indifference” in economics means something different
from the most common usage.
In order to establish the proper dependent odds within the
suspect group, the mathematician needed to apply those
independent odds to the particular group composition. One
way would be to conclude something like: “If all things
were equal, each person would have a 1/1000 chance of being
the killer. They’re not equal. The driver is only 1/10 as
likely to be the killer as any other individual in the
group. Therefore, his chances are 1/10,000. Let’s multiply
each of these by 10,000 to make the math simple. Therefore,
each commando has 10/x probability and the driver has 1/x
probability. X= 999 * 10 + 1 or 9991. The driver has about
.0001 odds of being the killer, while each commando has
about .001 chance.”
If the odds were as quoted in the original problem, and on
the same independent basis, we’d have to do the same
operation against the suspect pool. Bob, Carol, and Ted
each get 2 chances, while Alice gets 1. We divide by 7
(2+2+2+1) to get 2/7 chance for each of the first three and
1/7 for Alice. This places 6/7 of the probability weight on
the smokers.
Ron,
you can NEVER make such a probability statement except with regard to some specfic pool of people. USUALLY, that means it is made with regard to the general population. This is so common that we don’t even realize it.
The statement of the pathologist MUST relate to SOME pool of people.
It CANNOT stand alone.
The odds the pathologist will give you then are dependent on the number of trained knife-fighters in the general population.
If the general population has no knife-fighters, the pathologist will never see any knife-fighter wounds (only random luck wounds), and then the likelihood would be 0.
If there are only a few knife-fighters, most such wounds will be caused by random luck, and the odds of such wounds actually having been inflicted by a knife-fighter will be low.
If there are many trained fighter in the general population, the pathologist will in his practice see a greater number of such wounds in total, and in many cases those wounds will indeed by inflicted by knife fighters.
Again, the pathologist cannot give any odds for the wounds being inflicted by a knife-fighter except as they relate to a specific population.
It’s the same with HIV tests, for example. You don’t know what the odds are that a positive result is a false positive, unless you know something about the HIV-rate among the population the test subject came from.
The same test which will produce 1% false positives in Botswana will produce 50% false positives in Sweden.
@Ben 156:
As I’ve said before,
the statements:
1. The suspects in the case are B,C, T or A.
2. Odds are 2:1 that the culprit is a smoker.
can only be true at the same time if this statement is also true:
3. With regard to the suspect pool containing B, C, T and A, chances are 2/3 that the killer is a smoker.
1. and 2. cannot both be true at the same time unless 3. is true as well.
Given the truth of statement 1., statement 2. can relate to no other population than the suspect population under statement 1., because no one else could possibly have committed the murder.
Max (#66):
Sorry, I was using quote marks to delimit the clue, and wasn’t intending to hoodwink anybody into thinking it was a direct quote.
Quite the opposite, I was intentionally rewording to make the meaning I was using clear to anybody else who had interpreted the question differently.
You claim I have “badly misrepresented the problem as originally presented”. I don’t believe I have. Steve, who phrased the puzzle, agrees with me. But it doesn’t look like this is something either of us will persuade t’other of.
Clearly if you interpret the puzzle differently then the maths works out differently, but there’s no point in discussing the details of that if we don’t agree on what the clues mean.
I’m intrigued by the concept of being bad at misrepresenting something though!
@Guy: oops. I believe you are right. This is NOT like flipping a coin, it is drawing without replacement.
I assume I am welcome at your casino?
Ron:
Here’s my counterstory.
A fiber is found at the scene of an ax murder. The pathologist is called in. He reports that men and women wear these fibers equally often. He concludes that it’s 50-50 whether the criminal is a man or a woman.
Just then, your officers report that they’ve arrested a guy who was known to hate the victim, was carrying a bloody ax, and was seen using it to hack the victim to death by 46 eyewitnesses.
Your response, of course, is that the officers have to let him go. There is, after all, a 50% chance this crime was committed by a woman. With such enormous reasonable doubt, no DA will invite.
Your officers, being new at this game, suggest that maybe in light of the new information, the probabilities have changed. But you inform them that you had your police training from Captain Ron, at whose feet you digested the lesson that new evidence never affects established probabilities. So of course you let the guy go.
Ron:
If the odds were as quoted in the original problem, and on
the same independent basis, we’d have to do the same
operation against the suspect pool. Bob, Carol, and Ted
each get 2 chances, while Alice gets 1. We divide by 7
(2+2+2+1) to get 2/7 chance for each of the first three and
1/7 for Alice. This places 6/7 of the probability weight on
the smokers
Alright, then: Let’s go backward a little in this story.
Originally you had four equally likely suspects and no evidence
at all. Alice was one of them.
Question 1: Do you agree that at this point Alice is (given
your knowledge) 1/4 likely to be the culprit?
Now some evidence has come in which causes you to say the
odds for a smoker are 2-to-1.
Question 2: Do you understand that this is equivcalent to
saying that Alice is 1/3 likely to be the suspect?
So the evidence — call it E — ups the chances that it’s
Alice.
Now additional evidence comes in, proving the killer is female.
And we still have the original evidence, E, which upped
the chance that it’s Alice.
Question 3: Does the new evidence mean we should ignore
the old evidence? Or are we allowed, like good detectives,
to consider both? And if the latter, shouldn’t evidence
E, which pointed to Alice originally, still point to her?
@ Steve, you wrote:
“Initially, the probabilities for Bob, Carol, Ted and Alice are 2/9, 2/9, 2/9, 1/3.”
This seems to be the crux of some confusion. Because I’m ignorant of much of this probability stuff, I’m confused too. There appear to be two interpretations of the information given up to the point, but not including, where we learn that the culprit is definitely female.
Call them the And-And-And, and the Or-Or-And interpretations. They are both based on the following information:
Crime
4 Suspects (B, C, T, A), 3 (B, C, T) of whom are smokers.
1 culprit
Odds that “culprit is a smoker” = 2:1.
Interpretation 1:
Odds of being the culprit, smokers in parenthesis.
(B = 2/9 AND C = 2/9 AND T = 2/9) AND A = 1/3. (6/9) : 3/9 = (2):1.
Interpretation 2:
Odds of being the culprit, smokers in parenthesis.
(B = 2/3 OR C = 2/3 OR T = 2/3) AND A = 1/3. (2/3) : 1/3 = (2):1.
My question, Why is the And-And-And interpretation the proper one, and the Or-Or-And improper?
Advo (@160):
You make an excellent point. However, the pathologist MUST give the
odds with respect to the general population. If you round up the usual
suspects and can’t pin it on any of them, the tailored odds for that
pool are incorrect for your next suspect pool.
It’s up to the mathematician to adjust the general-population odds to a
particular distribution set. I think this is a good division of labor.
I wouldn’t trust the mathematician to examine the body and come up with
the conclusion, and I wouldn’t trust the pathologist to correctly adjust
the base population odds to whatever special circumstances exist. In
each case, they might be correct, but it’s outside their core
competencies.
Nearly all of the problems people are having with this puzzle are caused by different interpretations of the set-up. Here’s another example that I think is the same, but shouldn’t allow for different interpretations. (I apologize if someone else has already used this as a way to illustrate the problem.)
Suppose there are four urns, each containing 4 balls:
Urn Bob: 0 Red, 2 White, 2 Blue
Urn Carol: 1 Red, 1 White, 2 Blue
Urn Ted: 0 Red, 2 White, 2 Blue
Urn Alice: 1 Red, 0 White, 3 Blue
One of the urns is selected at random and you get to draw one ball from that urn. (You know the contents of each urn, you just don’t know which urn has been selected.)
1. Suppose you draw a Blue ball from the urn. What is the probability that the urn is Urn Alice? What is the probability that the urn is one of the smokers (Ted, Carol or Bob)?
2. Suppose you draw a Red ball from the urn. What is the probability that the urn is Urn Alice? What is the probability that the urn is one of the male Urns?
Now suppose you draw a ball from the urn, then replace it and randomly draw another.
3. Suppose the first ball is Blue and the second is Red. What is the probability that the urn is Urn Alice?
4. Suppose the first ball is Red and the second is Blue. What is the probability that the urn is Urn Alice?
——
Can you see that drawing a Blue ball is equivalent to the evidence that Steve saw? Can you see that drawing a Red ball is equivalent to the evidence that the Team saw?
And, now, because you have an explicit model for how the evidence comes about, you can write nice equations for all the terms when you apply Bayes’ Rule. That is, you should be able to write down all the values for things like the probability of seeing the evidence (Red and Blue balls) conditioned on each of the urns being the unknown urn.
Originally you had four equally likely suspects and no evidence
at all. Alice was one of them.
Question 1: Do you agree that at this point Alice is (given
your knowledge) 1/4 likely to be the culprit?
Agreed.
Now some evidence has come in which causes you to say the
odds for a smoker are 2-to-1.
Question 2: Do you understand that this is equivcalent to
saying that Alice is 1/3 likely to be the suspect?
I see it, now. Let’s go through this with implicit assumptions made
explicit. I read that as saying, “From the condition of the body, it’s
twice as likely as not that the murderer is smoker, ignoring all other
factors.”, while your statement was equivalent to “The odds of the
murderer from this particular pool of suspects being a smoker is 2:1”.
The “1/3 likely” and the rest of the solution as outlined by you follows
logically from the latter interpretation.
Steve Landsburg (#147):
My claim is that of course you will account for the fact that 2/3 of all murders are committed by smokers when you formulate the probability that this particular murder was committed by a smoker. But you cannot formulate that probability without knowledge of the suspect pool.
I am pretty confident that this is simply wrong. My claim is that if 2/3 of all murders are committed by smokers, then we may reasonably say that there is a 2/3 probability of any particular murder being committed by a smoker, even in the absence of knowledge about the suspect pool for that particular murder. Or we might say that it is not possible to have no knowledge of the suspect pool for a particular murder once we know that 2/3 of all murders are committed by smokers. Do you really disagree? If so, there are a LOT of bets I want to make with you.
Steve Landsburg (#149):
Note, incidentally, that your story is inconsistent on its own grounds. According to you, you believe ex ante that the probability for a smoker to have committed the crime is 2/3. You also presumably believe ex ante that the probability for Carol is less than 2/3 (because it might have been Bob or Ted).
When the new report comes in (“It’s a female!”), you’re willing to update the odds for Alice (from less-than-2/3 up to 2/3) but you’re not willing to update the odds for a smoker. In other words you’re trying to be both a Bayesian and not-a-Bayesian at the same time. I can’t imagine any theory of probability that could countenance both these positions at the same time.
There is no reason to suppose that the odds of the killer being a smoker are in any way related to the composition of the suspect pool unless it consists of all smokers or all nonsmokers. An analogy may help you to see why this is true.
Note, incidentally, that your story is inconsistent on its own grounds. According to you, you believe ex ante that the probability a human will accelerate to a speed of >0.1c is x. You also presumably believe ex ante that the probability a human will accelerate to a speed of >c is less than x (because a human might accelerate to a speed of >0.1c but not >c).
When the new report comes in (“A human accelerated to 0.09c!”), you’re willing to update the odds for a human accelerating to a speed of >0.1c (from x up to x+y) but you’re not willing to update the odds for a human accelerating to a speed of >c. In other words you’re trying to be both a Bayesian and not-a-Bayesian at the same time. I can’t imagine any theory of probability that could countenance both these positions at the same time.
^This is what you sound like to me.
Could you please clarify what you believe to be the relevant disanalogy between your story and my story?
Advo (#151):
alright. So now we return to post #122.
We have established that statement 3. invalidates statement 1. (in its restated version).
Correct. Apparently our God is fallible.
Once we know 3. (the culprit is a woman), statement 1. is no longer true, thus (in this specific case, since the smoker/non-smoker composition changes), statement 2. cannot be true any longer either.
Of course.
The fact that you thought this needed to be clarified makes it pretty apparent that you do not understand what I have been saying. I’m willing to chalk this up to my own inability to explain rather than your inability to understand, but I’m afraid I don’t know how to simplify what I’ve said any more than I already have.
Smylers (#162):
I’m intrigued by the concept of being bad at misrepresenting something though!
I can’t tell if you are unintentionally badly misinterpreting the phrase “badly misrepresented” or are just joking. I could probably use Bayes’ Theorem to refine my estimation of the probability of these two possibilities using priors and whatnot, but instead I’ll simply assume the latter since it’s more fun that way. =D
Max (171):
Could you please clarify what you believe to be the relevant disanalogy between your story and my story?
The relevant disanalogy is that I also have substantial evidence (not mentioned by you) that humans can’t accelerate to speeds > c.
As I said upfront in the post, the analysis here assumes there is no important evidence beyond what’s stated in the problem. In other words, the answer here assumes that I played fair in the statement of the problem.
Steve Landsburg (#174):
The relevant disanalogy is that I also have substantial evidence (not mentioned by you) that humans can’t accelerate to speeds > c.
But I also have substantial evidence (not mentioned by you) that the odds are 2-to-1 that the culprit is a smoker. So the stories seem to me analogous in this way as well.
As I said upfront in the post, the analysis here assumes there is no important evidence beyond what’s stated in the problem. In other words, the answer here assumes that I played fair in the statement of the problem.
As I have said in numerous comments on the post, there is important evidence beyond what’s stated in the problem, and so all of your analysis is wrong, or at least incomplete. You did not play fair in the statement of the problem.
The important evidence not stated in the problem is whether the odds given of the culprit being a smoker (2-to-1) were formed independently of our knowledge of the suspect pool. You insist that it is not possible to formulate such odds in the absence of knowledge of the suspect pool. I claim otherwise. This is the crux of our disagreement.
Please respond to comment #170.
Max (#174):
But I also have substantial evidence (not mentioned by you) that the odds are 2-to-1 that the culprit is a smoker.
You’re claiming that I never mentioned the odds are 2-to-1 that the culprit is a smoker? I wonder, then, how all of these other commenters happened to think that the odds were 2-to-1. Where could they have gotten that I idea if I’d never mentioned it?
I no longer believe you intend to be taken seriously, so I will no longer bother reading your comments.
I think what Yancy and others are missing is that, if you work out the odds for smoker/non-smoker at each step, the 2-1 odds for a smoker are a pretty significant reduction in the odds for a smoker. To get an increase in the odds for a smoker from the default position, the odds would have had to go to 4-1 or more after the new evidence.
At the start, we have 4 suspects, three of which are smokers and one is not. Since we have no more information than this, they each hold an equal 25% probability of being the killer just by default.
With 3 smokers and 1 non-smokers, that’s 3 to 1 odds that the killer was a smoker.
Now, new evidence comes about that REDUCES the odds that it’s a smoker down to 2-1. This is an INCREASE in the odds that April is the murderer, all the way up to 33%, because she’s the only non-smoker. Meanwhile the odds that Carol is the murderer have gone down to 22%
The statement of the problem makes this confusing, but then that’s the point of the whole exercise. In real life, we don’t get to pick and choose how we receive information, and sometimes it comes to us in very confusing ways.
I highly recommend Eleizer Yudkowsky’s “Intuitive Explanation of Bayes’ Theorem” at http://yudkowsky.net/rational/bayes
He uses a more complex but real-life example of breast cancer and mammograms to illustrate how easy it is to screw probabilities up, and how you can recognize and catch those traps with a proper application of Bayes.
Steve, in real life in real investigations, do you think the lab technicians in forensic crime labs are advised of the suspect pool before they write their reports?
Additionally, do you think the scientific statement “When we exposed many samples of type A and type B to treatment X, effect Y was observed for type A samples twice as frequently (on a per-sample basis) as it was observed for type B samples” requires any knowledge of the makeup of sample types in a later experiment to be useful in predicting that later experiment?
Ben:
Steve, in real life in real investigations, do you think the lab technicians in forensic crime labs are advised of the suspect pool before they write their reports?
No. Why would they be? It’s not their job to identify a suspect.
Additionally, do you think the scientific statement “When we exposed many samples of type A and type B to treatment X, effect Y was observed for type A samples twice as frequently (on a per-sample basis) as it was observed for type B samples” requires any knowledge of the makeup of sample types in a later experiment to be useful in predicting that later experiment?
If I understand your question, the answer is: a) To be useful, it needs no knowledge of the makeup of the sample types. b) However, it’s *more* useful if combined with such knowledge. c) Therefore if such knowledge is available, the detective — unless he’s extraordinarily incompetent — will surely make use of it.
Advo (@160):
One correction about false positive rate. The false positive
rate is defined as “The rate of occurrence of positive test results
in subjects known to be free of a disease or disorder for which an
individual is being tested.”
http://medical-dictionary.thefreedictionary.com/false-positive+rate
That rate is not dependent on your sample set. The false positive
rate in Botswana will be identical to that of Sweden.
I don’t know if anyone looked carefully at my post (#168), but I think you can use it to help resolve the confusion that has come up about the statement that “after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker.”
Here’s a way to set up this problem with balls in urns:
Urn Bob: 0 Red, 2 White, 2 Blue
Urn Carol: 1 Red, 1 White, 2 Blue
Urn Ted: 0 Red, 2 White, 2 Blue
Urn Alice: 1 Red, 0 White, 3 Blue
To make the connection to the ‘culprit’ problem, let’s say that the presence of a Red ball in the urn identifies a female, and the presence of a White ball in the urn identifies a smoker.
Suppose we obtain the ‘guilty’ urn. We know the contents of each urn, but we don’t know which one we have obtained. Let’s look at a sequence of situations and ‘evidence’.
1. Without any further evidence, we would provide a probability of 1/4 for each of the possible urns.
2. Without any further evidence, we would provide a probability of 3/4 that the ‘guilty’ urn corresponds to one of the smokers, and a probability of 1/2 that the ‘guilty’ urn corresponds to one of the males.
Now let’s examine some evidence . . .
3. If we draw one ball from the urn and it is White, then we will change our probability to 1 that the urn corresponds to one of the smokers.
4. If we draw one ball from the urn and it is Blue, then we will change our probability to 2/3 that the urn corresponds to one of the smokers.
5. If we draw one ball from the urn and it is Red, then we will change our probability to 1/2 that the urn corresponds to one of the smokers, and we’d update our probability to 0 that the urn corresponds to one of the males.
Do you see how the manner in which we change the probability depends on both the evidence we see and our knowledge of the possible ‘culprits’? When Steve said that “after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker,” he was saying that something like situation number 4 had happened.
If you drew two balls (with replacement) from the Urn, you’d have 9 possible outcomes (6 really, because the order doesn’t matter). It is a worthwhile exercise to determine the new probabilities for each urn for each of the 6 possible two-ball draws. The relevant situation is to draw a Blue and Red ball.
Steve
You said
“Really? Suppose I believe the odds that the culprit is a smoker are 2-to-1. Then I discover the suspect pool contains zero smokers. Does that mean that as a “good detective”, I have to go on believing the odds for a smoker are 2-to-1?
Or suppose I believe there’s a 50% chance that OJ did it and a 50% chance that Pierre Deligne did it. Then I discover that Deligne has an unshakable alibi, so he’s removed from the suspect pool. Would a “good detective” continue to believe that the probability for OJ is 50%?”
A good detective would not rule out the possibility that the subject pool is incomplete.
A person was murdered. Someone did it. We have no suspects in our suspect pool. Should I we reassess whether there was a murder or should we reassess whether our suspect pool is big enough?
Pat T: The fact that we have uncertainty is the reason we’re talking abobut probabilities in the first place. When we get new information, we update those probabilities. Your argument, as I understand it, is “But we still have uncertainty!”. Sure. That’s why we’re still using probabilities. But it’s important to use the right ones.
Ron:
That rate is not dependent on your sample set. The false positive
rate in Botswana will be identical to that of Sweden.
Right. But one should add (for the benefit of others, because I think you already get this) that the proability of being sick, conditional on having received a positive, is very different in Sweden than in Botswana.
Steve, ok, I feel like we’re one step away from agreement. You acknowledge that forensic lab technicians do not account for the suspect pool when they report their results. Similarly, a general scientific result applicable to the crime can be reported independent of the suspect pool. Therefore, I propose that we agree that somewhere in the forensic deductive process, there exists useful information that is independent of the suspect pool.
The proper next step in making use of this information is, of course, to add in knowledge about the suspect pool — any competent detective will surely incorporate suspect pool information.
The question is whether the information presented in the question is clearly and unambiguously, for any reasonable person at first glance, post-incorporation of the suspect pool. I suggest that it is not. I interpreted the odds sentence as describing the results of a forensic test. I understand that the best interpretation is probably that the odds sentence describes the ultimate conclusion after combining information from a forensic test with the makeup of the suspect pool. But either way, you can see how the former interpretation is at least not crazy, right?
I even provided a possible suspect-poolless forensic test in #59 that, when combined with knowledge of the suspect pool, leads to the 2:1 conclusion in the original problem, thus providing possible background that would harmonize both interpretations. But, you said that was crazy. Do you still believe it is crazy?
Steve:
But one should add (for the benefit of others, because I think
you already get this) that the proability of being sick, conditional
on having received a false positive, is very different in Sweden
than in Botswana.
I believe we’re in agreement, but I think you missed excising one
word. I believe you meant to say:
The probability of being sick, conditional on having received a
positive, is very different in Sweden than in Botswana.
Ron: Right you are. I’m going back to edit this. Thank you.
Ken B: I see Guy got there before me! I figured that there are 4 balls, 1,2,3,4. Ted could have 1,2; 1,3; 1,4; 2,3; 2,4 or 3,4. If 1 and 2 are gold then in 4 out of five of the options Ted would have a gold if he first gave me a Ping-Pong.
@Harold:
I can hear Alice’s defence attorney now: Alibi, schmalibi. If you can’t trust Ken B on combinatorics how can you trust him on Bob & Ted?
@Ron,
among 100 positive test results in Sweden, there’s going to be far fewer false positives than among 100 positive test results in Botswana.
Advo (@190):
Are you assuming a high HIV rate in Sweden and low in Botswana?
In a country with very low HIV rates, almost all positives are
false positives. Thus, the number of false positives per 100
positives would be high. In a country with a high HIV rate,
fewer of those 100 positives will be false ones.
Yes,
I switched the numbers around accidentially.
Steve, you said, “Your argument, as I understand it, is “But we still have uncertainty!”. ”
Had you told me that the 2 to 1 smoker odds were conditional on the 4 and only 4 suspects, I would’ve done the math right. Eliminating men didn’t give me a reason to reject the careful reading of evidence (and not suspects) that offered those odds of smoking.
Pat:
Had you told me that the 2 to 1 smoker odds were conditional on the 4 and only 4 suspects…..
But I did tell you this. Which part of “You are quite sure that one (and only one) of these suspects is the culprit. Moreover, after carefully examining the evidence, you’ve concluded that the odds are 2-to-1 that the culprit is a smoker” did you not understand?
—
PS: If your position is that it’s reasonable to think the detective might have completely ignored the suspect pool, then of course you’re saying that the detective is a madman, in which case we must conclude there might be no rhyme or reason to anything he does. In that case, of course, we can draw no conclusions whatsoever.
@194:
As far as I can tell Steve the complaint is that, after having identified the only possible suspects, three of whom smoke, you wrote “is a smoker” rather than “is one of the smokers.”
I wish I had said it as concisely as Ken B. Originally, I thought the post was a gotcha trying to trick people into Bayesian calculations rather than just keeping Alice at 33%. You fooled me by not fooling me, Steve.
To illustrate the point about dependence I will come clean. I was Steve’s crack investigator.
We knew it had to be one of the four as it happened in a boat. My investigation showed Bob & Ted were owing at the time, I saw them from shore.
I began listing deductions. Bob was innocent. So therefore of the three remaining suspects only one was non smoker. At this point I had to leave and lazily sent Steve just this last deduction. Later Roger confirmed the alibi for Bob & Ted.
Now knowing all this the odds are 50/50. So, is Steve wrong?
No, emphatically not. Steve followed the correct logic for combining uncertain information.
The interesting point is this. My story here bolsters Steve’s analysis even as it provides a better estimate. Why? It shows that estimates like the 2/3 always depend on a context of incomplete information so are always subject to such updating.
Of the remaining 3 only one was a man – why did you not send this information? Why did you not transmit all the information and say Bob was innocent? The reason why people find the argument unsatisfactory is that you have more information than you transmit. You know it is not Bob, yet chose to transmit only the statistics about smokers, which could mean that any of the smokers was innocent. In fact, you (or someone) must have had more information than was transmitted.
This does not make the problem wrong, but just a little unsatisfactory.