Up for a Saturday morning puzzle?
This is adapted from a more complicated version submitted by Jared Bronski in “The Riddler”. (Thanks to Matt Mooore who gifted me the book this week).
It is one more of those hats on your head and strategy formulation question.
There are three of you. Randomly a white or black hat is put on your head. Each one of you can see the color of the hat on the other two but not the one that is on your own head.
One by one, you will be asked to guess what color hat you have on your head.
Your options are (*) tell a color – Black or White or say (*) Pass.
At any point, if somebody calls a wrong color – you all lose. But if somebody calls a right color – you all win. And if everybody passes at the end of first round, then you all lose.
What strategy can you formulate beforehand to maximize the chances of winning? And what is that probability?
Now for the answer.
The correct answer should be 3/4 (0.75).
There are 8 possibilities
Note that the first person – if she says White or Black, there is a 50-50 chance that they all win or they all lose (she has no idea what she has). So, minimum there is 1/2 chance to win.
She also has another option – to say “Pass”. Can that be used to increase the probability? Sure. She can prearrange to tell her friends that when she says Pass, it means she saw something in particular; else she will go for the guess work and take a 50-50 chance.
Assuming she has no idea who will be called next, she can signal that she is seeing 2 identical color on the other two (or two different colors on the other two) by saying Pass. The second person sees the color of the hat on the third person and together with the info conveyed thru “Pass”, the second person can nail the answer (100% prob).
So, let’s see what we have. One strategy would be – “Whoever gets called first checks the color on the other two. If they are same colors, he or she says Pass. If they are different colors, go ahead and take a guess on what you have on your head. If the first person passes, the second person looks at the third person’s hat color and calls out that color”.
The only way they all lose is if
(*) the first person saw dissimilar colors on the other two (prob 1/2) AND guessed his/her own color wrong (prob 1/2). Thus the prob of losing is 1/4
Put another way, if he/she saw the other two to be same (prob 1/2), the call is Pass. The second person knows that he/she and the third person has the same color. He/she just calls whatever he/she sees on thee third person’s hat. (prob 1).
However, if the first person saw dissimilar colors (prob 1/2), he/she takes a guess (prob 1/2). Therefore all can win 1/4 of the time.
So prob of winning = 1/2*1 + 1/2*1/2 = 3/4