In Craps, one of the bets you can do is 'pass line', which means player should roll 7 or 11 with two dices. I was thinking on how the probabilities on the dice should be skewed (assuming both dice are identical) to maximize the win on that bet and I came up with this solution.
We find all combinations which would give a positive outcome (which would be 1-6, 2-5, 3-4, 5-6 and it's swapped counterparts). Then compute the frequencies at which the numbers on those dice appear in winning combinations.Finally, divide those frequencies by the total number of appearances, which would give us the probabilities for each number on the die. The probability of getting 7 or 11 on that die is 28.1%, which is larger than for the standard die(22.2%).
Is this die actually optimal for such bet? If not, how would one find the probabilities for the optimal die?
Let $p_i \ge 0$ be the probability for side $i \in \{1,\dots,6\}$. The problem is to maximize $\sum\limits_{\substack{1 \le i, j \le n\\i+j \in \{7,11\}}} p_i p_j$ subject to $\sum\limits_{i=1}^6 p_i = 1$. Some of the optimal solutions $(p_1,\dots,p_6)$ are: \begin{align} (1/2, 0, 0, 0, 0, 1/2) \\ (0,1/2, 0, 0, 1/2, 0) \\ (0,0,1/2, 1/2, 0, 0) \\ (0,0,0,0,1/2, 1/2) \end{align} But there are others, where $0 < x < 1/2$: \begin{align} (0,x,0,0,1/2,1/2-x) \\ (x,0,0,0,1/2-x,1/2) \end{align}