"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=1−p= 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=p⋅rk+1+q⋅rk−1, with 1≤k≤G−1
"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+1−rk=(qp)(rk−rk−1)
=> r_{k+1} - r_k = (q/p)^k (r_1 - r_0)⇒rk+1−rk=(qp)k(r1−r0)
"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_0⇒rk+1−rk=r1−r0
r_G - r_0 = -1 = sum_{k=0}^{G-1} (r_{k+1} - r_k)rG−r0=−1=G−1∑k=0(rk+1−rk)
= sum_{k=0}^{G-1} (r_1 - r_0)=G−1∑k=0(r1−r0)
=> r_1 - r_0 = -1/G⇒r1−r0=−1G
"For "r_k" we have"For rk we have
r_k - r_0 = sum_{i=0}^{k-1} (r_{i+1} - r_i)rk−r0=k−1∑i=0(ri+1−ri)
= k * (r_1 - r_0)=k⋅(r1−r0)
= - k/G=−kG
=> r_k = r_0 - k/G = 1 - k/G = (G - k)/G⇒rk=r0−kG=1−kG=G−kG
"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+b⇒k=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)G−kG=a+b−aa+b=ba+b
"The odds that he wins are"The odds that he wins are
1 - b/(a+b) = a/(a+b) => " Answer D)"1−ba+b=aa+b⇒ Answer D)