## A Mind-Expanding Probability Problem

Three mathematicians play a three-stage game. In the first stage, they talk to one another about the strategies they are going to use to play the next two stages.

Then, in the second stage, they each put on a hat. Each hat will either be red or blue, with equal probability, and the colors of the three hats are independent. Each mathematician does not see his own hat, but he can see the hats of his two friends. The mathematicians may not communicate to each other any information about the hats, or discuss any strategies to use in the game at this point.

Finally, in the third stage, each mathematician writes down on a slip of paper one word. Each may write ‘Red’, ‘Blue’, or ‘Pass’. He may not show this slip to his friends or communicate any information to them in any way.

Now the results come in. All strips of paper are shown. If at least one mathematician wrote down the color of his own hat, and no mathematician wrote down a color other than the color of his hat, then all mathematicians are awarded an equal sum of money. Otherwise, meaning if no mathematician correctly identified the color of his hat, or if even one identified incorrectly, then no mathematicians get any money at all.

What strategy should the mathematicians agree upon in the first stage so that they maximize their average earnings?

Agree to show a different emotion depending on what other hats you see, if you see red and blue, be happy, if you see blue and blue, be sad, if you see red and red be angry!!

based on each person’s expressions…yadda yadda yadda

If a mathematician sees two hats of same color, he writes down the remaining color as the color of his hat. If he sees two hats of two different colors, they write he writes down pass. This ensures only one mathematician writes down an answer in each game. And there is a 75% chance of him being right.

If we assign a win the point value of 1, and a loss the point value of 0, I’m seeing a maximum expected value of .75 for the game.

Strategy: Pass if you see different colors, Guess the opposite color if you see the same one.

Why? There is a 1/4 chance of all the hats being the same color (1/8 all red, 1/8 all blue). In this case, all three mathematicians guess incorrectly at the same time.

There is a 3/4 chance of there being one odd man out (RRB, RBR, BRR, RBB, BRB, BBR). In each of these cases precisely two people will see different colored hats and pass, and precisely one person will see the same colored hats and correctly guess that his hat was the different color.

A 3/4 chance of winning the whole prize + a 1/4 chance of nothing yields a .75 expected value overall, which is much better than the .125 expected value when everyone guesses randomly or the .5 expected value when one person is nominated to guess randomly.

I’d like to know if there’s a way to do better, but I’m skeptical if there is.

It’s interesting that the optimal strategy here uses the gambler’s fallacy to maximize payoff.

The best strategy they could possibly agree upon would be:

First off, if they see two hats of the same colour, then guess the opposite one. Secondly, if they see two different colour hats, then pass.

Correct. The 75% winning strategy is indeed the optimal one.