Probability in modified game of dice w/ dodecahedra

Hi, I saw this question on a forum:

Suppose a modified version of the dice game craps is played with two regular dodecahedra.
Each die has its sides numbered from 1 to 12 so that after each throw of the dice the sum of the numbers on the top two surfaces of the dice would range from 2 to 24.
If the player gets the sum 13 or 23 on the first throw (a natural), they win. If they get 2, 3, or 24 on the first throw (craps), they lose.
If they get any other sum (the point), they must throw the dice again, and keep throwing until they get the sum of 13 (when they lose), or throw their point again (when they win).

What is the probability at the start of any game that the dice thrower will win?

How does the probability of winning differ from the probability of winning on the first throw? That is, how do multiple rounds affect the chances of winning?

Re: Probability in craps

In other words, what is the probability that two 12 -sided die will sum to 13 or 23?.

If so, we can use the generating function \(\displaystyle \left(\sum_{k=1}^{12}x^{k}\right)^{2}\), expand and look at the coefficients of the x terms with exponents 13 and 23.

Doing so, we see we have \(\displaystyle 12x^{13} \;\ and \;\ 2x^{23}\)

12+2=14.

There are 12^2=144 possible combinations, so the probability of the two 12-sided die summing to 13 or 23 is \(\displaystyle \frac{14}{144}=\frac{7}{72}\)

Re: Probability in craps

galactus said:

In other words, what is the probability that two 12 -sided die will sum to 13 or 23?.

If so, we can use the generating function \(\displaystyle \left(\sum_{k=1}^{12}x^{k}\right)^{2}\), expand and look at the coefficients of the x terms with exponents 13 and 23.

Doing so, we see we have \(\displaystyle 12x^{13} \;\ and \;\ 2x^{23}\)

12+2=14.

There are 12^2=144 possible combinations, so the probability of the two 12-sided die summing to 13 or 23 is \(\displaystyle \frac{14}{144}=\frac{7}{72}\)

Click to expand...

I had gotten that far by counting total combinations versus ones that produce 13 or 23, but that only counts as the probability that the thrower will win on the first throw, right? Doesn't the probability of winning at the start of the game incorporate the chance of missing initially and throwing again?