There is a game with two players and they throw a pair of dice. Player 1 wins if the sum is 7 and player 2 wins if the sum is 6. The stakes are split if neither player wins. What is the expectation of each player? Suppose the game continues until someone wins. What is the probability that each will win if player 2 starts?
[Math] Probability of winning a dice game.
probabilitystatistics
Best Answer
Your probabilities are correct: in a given round, player 1 has probability $p = 1/6$ of winning and player $2$ has probability $q = 5/36$ of winning. The expectation of each player (assuming you mean winnings) is $pd$ for player 1 and $qd$ for player 2, where $d$ is the amount of money in the pot.
If we now let the game continue until someone wins, a third possibility arises: each round, the probability of passing to the next round is either $1-p$ or $1-q$, depending on whose turn it is.
Let $A$ mean "player 1 wins", let $B$ mean "player 2 wins", and let $\bar{A}$ mean "player 1 loses" and let $\bar{B}$ mean "player 2 loses" (on a round-by-round basis). From your clarification in your comment, it follows that if player 2 starts, the first round's possible outcomes are $B$ or $\bar{B}$, the second round's possible outcomes are $A$ or $\bar{A}$, and so on, alternately. Thus, the possible game sequences are $\{B, \bar{B}A, \bar{B}\bar{A}B, \bar{B}\bar{A}\bar{B}A, \bar{B}\bar{A}\bar{B}\bar{A}B, \dotsc\}$.
The probability of a sequence like $\bar{B}\bar{A}\bar{B}A$ is the product of the probabilities: $$(1-q)(1-p)(1-q)p.$$
Now, player 2 wins iff the game sequence was one of the following: $\{B, \bar{B}\bar{A}B, \bar{B}\bar{A}\bar{B}\bar{A}B, \dotsc\}$. Thus, the probability of player 2 winning is $$\sum_{k=0}^\infty [(1-q)(1-p)]^kq = \frac{q}{1-(1-p)(1-q)} = \frac{q}{p + q - pq}$$ by the geometric series formula.
Player 1 wins iff the game sequence was one of the following: $\{\bar{B}A, \bar{B}\bar{A}\bar{B}A, \bar{B}\bar{A}\bar{B}\bar{A}\bar{B}A, \dotsc\}$. Thus, the probability of player 1 winning is $$(1-q)\sum_{k=0}^\infty[(1-p)(1-q)]^k p = \frac{p(1-q)}{p+q-pq}.$$