# Probability question in textbook

"In the game of craps, a player rolls two dice. They win at once if the total is 7 or 11, and lose at once if the total is 2, 3, or 12. Otherwise, they continue rolling the dice until they either win by throwing their initial total again, or lose by rolling 7. Show that the probability they win is 0.493."

I used this to calculate the probability:

$$P(4)\frac{P(4)}{P(4)+P(7)}+P(5)\frac{P(5)}{P(5)+P(7)}+P(6)\frac{P(6)}{P(6)+P(7)}+P(7)+P(8)\frac{P(8)}{P(8)+P(7)}$$ $$+P(9)\frac{P(9)}{P(9)+P(7)}+P(10)\frac{P(10)}{P(10)+P(7)}+P(11)$$

where $P(k)$ denotes the probability of getting a sum of $k$ with one throw of the two dice.

Somehow, though, this is wrong (doesn't give the right answer). Why?

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Direct answer to the question: they're allowed to get more than two rolls: it looks like your calculation implicitly assumes that the second roll is definitely either going to be a seven or the same as the first roll. –  Eric Stucky Jan 21 '13 at 2:23
Isn't the probability of $A$ happening before $B$ does, when the thing is repeated indefinitely until one of $A$ or $B$ happens, $\frac{P(A)}{P(A)+P(B)}$? –  user54609 Jan 21 '13 at 2:34
@EricStucky: I think the argument is correct. What is needed is just $P(A|A \cup B)$. Or we can write out the expression completely. Say we have $3$ disjoint possibilities $A$ (win), $B$ (lose) and $C$ (re-throw). Probability of winning is $P(A) + P(C)P(A) + [P(C)]^2P(A) + \ldots = \frac{P(A)}{1-P(C)} = \frac{P(A)}{P(A)+P(B)}$. –  polkjh Jan 21 '13 at 2:48
@EricDong: The expression looks correct and it is evaluating to the required probability. Can you check your calculations again? –  polkjh Jan 21 '13 at 2:52
I agree with polkjh's assertion that the expression looks correct. The answer is supposed to be $\frac{244}{495}=0.49292929\cdots$. Are you getting something different? A detailed statement of the game and complete solution are available on-line. –  Dilip Sarwate Jan 21 '13 at 3:02

This is an error in the textbook. It should say that the probability is approximately $0.493$. Your calculations are correct.