Acting As a Team I

Problem

Acting As a Team I

Strategy 1

Each player chooses the color randomly. The probability of success is $\displaystyle P=\left(\frac{1}{2}\right)^3=\frac{1}{8}.$

Strategy 2

There are eight possible combinations of hat colors. Each player has to choose between $R(ed),$ $B(lue),$ or $P(ass),$ making $27$ possible team answers. Each player makes a random choice, now between three possibilities.

If all pass, the team loses. With two passes, there are $6$ possible answers, each with probability of $\displaystyle \frac{1}{2}$ of a correct guess. With one pass, there are $12$ possible answers, each with probability of $\displaystyle \frac{1}{4}$ of a correct guess. With no passes, there are $8$ possible answers, each with probability of $\displaystyle \frac{1}{8}$ of being right. The total comes to

$\displaystyle\begin{align}P&=\frac{1}{27}\left(0\cdot 1+\frac{1}{2}\cdot 6+\frac{1}{4}\cdot 12+\frac{1}{8}\cdot 8\right)\\ &=\frac{3+3+1}{27}=\frac{7}{27}\gt \frac{7}{28}=\frac{1}{4}. \end{align}$

Strategy 3

Strategies 1 and 2 are stupid: the fewer people make a guess, the higher is the probability of success. However, in order to win at least one guess ought to be made. The probability of success is $\displaystyle \frac{1}{2}.$

Elwin Berlekamp's Strategy

  1. If you see two hats of the same color, guess the other color.
  2. If you see two hats of different colors, pass.

The results of the strategy are summarized in the table below:

$\begin{array}{ccccccccc} A&B&C&&A&B&C&&Outcome\\ \hline R&R&R&&GBW&GBW&GBW&&lose\\ R&R&B&&pass&pass&GBC&&win\\ R&B&R&&pass&GBC&pass&&win\\ R&B&B&&GRC&pass&pass&&win\\ B&R&R&&GRC&pass&pass&&win\\ B&R&B&&pass&GRC&pass&&win\\ B&B&R&&pass&pass&GRC&&win\\ B&B&B&&GRW&GRW&GRW&&lose \end{array}$

In the table, $GBW=guess\;blue:\;wrong,$ $GBC=guess\;blue:\;correct,$ $GRW=guess\;red:\;wrong,$ $GRC=guess\;red:\;correct.$ The strategy gives six team wins out of eight: $\displaystyle \frac{6}{8}=\frac{3}{4}.$

Acknowledgment

This is one of the problems from Chapter 6 of J. Havil's Impossible? Surprising Solutions to Counterintuitive Conundrums (Princeton University Press, 2008). Previously, the problem appeared in P. Winkler's Mathematical Puzzles: A Connoisseur's Collection (A K Peters, 2004)

 

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