Expected value of dice game (with options), follow up This is a follow up to this question: Expected value of dice game (with options)
Two people throw a regular $6$-sided dice, and the person with the larger number wins $1$ dollar (and the person with the lower number neither winning nor losing anything), with nobody winning anything in the case of a tie.
The expected value of playing this game is going to be$${{15}\over{36}} = {5\over{12}}.$$If I have the option of paying $25$ cents to add $2$ to the number I get, when would I do so, and what's my new expected value? I'd pay to add $2$ only when I don't roll a $6$, so the expected value when playing then becomes$$\frac{1}{6}\left(\frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6} + \frac{5}{6}\right) - \frac{5}{6} \times \frac{1}{4} = \frac{35}{72}.$$Question. Now let's assume my opponent also has the option to pay $25$ cents to add $2$ to the number he gets, but neither of us will know each others' initial rolls nor whether or not we choose to pay a dollar. We will only know each others' final numbers. When then should I choose to pay $25$ cents to add $2$ to my number, and what's the new expected value of playing this game?
 A: This new modification moves us into the realm of game theory. Your best strategy will depend on how you think your opponent will behave. Your "strategy" in this game can be described via a list of probabilities $(p_1, p_2, p_3, p_4, p_5, p_6)$. To execute this strategy, first suppose you rolled $k$ on your die. Then you should pay to upgrade with probability $p_k$, and skip paying with probability $1 - p_k$. Let's also say your opponent's strategy is given by $(q_1, q_2, \cdots, q_6)$.
In a game theory setting, there won't be any such thing as a universally optimal strategy. However, a reasonable goal can be to find a Nash equilibrium, which basically means a combination of strategies where neither player can gain by changing their plan if they assume the other player won't change. Here's how we can solve for Nash equilibria in this case.
Let $W$ be your winnings as a random variable, given that both players use these strategies. Let $O$ be your opponent's final score (die value, plus bonus if he pays for it). Then $P[O=k] = \frac{1}{6} (q_{k-2} + 1 - q_k)$ where we take values $q_j$ to be 0 if $j < 1$ or $j > 6$.
Now we can calculate your expected winnings $E[W]$ as follows:
$$\begin{align}
E[W] &= \sum_{k=1}^6 \frac 1 6 \left[ p_k (E[W | \text{roll $k$ then pay to upgrade}]-1/4) + (1-p_k) E[W | \text{roll $k$, don't pay}] \right] \\
&= \frac 1 6 \sum_{k=1}^6 p_k (P[O<k] - 1/4) + (1-p_k) P[O<k+2].
\end{align}$$
At this point I've written enough formulas to fully define the expected value as a function of the $p_j$ and $q_j$. The expression doesn't look like much fun to work with by hand though, so I'll move to using Mathematica for the computations from here on out.
First, I compute an expression for the expected value.
q[0] = 0; q[-1] = 0; q[7] = 0;
o[k_] := 1/6*(q[k - 2] + 1 - q[k]);
e = 1/6*Sum[
    p[k]*(Sum[o[j], {j, 1, k + 1}] - 1/4) + (1 - p[k])*
      Sum[o[j], {j, 1, k - 1}]
    , {k, 1, 6}] // Simplify


We'll also want the derivatives $\frac{dE[W]}{dp_j}$:
de = D[e, {Table[p[k], {k, 1, 6}]}];
de // MatrixForm


My goal is to find a combination where $p_j = q_j$ for each $j$ AND neither player has incentive to switch strategies. To make sure nobody wants to switch, for each $j$ we need
$$
\left(\frac{dE[W]}{dp_j} = 0\right) \text{ OR }
\left(\frac{dE[W]}{dp_j} < 0 \text{ and } p_j = 0 \right) \text{ OR }
\left(\frac{dE[W]}{dp_j} > 0 \text{ and } p_j = 1 \right).
$$
We can plug these requirements directly into Mathematica:
Reduce[
 Table[(de[[k]] == 0 && 0 <= q[k] <= 1) || (de[[k]] > 0 && 
     q[k] == 1) || (de[[k]] < 0 && q[k] == 0), {k, 1, 6}]
 , Table[q[k], {k, 1, 6}]]


Conclusion
We conclude that both players can use the strategy $\left( \frac{1}{6}, \frac 1 3, \frac 1 3, \frac 2 3, \frac 1 2, 1 \right)$ and then neither player will have incentive to change to a different strategy.
This combination of strategies results in expected winnings of $\frac{73}{216}$ for each player. They both would have been better off if the option to pay for +2 were not given in the first place! That isn't really surprising; the total winnings per game will be significantly lower in the new version because some money drains away for the bonus payments, and the game is symmetric so both players will share the burden equally.
