Question

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)

Answer 1

`"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)"`