## Study math, get rich July 6, 2010

Posted by Ezra Resnick in Math, Puzzles.

You receive a letter informing you that you have been selected as one of 32 lucky people around the world to participate in an experiment, and you may win a million dollars! The rules are simple: by midnight tonight, if exactly one of the 32 participants calls a certain phone number (it doesn’t matter who), you will each receive a million dollars. In any other event (more than one person calls the number, or no one calls), none of you will get anything. Assume the offer is genuine and all participants received the letter on the same day, but you have no way of tracking down the others. What should you do?

The solution is rather straightforward. In order to maximize the chances of exactly one participant making the phone call, each participant should decide to make the call with probability 1/32. For instance, you could roll a die with 32 faces (labeled 1 through 32) and make the phone call only if the number 1 comes up. Another option is to toss five coins and make the call only if they all come up heads (25=32). Problem solved. A more interesting question, however, is this: assuming that all the participants use the optimal strategy, what are the odds that they will actually win the money? And what would be the odds of winning if there were a thousand participants? Or a million?

What I like about problems like this is that the mathematics of the solution are quite simple, and yet the result is counter-intuitive, demonstrating (yet again) how bad human intuition about probability is. In this case, I would have guessed that the odds of winning are pretty small, and that they decrease significantly as the number of participants grows. With a million participants, the chances that all the dice rolls will work out so that exactly one participant makes the call seem pretty slim. Well – let’s think it through. Assume for the moment that there are only two participants, so they each toss a coin and make the phone call only if the result is heads. What are the odds of their winning the money? There are four (equally probable) outcomes: either they both get heads, or they both get tails, or the first participant gets heads and the second gets tails, or the first participant gets tails and the second gets heads. In the first of the four outcomes, they will both make the phone call, and in the second, neither of them will – in both cases, they lose. In either of the latter outcomes, however, exactly one of them makes the call and they win. So the odds of winning in this case are 50%. Not bad, but surely the odds decrease rapidly as we add participants?

Assume there are now three participants. If they all follow the strategy we described, there are 27 (33) possible outcomes, and in 12 of them exactly one participant makes the call, so the odds of winning are about 44%. In the general case, let n denote the number of participants, and assume they each use a die with faces labeled 1 through n. The chances of winning due to participant #1 being the only one to make the phone call are 1/n times ((n-1)/n)n-1 (participant #1 must roll a 1, and all the other n-1 participants must roll any of the n-1 values other than 1). The odds of winning due to participant #2 being the only one to make the phone call are the same as for participant #1, as they are for any other participant, so for the total probability of winning we must multiply the previous expression by n, giving the final answer: ((n-1)/n)n-1.

Plugging n=2 or n=3 into this formula gives the results mentioned above; plugging in n=32 gives a winning probability of about 37% – still pretty good! What about larger values of n? Those who know some calculus will recognize that the above formula converges to 1/e as n grows (where e is Euler’s number, namely 2.718…), so in fact, the odds of winning remain close to 37% no matter how large the number of participants. (Interestingly, the number 1/e pops up in many other, seemingly unrelated, probabilistic questions.)

So remember (especially when visiting a casino): don’t trust your intuitions when it comes to probability!