Answer:
The probability of winning directly is, as you calculated, 8/36, and the probability of losing directly is (1+2+1)/36=4/36.
For the remaining cases, you need to sum over all remaining rolls. Let p be the probability of rolling your initial roll, and q=6/36=1/6 the probability of rolling a 7. Then the probability of rolling your initial roll before rolling a 7 is p/(p+q), and the probability of rolling a 7 before rolling your initial roll is q/(p+q). Thus, taking into account the probability of initially rolling that roll, each roll that doesn't win or lose directly yields a contribution p2/(p+q) to your winning probability.
For p=5/36, that's
(536)25+636=2511⋅36,
and likewise 16/(10⋅36) and 9/(9⋅36) for p=4/36 and p=3/36, respectively. Each of those cases occurs twice (once above 7 and once below), so your overall winning probability is
836+236(2511+1610+99)=244495=12−7990≈12−0.007.
Step-by-step explanation:
Suppose you throw a 4 and let p(4) your winning probability. At your next roll you have a probability 3/36 of winning (you throw a 4), a probability 6/36 of losing (you throw a 7) and a probability 27/36 of repeating the whole process anew (you throw any other number). Then:
p(4)=336+2736p(4),so thatp(4)=13.
Repeat this reasoning for the other outcomes and then compute the total probability of winning as:
ptot=836+336p(4)+436p(5)+…
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