# Probability of winning a dice game.

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?

• In the first game, do they toss simultaneously? And if so, what happens if they both "win"? – André Nicolas Dec 2 '14 at 7:14
• @AndréNicolas I think the game is set up so that each one follow their own turns. – Kururugi Suzaku Dec 2 '14 at 7:17
• The question will presumably be answered by tomorrow. If not, I will write a solution (it is late). – André Nicolas Dec 2 '14 at 7:22
• @AndréNicolas Thanks. I will be eagerly waiting for solution. – Kururugi Suzaku Dec 2 '14 at 7:28
• What is the significance of player 2 starting (first)? Do the players plays in rounds with the property that a player can win if and only if he/she is the thrower, or do the players throw the two dice together (in which case, who "starts" doesn't matter)? – Unit Dec 2 '14 at 7:47

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 $$(1-q)\sum_{k=0}^\infty[(1-p)(1-q)]^k p = \frac{p(1-q)}{p+q-pq}.$$

• I think you're making a mistake with your $X$'s. You say in the first round possible outcomes are $B$ and $X$ and in the second round $A$ and $X$. That means these $X$'s are not the same. They're $(1-q)$ and $(1-p)$, respectively. – SPK.z Dec 2 '14 at 8:45
• Yes, you're right! Let me fix that! – Unit Dec 2 '14 at 8:46
• @Unit Thanks a lot for your explanation and detail solution. Just one more question, according to the geometric series the probability of winning for player 2 is 30/61 and for player 1 is 31/61, right ? – Kururugi Suzaku Dec 2 '14 at 17:03
• Yes, plugging in $p = 1/6$ and $q = 5/36$ into the above formulas yields $31/61$ for player 1 and $30/61$ for player 2. – Unit Dec 2 '14 at 17:20
• @Unit Thanks a lot for all the help. – Kururugi Suzaku Dec 3 '14 at 3:07

I know that for player 1 there are 6 ways to get a 7, so he has 6/36 = 1/6 probability of getting that combination. Player 2 has 5/36 probability of getting a combination of 6. However, how do I proceed from here? Thanks.

For the first question, I will assume that Player 1 goes first, and if she tosses a total of $7$ she has won, the game is over. So the probability she wins is $\frac{6}{36}$.

In order for Player 2 to win, Player 1 must not win, and then Player 2 must toss a total of $6$. The probability that Player 2 wins is therefore $\frac{30}{36}\cdot \frac{5}{36}$.

The game is a lot fairer if Player 2 tosses first. Then her probability of winning is $\frac{5}{36}$, and the probability Player 1 wins is $\frac{31}{36}\cdot\frac{6}{36}$.

In general, let $a$ be the probability a certain player A wins, and b the probability the other player B wins. Note that a draw is also possible, indeed quite likely.

If the players bet a dollar on the outcome, then A's expected net winnings are $(a)(1)+(b)(-1)+(1-a-b)(0)$, or simply $a-b$.

The second game is a lot more clearly specified. Let $p$ be the probability Player 2 (who goes first) ultimately wins.

She wins if she gets a sum of $6$ immediately, or if she doesn't but still ultimately wins. That happens if she strikes out on her first toss, and Player 1 also does, and Player 2 ultimately wins. Given they both struck out, the probability Player 2 ultimately wins is $p$, since in essence the game starts again. Thus $$p=\frac{5}{36}+\frac{31}{36}\cdot\frac{30}{36}\cdot p.$$ Solve this linear equation for $p$.

Alternately, we can produce an infinite geometric series for the probability Player 2 ultimately wins. The probability Player 1 ultimately wins is $1-p$.