The question is related to this question. I'll repeat the question for the sake of completeness:
Roll a die repeatedly. Say that you stop when the sum goes above 63. What is the probability that the second to last value was X. Make a market on this probability. Ie what is your 90 percent confidence interval.
I was trying to solve it with another approach. Let $T$ denotes the event of termination of the game. Then $$ 1=P(T) = P(T|62)P(62) + ....+P(T|57)P(57) $$ by the law of total probability. So the question is to find out what is $P(T|X)$. To obtain $P(T|X)$, we need first to know what is $P(X)$ But I can't find a way to calculate the P(X). Is there any smart way to do it?
Alternatively, is there any rigorous mathematical proof to solve this problem?
Roughly speaking the probability you hit any value is $\frac 27$ because the average roll is $\frac 72$. There is a small perturbation which you can assess from the recursion on the chance you hit any value. I believe it is very small by this point. Then if you hit $63$ the chance you go over from there is $1$. If you hit $62$ the chance you go over from there is $\frac 56$ and so on. The chance that the next to last value was $63$ is then $\frac 6{21}$, the chance of $62$ is $\frac 5{21}$ and so on. Given that, you should be able to "make a market". I don't know how that gives a specific answer. I could come up with a number of fair bets.