In a dice game player continues to throw two dice until they

In a dice game, player continues to throw two dice until they either win or lose. The player wins on the first throw if the sum of two upturned faces is either 7 or 11, and loses is the sum is 2, 3, or 12. Otherwise, the sum of two upturned faces becomes the player\'s \"point.\" The player continues to throw until the first succeeding throw on which he makes his \"point\" (in which case he wins), or until he throw a 7 (in which case he loses). What is the probability that the player will win the game?

Solution

78.205% of winning

So the probability that he wins on his first throw is:
36 total possible outcomes
{(1,6) (2,5) (3,4)}x2 = 6 winning outcomes

so 1/6 chance of winning on your first throw

the probability of getting to throw dice to win relies on your not winning on the first throw and not losing on the first throw, so:
{(1,1) (1,2) (6,6)}x2 = 6 losing outcomes
so a 1/6 chance of losing on the first throw
and a 2/3 chance of throwing for a possible win after the 1st throw

The player\'s point can only be 4,5,6,8,9,10 as any other roll would have won or lost on the first throw.

You can roll a 7 in 2 ways (3,4).

You can roll for 4,5,6,8,9,10 by:

{(1,3) (2,2) (1,4) (2,3) (1,5) (2,4) (3,3) (2,6) (3,5) (4,4) (3,6) (4,5)}x2 = 24 outcomes

so 24:2 that you will win or 92.307% on non-first throws

you have a 2/3 chance of getting to throw non-first throws so,

92.307% * 2/3 = 61.5385% + 1/6 for winning on the first throw = 78.205% of winning this game