Players A and B play a sequence of independent games. Player A throws a die first and wins on a "six." If A fails, then player B throws and wins on a "five" or "six." If B fails, then A throws and wins on a "four," "five," or "six." And so on. How to find the probability of each player winning the sequence?
2026-03-27 19:33:21.1774640001
How to compute of each player winning this sequence of games?
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Hint: There are exactly $3$ ways that $A$ can win. Compute each case's probability separately, then add them all together to get your answer. (NOTE: Each time someone rolls the die, I call that a "Turn", for a maximum of $6$ possible Turns).
Case 1: $A$ wins on Turn $1$.
Case 2: $A$ fails on Turn $1$, $B$ fails on Turn $2$, and $A$ wins on Turn $3$.
Case 3: $A$ fails on Turn $1$, $B$ fails on Turn $2$, $A$ fails on Turn $3$, $B$ fails on Turn $4$, and $A$ wins on Turn $5$.