Two players each hold from one to five coins. S wins all if the same number; but D, if a different number. What are their optimal strategies?

Question

It seems to me that S will play a low number of coins because the two players will likely reveal a different number of coins; and D, a moderate number of coins to avoid those low numbers while not needlessly risk too much on the high numbers to win. Of course, if S plays more of the low numbers, than D may resort to more of the high numbers more often. Am I in the ballpark?

Answer 1

I'll assume "wins all" means that the winning player wins the coins shown by the opponent.

Thus, the game is zero sum.

Claims:

• The unique optimal strategy for $S$ is to always show $1$ coin.
• An optimal strategy for $D$ is to show $1,2,3,4,5$ coins with probabilities $q_1,q_2,q_3,q_4,q_5$, respectively, where $$ q_1 = 0,\;\;\; q_2 = \frac{17}{154},\;\;\; q_3 = \frac{37}{154},\;\;\; q_4 = \frac{47}{154},\;\;\; q_5 = \frac{53}{154} $$

To show that the "always show $1$ coin" is a unique optimal strategy for $S$, suppose instead that $S$ shows $1,2,3,4,5$ coins with probabilities $p_1,p_2,p_3,p_4,p_5$, respectively, where $p_1 < 1$.

From $p_1 < 1$, we get $p_2+p_3+p_4+p_5 > 0$.

Then if $D$ plays the strategy $q_1,q_2,q_3,q_4,q_5$, as defined above, the expected value for $S$ is \begin{align*} &-p_1 +2p_2\left({\small{\frac{17}{154}}}-{\small{\frac{137}{154}}}\right) +3p_3\left({\small{\frac{37}{154}}}-{\small{\frac{117}{154}}}\right) +4p_4\left({\small{\frac{47}{154}}}-{\small{\frac{107}{154}}}\right) +5p_5\left({\small{\frac{53}{154}}}-{\small{\frac{101}{154}}}\right) \\[4pt] =\;&-p_1-\left({\small{\frac{120}{77}}}\right)\left(p_2+p_3+p_4+p_5\right)\\[4pt] =\;&-(p_1+p_2+p_3+p_4+p_5)-\left({\small{\frac{43}{77}}}\right) (p_2+p_3+p_4+p_5)\\[4pt] =\;&-1-\left({\small{\frac{43}{77}}}\right)(p_2+p_3+p_4+p_5)\\[4pt] < \;&-1\\[4pt] \end{align*} which shows that any mixed strategy for $S$ with $p_1 < 1$ is inferior to the pure "always show $1$ coin" strategy.

But if $S$ uses the "always show $1$ coin" strategy, then no strategy for $D$ can possibly yield a win of more than $1$ coin.

Thus, the specified strategies for $S$ and $D$, are optimal, as claimed.