Problem 201 Solution

Question

Two dice are rolled until either a total of 12 or two consecutive totals of 7. What is the probability the 12 is rolled first?

Let's call p the probability that a 12 is rolled first from the first roll, or at any time when the last roll was not a 7.

Let's call q the probability that a 12 is rolled first when the last roll was a 7.

The probability of rolling a 12 first from the first roll can be expressed as:

p = (1/36) + (1/6)*q + (29/36)*p
36p = 1 + 6q + 29p
7p = 1 + 6q

The probability of rolling a 12 first when the last roll was a 7 can be expressed as:

q = (1/36) + (6/36)*0 + (29/36)*p 36q = 1 + 29p
29p = 36q - 1

So, we have two equations and two unknowns: (1) 7p = 1 + 6q
(2) 29p = 36q - 1

Multiplying equation (1) by 6:

(3) 42p = 6 + 36q Multiplying equation (2) by -1

(4) -29p = -36q + 1

Adding equations (3) and (4):

13p = 7
p = 7/13

Substituting 7/13 for p in equation (1):

49/13 = 1 + 6q
49 = 13 + 78 q
78q = 36
q = 36/78
q = 6/13

So, from the first roll the probability that a 12 will be rolled first is 7/13. If any point a 7 is rolled, the probability of a 12 being rolled before two sevens is 6/13.