roll a dice repeatedly until the sum goes above 63

Question

I saw this question online but I don't know if my solution is right or not. Here is the original question:

Roll a die repeatedly. Say that you stop when the sum goes above 63. What is the probability that the second to last sum value (total) was X. Make a market on this probability. Ie what is your 90 percent confidence interval.

This is my solution:

X=63, 6 possible last roll as 1,2,3,4,5,6;

X=62, 5 possible last roll as 2,3,4,5,6;

X=61, 4 possible last roll as 3,4,5,6;

X=60, 3 possible last roll as 4,5,6;

X=59, 2 possible last roll as 5,6;

X=58, 1 possible last roll as 6.

BUT I don't know if it's right to think backward instead of considering the possible second to last value first? i.e. how many possibility to get a 63 regardless of last roll.

Thank you very much!

Answer 1

You can calculate the odds quite easily using excel, and this is what happens, suppose you start rolling a dice and keep adding the values to infinity, the probability of seeing a certain sum $X$ will converge to $\frac{6}{21}$ as $X$ goes to infinity, with values around 63 having up 8 decimals of accuracy. So for practical purposes your method will work for large numbers like 63.