I am trying to derive probabilities of winning the game craps on the ith throw of 2 dice (1 is fair and 1 is not). Suppose we are on the first roll, and we do not win or lose (no 7,11,2,3,12), instead we get the sum X with roll probability p.
Given that we have already made it to this point (Being on the second throw), what’s the chance of winning going forward with each throw until the game ends in either losing or winning?
Preferably, a Python Code solution in addition to the theory would be super helpful.
Now, I am stuck on deriving the chance of winning going forward at the 2nd throw all the way until the game is either won or lost. I hope somehow can guide. Thanks!
Determine the probability for each possible sum: $$\begin{array}{c|c}\text{Sum} & \text{Prob} \\ \hline 2 & 0 \\ \hline 3 & \dfrac{1}{24} \\ \hline 4 & \dfrac{1}{12} \\ \hline 5 & \dfrac{1}{8} \\ \hline 6 & \dfrac{1}{6} \\ \hline 7 & \dfrac{1}{6} \\ \hline 8 & \dfrac{1}{6} \\ \hline 9 & \dfrac{1}{8} \\ \hline 10 & \dfrac{1}{12} \\ \hline 11 & \dfrac{1}{24} \\ \hline 12 & 0\end{array}$$
Next, we determine the probability of the point:
Let $P=p$ be the event that the point is $p$, where $p \in \{4,5,6,8,9,10\}$.
Thus, we have:
$$P(P\in \{4,5,6,8,9,10\}) = \dfrac{2}{12}+\dfrac{2}{8}+\dfrac{2}{6} = \dfrac{3}{4}$$
And then
$$P(P=p|P\in \{4,5,6,8,9,10\}) = \dfrac{P(P=p)}{P(P\in \{4,5,6,8,9,10\})}$$
for all $p \in \{4,5,6,8,9,10\}$. Which yields:
$$\begin{array}{c|c}\text{Point} & \text{Prob} \\ \hline 4 & \dfrac{1}{9} \\ \hline 5 & \dfrac{1}{6} \\ \hline 6 & \dfrac{2}{9} \\ \hline 8 & \dfrac{2}{9} \\ \hline 9 & \dfrac{1}{6} \\ \hline 10 & \dfrac{1}{9}\end{array}$$
Let $P(W^*)$ be the probability that you eventually win. Let $P(R=r)$ be the probability of rolling a certain number on your next roll. Then:
$$P(W^*|P=p) = P(R=p)+(1-P(R=p)-P(R=7))P(W^*|P=p)$$
Solving for $P(W^*|P=p)$ gives:
$$P(W^*|P=p) = \dfrac{P(R=p)}{P(R=p)+P(R=7)}$$
Next, we can use the law of total probability:
$$\begin{align*}P(W^*|P \in \{4,5,6,8,9,10\}) & = P(W^*|P=4)P(P=4|P \in \{4,5,6,8,9,10\}) \\ & + P(W^*|P=5)P(P=5|P \in \{4,5,6,8,9,10\}) \\ & + P(W^*|P=6)P(P=6|P \in \{4,5,6,8,9,10\}) \\ & + P(W^*|P=8)P(P=8|P \in \{4,5,6,8,9,10\}) \\ & + P(W^*|P=9)P(P=9|P \in \{4,5,6,8,9,10\}) \\ & + P(W^*|P=10)P(P=10|P \in \{4,5,6,8,9,10\}) \\ & = 2\left(\dfrac{\tfrac{1}{12}}{\tfrac{1}{12}+\tfrac{1}{6}}\right)\left(\dfrac{1}{9}\right) + 2\left(\dfrac{\tfrac{1}{8}}{\tfrac{1}{8}+\tfrac{1}{6}}\right)\left(\dfrac{1}{6}\right)+2\left(\dfrac{\tfrac{1}{6}}{\tfrac{1}{6}+\tfrac{1}{6}}\right)\left(\dfrac{2}{9}\right) \\ & = \dfrac{83}{189}\end{align*}$$
To verify our answer, we will want to check the probability that you eventually lose. It should be the complement of our probability that you eventually win.
$$P(L^*|P=p) = P(R=7)+(1-P(R=7)-P(R=p))P(L^*|P=p)$$
And solving for $P(L^*|P=p)$ gives:
$$P(L^*|P=p) = \dfrac{P(R=7)}{P(R=7)+P(R=p)}$$
$$\begin{align*}P(L^*|P \in \{4,5,6,8,9,10\}) & = P(L^*|P=4)P(P=4|P \in \{4,5,6,8,9,10\}) \\ & + P(L^*|P=5)P(P=5|P \in \{4,5,6,8,9,10\}) \\ & + P(L^*|P=6)P(P=6|P \in \{4,5,6,8,9,10\}) \\ & + P(L^*|P=8)P(P=8|P \in \{4,5,6,8,9,10\}) \\ & + P(L^*|P=9)P(P=9|P \in \{4,5,6,8,9,10\}) \\ & + P(L^*|P=10)P(P=10|P \in \{4,5,6,8,9,10\}) \\ & = 2\left(\dfrac{\tfrac{1}{6}}{\tfrac{1}{12}+\tfrac{1}{6}}\right)\left(\dfrac{1}{9}\right) + 2\left(\dfrac{\tfrac{1}{6}}{\tfrac{1}{8}+\tfrac{1}{6}}\right)\left(\dfrac{1}{6}\right)+2\left(\dfrac{\tfrac{1}{6}}{\tfrac{1}{6}+\tfrac{1}{6}}\right)\left(\dfrac{2}{9}\right) \\ & = \dfrac{106}{189}\end{align*}$$
And by the law of total probability, we have:
$$P(W^*|P \in \{4,5,6,8,9,10\})+P(L^*|P \in \{4,5,6,8,9,10\}) = \dfrac{83}{189}+\dfrac{106}{189} = 1$$
For the probability that you win on the $n$-th throw of the dice, $P(W=n|P\in \{4,5,6,8,9,10\}$, we first calculate $P(W=n|P=p)$. You can win on the $n$-th throw if every roll from $k=2$ to $k=n-1$, you roll some number other than $p$ or $7$. Thus:
$$P(W=n|P=p) = (1-P(R=7)-P(R=p))^{n-2}(P(R=p))$$
Writing this out, we obtain the formulas:
$$\begin{align*}P(W=n|P=4) & = \left(\dfrac{1}{12}\right)\left(\dfrac{3}{4}\right)^{n-2} \\ P(W=n|P=5) & = \left(\dfrac{1}{8}\right)\left(\dfrac{17}{24}\right)^{n-2} \\ P(W=n|P=6) & = \left(\dfrac{1}{6}\right)\left(\dfrac{2}{3}\right)^{n-2} \\ P(W=n|P=8) & = \left(\dfrac{1}{6}\right)\left(\dfrac{2}{3}\right)^{n-2} \\ P(W=n|P=9) & = \left(\dfrac{1}{8}\right)\left(\dfrac{17}{24}\right)^{n-2} \\ P(W=n|P=10) & = \left(\dfrac{1}{12}\right)\left(\dfrac{3}{4}\right)^{n-2}\end{align*}$$
Then, to get the overall probability, let $A = \{4,5,6,8,9,10\}$. We use the formula:
$$\begin{align*}P(W=n|P\in A) & = P(W=n|P=4)P(P=4|P\in A) \\ & + P(W=n|P=5)P(P=5|P\in A) \\ & + P(W=n|P=6)P(P=6|P\in A) \\ & + P(W=n|P=8)P(P=8|P\in A) \\ & + P(W=n|P=9)P(P=9|P\in A) \\ & + P(W=n|P=10)P(P=10|P\in A)\end{align*}$$
Plugging in and simplifying, we obtain:
$$\boxed{P(W=n|P\in A) = \left(\dfrac{1}{54}\right)\left(\dfrac{3}{4}\right)^{n-2}+\left(\dfrac{1}{24}\right)\left(\dfrac{17}{24}\right)^{n-2} + \left(\dfrac{2}{27}\right)\left(\dfrac{2}{3}\right)^{n-2}}$$
To check this, we can verify:
$$\sum_{n=2}^\infty P(W=n|P\in A) = \left(\dfrac{1}{54}\right)\left(\dfrac{1}{1-\tfrac{3}{4}}\right)+\left(\dfrac{1}{24}\right)\left(\dfrac{1}{1-\tfrac{17}{24}}\right)+\left(\dfrac{2}{27}\right)\left(\dfrac{1}{1-\tfrac{2}{3}}\right) = \dfrac{83}{189}$$
Which matches what we found above.