最优放弹珠 I

单次抽取下 A 的期望收益为

$$\frac{a}{a+b}(100-a).$$

由于两次抽取独立且策略不会因次数改变，最优策略与单次相同，最后把单次期望乘以 2。

连续最优反应可得到对称均衡在 $a=b\approx 100/3$。在整数限制下取 $a=b=33$ 可满足最优性近似。

此时单次期望收益为

$$\frac{33}{66}\cdot (100-33)=\frac{1}{2}\cdot 67=33.5,$$

两次总期望约为 $2\times 33.5=67.0$。

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Original Explanation

We can first make some simplifications to the game. Firstly, if the strategy is optimal, then if this game were to be repeated many times, they would not change their strategy. Therefore, the optimal strategy for the game where there are $2$ consecutive marble draws is the same as the optimal strategy for the game with one draw. Then, we just multiply the expected profit by $2$ to represent the two draws. Furthermore, as this game is symmetric for the two players, their optimal strategy will be the same. This point will be important later.

Let $A(a,b)$ be the expected profit that $A$ obtains with player $A$ putting in $a$ balls and $B$ putting in $b$ balls. Namely, for the one draw game,

$$A(a,b) = \frac{a}{a+b} \cdot (100 - a)$$

As player $a$ draws his ball with probability $\frac{a}{a+b}$ and $100-a$ is the payout. Let's fix $b$ and find the $a$ that is the best response to this $b$. In other words, given $b$, what $a$ optimizes $A(a,b)$? To do this, we take the partial derivative of $A(a,b)$ in $a$ and treat $a$ as continuous for now. We will then account for discreteness at the end.

This yields that

$$\frac{\partial}{\partial a}A(a,b) = -\frac{a^2 + 2ba - 100b}{(a+b)^2} = 0 \iff a^2 + 2ba - 100b = 0$$

Solving the above with the quadratic equation yields that $a^* = \frac{-2b \pm \sqrt{4b^2 + 400b}}{2}$. However, the $-$ root results in a negative value, so $a^* = \sqrt{b^2 + 100b} - b$ is the best response for player $A$ if player $B$ puts $b$ marbles in. Similarly, as this game is symmetric, the optimal response for player $B$ if player $A$ puts $a$ marbles in is $b^* = \sqrt{a^2 - 100a} - a$.

To find the optimal strategy for each player, this means that we need to find the combo $(a^*,b^*)$ such that neither of the players can do better by adjusting their strategy. We already are aware from before that $a^* = b^*$ by the symmetry of the game. Therefore, to solve for this, we just substitute in $b$ as $b^* = a^*$ in the first equation. This yields we can say that

$$a^* = \sqrt{(a^*)^2 + 100a^*} - a^* \iff 4(a^*)^2 = (a^*)^2 + 100a^* \iff a^*(3a^* - 100) = 0 \iff a^* = 0,\frac{100}{3}$$

As $0$ is not possible, we conclude that $a^* = b^* = \frac{100}{3}$ is the optimal strategy. However, this is not actually possible, as our marbles must be an integer value. Therefore, we should test $(33,33)$ and $(34,34)$ to see if they are Nash equilibria.

For $(33,33)$, the expected payout for one draw for each player is $\frac{67}{2}$. One can check that by varying $a$ and keeping $b$ fixed at $33$, player $A$ can't do any better. Therefore, $(33,33)$ is a Nash equilibrium. For $(34,34)$, the expected payout is $33$. However, one can also verify that the expected payout for $(33,34)$ is also $33$. However, this can't be an equilibrium, as $b$ should change to $33$ marbles to yield higher expected payout. Thus, while $(34,34)$ is also a Nash equilibrium, $(33,33)$ is preferable because of the higher expected payout. This means that the optimal strategy is for both players to place $33$ marbles and have a total expected payout of $\frac{67}{2} \cdot 2 = 67$.