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The probability of winning on a single roll of dice is $P$. Alice and Bob play the following game. Alice rolls the dice first, and if she fails to win, she passes the dice to Bob, who then attempts to win on his roll. They continue to pass the dice back and forth until one of them wins. What are their respective probabilities of winning?

So, since Alice goes first her probability of winning on her first turn:

$$ P(A_{1}) = P $$

Then Bob's chance of winning on his first turn : $$ P(B_{1}) = A^{c}_{1} \cap P = (1 - P)P $$

where Bob depends on Alice losing the first turn and the probability of rolling the dice. On Alice's second turn, she depends on losing the first turn, Bob losing the second turn, and her probability of winning.

$$ P(A_{2}) = A^{c}_{1} \cap B^{c}_{1} \cap P = (1-P)(1-P)PP $$

and so on and so on and to find the probability of winning for Alice is:

$$ P(A) = \sum_{i=1}^{\infty}P(A_{i}) $$

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One way to get the solution is by symmetry. That is, once a player doesn't win on their turn the game effectively restarts, but now with the other player beginning.

Thus we get $P(B) = (1-p)P(A)$, and since we must have $P(A) + P(B) = 1$ (someone's gonna win eventually as long as p > 0),

$$P(A) = \frac{P(B)}{1-p}$$ $$P(A) = \frac{1-P(A)}{1-p}$$ $$P(A) = \frac{1}{2-p}$$

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  • $\begingroup$ Is this solution scalable? If I were to have K number of players, would this approach also work? $\endgroup$
    – Tigertron
    Commented Apr 25, 2020 at 4:01

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