What is the probability of winning in the following infinitely repeated game?

Two players A & B play a fair game (probability of winning is 1/2 for both players) repeatedly for 1 Rupee per game. if originally A has 'a' rupees and B has b (a>b), what is A's chance of winning all of B's money, assuming the play goes on until one person has lost all his money?

Options are:
A) 1
B) 0
C) b/(a+b)
D) a/(a+b)

1 Answer
May 8, 2018

"Answer D)"

Explanation:

"It is the only logical answer, the others are impossible."

"This is the gambler's ruin problem."
"A gambler starts with k dollar."
"He plays until he reaches G dollar or fall backs to 0."
p = " chance that he wins 1 dollar in one game."
q = 1 - p =" chance that he loses 1 dollar in one game."
"Call "r_k" the probability (chance) that he gets ruined."
"Then we have"

r_0 = 1
r_G = 0
r_k = p * r_{k+1} + q * r_{k-1} , " with "1 <= k <= G-1
"We can rewrite this equation due to p+q=1 as follows :"
r_{k+1} - r_k = (q/p) (r_k - r_{k-1})
=> r_{k+1} - r_k = (q/p)^k (r_1 - r_0)

"Now here we have the case "p=q=1/2.

=> r_{k+1} - r_k = r_1 - r_0

r_G - r_0 = -1 = sum_{k=0}^{G-1} (r_{k+1} - r_k)
= sum_{k=0}^{G-1} (r_1 - r_0)
=> r_1 - r_0 = -1/G

"For "r_k" we have"

r_k - r_0 = sum_{i=0}^{k-1} (r_{i+1} - r_i)
= k * (r_1 - r_0)
= - k/G
=> r_k = r_0 - k/G = 1 - k/G = (G - k)/G

"So player A starts here with k = a dollar and plays until"
"he gets ruined or has a+b dollar."
=> k = a, " and "G = a+b
"So the odds that he gets ruined are"
(G - k)/G = (a+b-a)/(a+b) = b/(a+b)
"The odds that he wins are"
1 - b/(a+b) = a/(a+b) => " Answer D)"