Two gamblers A and B play a game throwing two ordinary dice. A wins if he obtains the sum 6 before B obtains 7, and B wins if he obtains the sum 7 before A obtains 6. Which of the players has a better chance of winning if player A starts the game?

Source: Miodrag S. Petkovic, Famous Puzzles of Great Mathematicians, American Mathematical Society, 2009, Problem 8.2.

Level: Senior

Let A and B denote the sum of the numbers on two ordinary dice are 6 and 7, respectively. Then P(A) = 5/36, P(not A) = 1-P(A) = 31/36, P(B) = 6/36 = 1/6, and P(not B) = 1-P(B) = 5/6. A will win the game if the event
S = A or (not A and not B and A) or (not A and not B and not A and not B and A) or ...
happens. Since the events separated by the "or" (union) operations are mutually exclusive, P(S) is the sum of the probability of these events. Each of these events is composed by one or more sub-events connected by the "and" (intersection) operation. Since these sub-events are mutually independent, so the probability of each "anded" event is the product of the probabilities of these sub-events. Therefore
P(S)
= P(A)+P(not A)*P(not B)*P(A)+P(not A)*P(not B)*P(not A)*P(not B)*P(A)+...
= 5/36+(31/36)(5/6)(5/36)+[(31/36)(5/6)]^2*(5/36)+...
= (5/36)*{1+[(31/36)(5/6)]+[(31/36)(5/6)]^2+...}
= (5/36)/[1-(31/36)(5/6)]
= (5/36)/(61/216)
= 30/61.
Similarly, B will win the game if the event
T = (not A and B) or (not A and not B and not A and B) or (not A and not B and not A and not B and not A and B) or ...
happens. Thus
P(T)
= P(not A)*P(B)+P(not A)*P(not B)*P(not A)*P(B)+P(not A)*P(not B)*P(not A)*P(not B)*P(not A)*P(B) +...
= (31/36)(1/6)+(31/36)(5/6)(31/36)(1/6)+(31/36)(5/6)(31/36)(5/6)(31/36)(1/6)+...
= (31/36)(1/6)*{1+[(5/6)(31/36)]+[(5/6)(31/36)]^2+...}
= (31/36)(1/6)/[1-(5/6)(31/36)]
= (31/36)(1/6)/(216/61)
= 31/61.
Since P(S) < P(T), hence B has a slightly better chance of winning. Note that P(S)+P(T) = 1 since either A or B wins.