Player A repeatedly rolls a fair eight-sided die (faces 1-8) and…

Question

Player A repeatedly rolls a fair eight-sided die (faces 1-8) and player B repeatedly rolls a fair six-sided die (faces 1-6). In each round both roll once; the round is decisive if their numbers differ, and is replayed if they tie. The player with the higher number wins the decisive round. The match goes to whoever first wins 3 decisive rounds. Let the probability that A wins the match be m/n in lowest terms. Find m+n.

Accepted Answer

First find the probability A wins a single decisive round. Over all 8*6 = 48 pairs (a,b): A>B in pa pairs, B>A in pb pairs, ties in the rest.

Count a>b: for each b in 1..6, the number of a in 1..8 with a>b is (8-b). Sum over b=1..6: 7+6+5+4+3+2 = 27. Count b>a: for each a in 1..8, number of b in 1..6 with b>a is max(0,6-a). Sum: 5+4+3+2+1+0+0+0 = 15. Ties = 6. (27+15+6 = 48 checks out.)

Conditional on a decisive round, P(A wins) = 27/(27+15) = 27/42 = 9/14. So p = 9/14, q = 5/14.

Match is first-to-3 (best of 5). A wins iff A reaches 3 wins before B. Sum over the number of B-wins (0,1,2) at the moment A scores its 3rd win; the final round is A's 3rd win, and before it there are 2 A-wins and losB B-wins arranged in C(2+losB, losB) ways:

P = sum_{losB=0}^{2} C(2+losB,losB) p^3 q^losB
 = p^3 [1 + 3q + 6q^2]
 = (9/14)^3 [1 + 3(5/14) + 6(25/196)].

Compute: 1 + 15/14 + 150/196 = 196/196 + 210/196 + 150/196 = 556/196 = 139/49.
(9/14)^3 = 729/2744. Multiply: (729/2744)(139/49) = 101331/134456 (already in lowest terms).

m+n = 101331 + 134456 = 235787.

Final answer: 235787

Player A repeatedly rolls a fair eight-sided die (faces 1-8) and…

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