Probability of first actor winning a “first to roll seven with two dice” contest?

Two players P and Q take turns, in which they each roll two fair and independent dice. P rolls the dice first.

The first player who gets a sum of seven wins the game. What is the probability that player P wins the game?

• Does P go first? – Michael Hardy Apr 25 '14 at 15:22
• No infinite series are needed here. My own answer below and that of André Nicolas make that clear. I hope mine is a bit simpler. – Michael Hardy Apr 25 '14 at 15:27
• . . . and now I see André Nicolas has edited his answer and it's no different from mine. Great minds think alike. – Michael Hardy Apr 25 '14 at 15:28
• Five answers so far and André Nicolas and I are the only ones to notice that no infinite series are needed. – Michael Hardy Apr 25 '14 at 15:29
• Here's a similar question. Three players compete: A, B, and C. In each round one player sits on the bench and a coin is tossed deciding which of the other two wins that round. The loser is the one who sits on the bench in the next round. The first player to win two consecutive rounds is the Winner. What is the probability that C is the Winner? You can solve this without infinite series by essentially the same method that works for this question. – Michael Hardy Apr 25 '14 at 15:32

I'm going to assume P goes first. The question should have said so, unless something else was intended, in which case it's unclear.

Let (lower-case) $p$ be the probability that P ultimately wins. Then \begin{align} p & = \Pr(\text{P wins on 1st trial}) + \Pr(\text{P loses on first trial and ultimately wins)} \\[8pt] & = \frac 1 6 + \frac 5 6 (1-p). \end{align} So solve the following: $$p =\frac 1 6 + \frac 5 6 (1-p).$$

• I think for someone who might actually ask this question the step $\operatorname{Pr}(\text{P loses on first trial and ultimately wins}) = \frac{5}{6}(1-p)$ might be difficult to understand. One way to see this is the following: let $p$ be the probability that the current player wins. clearly $p=\frac{1}{6} + \frac{5}{6}\operatorname{Pr}(\text{the other player loses given that it is his turn})$. As the other player is the "current player" at that point, his probability to lose is equal to $(1-p)$, thus the recursion. – example Apr 26 '14 at 9:28
• Ahhh What a answer . Awesome – laura Nov 20 '17 at 7:25

Let $p$ be the probability that the first player $P$ (ultimately) wins, and let $q$ be the probability that $Q$ ultimately wins. It is clear that with probability $1$, the game terminates, so $p+q=1$.

We condition on $P$'s first throw. With probability $\frac{1}{6}$, she gets a sum of $7$, and wins immediately.

Another way she can ultimately win is if she tosses something other than $7$, but $Q$ does not ultimately win. The probability $P$'s first toss is not a $7$ is $\frac{5}{6}$. Given that this has happened, the probability that $Q$ does not win is $1-p$. Thus $$p=\frac{1}{6}+\frac{5}{6}(1-p).$$ We have a linear equation for $p$. Solve.

Note that the probability of a seven is $1/6$. So the answer is

$$(1/6) + (5/6)\times (5/6) \times (1/6) + \dots= \frac{1}{6}\sum_{j=0}^\infty \Bigl(\frac{25}{36}\Bigr)^j = \frac{6}{11}.$$

Suppose that player $P$ begins the game. Probability that player $P$ wins after the first try is $1/6$, probability that player $P$ does not win after the first try is $5/6$. If player $P$ does not win after the first try, the probability that player $P$ can try to win again is $5/6$. So probability that player $P$ will win after the second try is $\frac16(\frac56\frac 56)$. The probability that player $P$ will win after the third try is $\frac16(\frac56\frac 56)^2$. The probability that player $P$ will win is then equal to $$\sum_{n=0}^\infty\frac16\biggl(\frac56\cdot\frac56\biggr)^n=\frac16\sum_{n=0}^\infty\biggl(\frac{25}{36}\biggr)^n=\frac6{11}.$$

Think about this problem as if it were a single player rolling dice, and we stop the game when she rolls a $7$. This is a geometric distribution, with the probability of "success" being $p=1/6$, so we can write the pmf describing the number of rolls $k$ until the first $7$ appears: $$P(X=k)=(1-p)^{k-1}p$$ Now, first player wins the game if the game stops on the first, third, fifth roll and so on, in other words $k$ must be odd. Let $k=2n-1, n=1,2,\dots$ Now we have $$P(P \text{ wins})=\sum_{n\ge 1}(1-p)^{2n-2}p=\dfrac{1}{2-p}$$ Substitute $p=1/6$ and get that your answer is $\dfrac{6}{11}$ (assuming P rolls first; otherwise it's just the complement). But the nice intuition here is that rolling first is advantageous!

The probability that $P$ eventually wins the game can be represented as the sum of an infinite geometric series.

For example,

The probability of $P$ winning in the first round itself is $\dfrac{1}{6}$. (Since out of the $36$ possible outcomes, there are $6$ outcomes in which the sum of the dice is equal to $7$.)

The probability of $P$ winning in the second round is $\dfrac{5}{6} \times \dfrac{5}{6}\times \dfrac{1}{6}$. This means that for $P$ to win th second round, he has to lose the first round, then $Q$ has to lose his turn and finally $P$ wins. This goes on.

$$\large\dfrac{1}{6} + \left( \dfrac{5}{6}^{2} \times \dfrac{1}{6} \right) + \left( \dfrac{5}{6}^{4} \times \dfrac{1}{6} \right) + \dots = \displaystyle\sum\limits_{i=0}^{\infty} \dfrac{1}{6} \times \dfrac{5}{6}^{2n}$$
$$\large\dfrac{\dfrac{1}{6}}{\left( 1 - \dfrac{5}{6}^2 \right)} = \boxed{\dfrac{6}{11}}$$