Taking turns to roll the first six

Question

Two players take turns rolling a fair die. Player A rolls first, then B, then A, and so on. The first player to roll a six wins immediately.

What is the probability that player A (who goes first) wins?

Accepted Answer

Let p = 	frac16 be the chance of a six on a roll and q = 	frac56 the chance of missing. Player A wins if A gets a six on the first roll, or both miss their first rolls and then A is effectively "first" again. Let w = P(	ext{A wins}). On A's turn: A rolls a six (prob p) and wins; otherwise (prob q) B rolls. If B rolls a six (prob p) A has lost; if B also misses (prob q) the situation resets to A rolling first with the same win probability w: w = p + q\big(q\,w\big) = p + q^{2} w. Solving, w(1 - q^{2}) = p \implies w = \frac{p}{1 - q^{2}} = \frac{p}{(1-q)(1+q)} = \frac{p}{p(1+q)} = \frac{1}{…

Taking turns to roll the first six

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