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.

Continue reading ‘A Bayesian Solution’

The solution to yesterday’s mortgage puzzle:

Commenters pointed to several reasons why biweekly payments of (say) $500 will pay your mortgage off so much faster than monthly payments of $1000, but of these by far the most important is that biweekly payments of $500 add up to 500 x 26 = 13,000 dollars, whereas monthly payments of $1000 add up to 1000 x 12 = 12,000 dollars. With the biweekly payments, you make the equivalent of 13 monthly payments every year.

In other words, the key observation is that two weeks is not half a month.

My colleague Michael Wolkoff posed this puzzle to me many years ago, and I’m embarrassed to admit I failed to solve it before Michael gave me the solution. I was reminded of it yesterday when I got a biweekly-plan offer in the mail.

Monday’s puzzle was open to various interpretations, but under what seems to me to be the most straightforward interpretation, if the number of runners you pass is the same as the number who pass you, you’re the mean runner, not the median.

You can find plenty of correct analysis in Monday’s comment section (see in particular Harold’s perfect comment #39), but here’s a more longwinded explanation:

First, suppose you randomly sample a large number of other runners and discover that half of them are faster than you and half are slower. Then you’re entitled to conclude that you’re the median runner (or, if we’re being careful, you’re entitled to conclude that you’re probably close to the median, since there’s always a chance your sample was unrepresentative).

Now in the problem as given it’s certainly true that half the runners you encounter are faster than you and half are slower. So you might be tempted to use the above reasoning and conclude that you’re the median runner. But that won’t work, because the runners you encounter are not a random sample.

So let’s start over. We might as well assume that you’re the center of the universe, so you’re completely motionless. Everyone who’s faster than you is running forward and everyone who’s slower than you is running backward. People “pass” you when they run past you in the forward direction, and you “pass” them when they run past you in the backward direction.

Continue reading ‘Wednesday Solution’

Remember the bullet problem from two weeks ago? If not, I’ll give you a few moments to refresh your memory.

Okay. Are we ready?

I am pretty well convinced that 16 bags suffice, as argued by Mike H, Jeffrey, Brian, and Jacopo, and generalized by Categories+Sheaves.

The explanation I liked best was Jeffrey’s. Let me try to illustrate it. (Warning: There’s no way, I think, to make this instantly clear. It will take a little work to understand it. Only you can assess the opportunity cost of your own time!)

The gray rectangle is the room you’re in.

The blue dot is you. The red dot is the shooter.

With mirrors on all the walls, you’ll perceive yourself as standing in an infinite grid. The graph shows 16 of the grid rectangles, but the pattern continues forever in every direction.

You’re standing at some point (a,b). The shooter is at some point (p,q). You’ll see copies of that shooter at every point with coordinates (2m ± p, 2n ± q) where m and n range over all integers. Sixteen of those infinitely many points are shown in the picture.

Continue reading ‘Tuesday Solution’

This has been making the rounds lately; I’m not sure where it first came from.

You’re in a rectangular room. Elsewhere in the room is a man with a gun, who shoots a bullet in a random direction. The bullet careens around the room, bouncing off walls, until it hits either you or one of the various punching bags you’ve placed around the room for purposes of absorbing the bullet. The punching bags must be positioned before you know the random direction of the bullet (though you do know both your own location and the bad guy’s location, neither of which you can change). How many punching bags do you need to guarantee your survival?

This being a math problem, you should treat the room as two dimensional, and yourself, the bullet and the punching bags as points.