# A probability puzzle

To be totally clear: the question is not assuming that there is ONLY one student whose neighbors both flipped heads, just that there is AT LEAST one such student. You can imagine that the teacher first asks for all students whose neighbors both flipped heads to step forward, then randomly selected one of the students that had stepped forward.

It seemed initially obvious to me that the teacher was correct. There are exactly as many possible worlds in which the three students are HTH as there worlds in which they are HHH, right? Knowing how your neighbors’ coins landed shouldn’t give you any information about how your own coin landed, and to think otherwise seems akin to the Gambler’s fallacy.

But in fact, the teacher is wrong! It is in fact more likely that the student flipped tails than heads! Why? Let’s simplify the problem.

Suppose there are just three students standing in a circle (/triangle). There are eight possible ways that their coins might have landed, namely:

HHH
HHT
HTH
HTT
THH
THT
TTH
TTT

Now, the teacher asks all those students whose neighbors both have “H” to step forward, and AT LEAST ONE steps forward. What does this tell us about the possible world we’re in? Well, it rules out all of the worlds in which no student could be surrounded by both ‘H’, namely… TTT, TTH, THT, and HTT. We’re left with the following…

HHH
HHT
HTH
THH

One thing to notice is that we’re left with mostly worlds with lots of heads. The expected total of heads is 2.25, while the expected total of tails is just 0.75. So maybe we should expect that the student is actually more likely to have heads than tails!

But this is wrong. What we want to see is what proportion of those surrounded by heads are heads in each possible world.

HHH: 3/3 have H (100%)
HHT: 0/1 have H (0%)
HTH: 0/1 have H (0%)
THH: 0/1 have H (0%)

Since each of these worlds is equally likely, what we end up with is a 25% chance of 100% heads, and a 75% chance of 0% heads. In other words, our credence in the student having heads should be just 25%!

Now, what about for N students? I wrote a program that does a brute-force calculation of the final answer for any N, and here’s what you get:

 N cr(heads) ~ 3 1/4 0.25 4 3/7 0.4286 5 4/9 0.4444 6 13/32 0.4063 7 1213/2970 0.4084 8 6479/15260 0.4209 9 10763/25284 0.4246 10 998993/2329740 0.4257 11 24461/56580 0.4323 12 11567641/26580015 0.4352 13 1122812/2564595 0.4378 14 20767139/47153106 0.4404 15 114861079/259324065 0.4430 16 2557308958/5743282545 0.4453 17 70667521/157922688 0.4475

These numbers are not very pretty, though they appear to be gradually converging (I’d guess to 50%).

Can anybody see any patterns here? Or some simple intuitive way to arrive at these numbers?

## One thought on “A probability puzzle”

1. sj2mi says:

Nice puzzle. Reminiscent of the question about Pr two daughters given at least one for two child families. If you select the person first and then keep making the group toss a coin until their neighbors both toss H then obviously the answer is 1/2.