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)"Answer D)

Explanation:

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

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

r_0 = 1r0=1
r_G = 0rG=0
r_k = p * r_{k+1} + q * r_{k-1} , " with "1 <= k <= G-1rk=prk+1+qrk1, with 1kG1
"We can rewrite this equation due to p+q=1 as follows :"We can rewrite this equation due to p+q=1 as follows :
r_{k+1} - r_k = (q/p) (r_k - r_{k-1})rk+1rk=(qp)(rkrk1)
=> r_{k+1} - r_k = (q/p)^k (r_1 - r_0)rk+1rk=(qp)k(r1r0)

"Now here we have the case "p=q=1/2.Now here we have the case p=q=12.

=> r_{k+1} - r_k = r_1 - r_0rk+1rk=r1r0

r_G - r_0 = -1 = sum_{k=0}^{G-1} (r_{k+1} - r_k)rGr0=1=G1k=0(rk+1rk)
= sum_{k=0}^{G-1} (r_1 - r_0)=G1k=0(r1r0)
=> r_1 - r_0 = -1/Gr1r0=1G

"For "r_k" we have"For rk we have

r_k - r_0 = sum_{i=0}^{k-1} (r_{i+1} - r_i)rkr0=k1i=0(ri+1ri)
= k * (r_1 - r_0)=k(r1r0)
= - k/G=kG
=> r_k = r_0 - k/G = 1 - k/G = (G - k)/Grk=r0kG=1kG=GkG

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