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A gambler starts with 3 rupees. In each round he wins 1 rupee with…
Question
A gambler starts with 3 rupees. In each round he wins 1 rupee with probability 2/5 and loses 1 rupee with probability 3/5, independently. He stops when his fortune reaches 0 (ruin) or 7 (target). Let p be the probability he reaches 7 before 0. Write p = m/n in lowest terms and find m+n.
✓ Verified answer: 2363checked by our engine — not a guess
Step-by-step solution
This is the classic gambler's ruin with win probability p = 2/5 and loss probability q = 3/5, so the ratio r = q/p = (3/5)/(2/5) = 3/2.
For absorbing barriers at 0 and N = 7, the probability of reaching N before 0 starting from i is:
P_i = (1 - r^i) / (1 - r^N), since p != q.
With i = 3, N = 7, r = 3/2:
r^3 = 27/8, r^7 = 2187/128.
Numerator: 1 - 27/8 = -19/8.
Denominator: 1 - 2187/128 = -2059/128.
P_3 = (-19/8)/(-2059/128) = (19/8)(128/2059) = 19*16/2059 = 304/2059.
2059 = 29*71, and 304 = 16*19, so the fraction is already in lowest terms.
m+n = 304 + 2059 = 2363.
Final answer2363
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