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? 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?
 A: 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.
