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

#"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)"#