How to show that a game is fair when two people take turns rolling a die

Question

Book problem: Annie and Bill each have a die, which they throw alternately. Bill scores if he throws a multiple of two, and Annie scores if she throws a multiple of three. The winner is the first person to score. Show that the game is fair if Annie rolls first.

My answer: I think the game would be fair if they each had the same probability of winning. But Annie wins on an odd number of rolls if she goes first and Bill wins on an even number of rolls is Annie goes first.

Suppose Annie goes first and wins on the $k$th roll so that $k$ is odd. Then both Annie and Bill need to lose $(k - 1)/2$ times and then Annie needs to win. Since Annie wins with a 3 or 6 and Bill with a 2, 4, or 6 we have that the probability of Annie winning is

$$ \left(\frac{4}{6}\right)^{\frac{k-1}{2}}\left(\frac{3}{6}\right)^{\frac{k-1}{2}}\left(\frac{2}{6}\right) $$

Similarly Bill wins when $k$ is even with probability

$$ \left(\frac{4}{6}\right)^{\frac{k}{2}}\left(\frac{3}{6}\right)^{\frac{k}{2}} $$

Now notice that when $k = 1$ Annie wins with probability 1/3. Likewise, Bill wins with probability 1/3 when $k = 2$. When $k = 3$ or $k = 4$ we get that they both win with the same probability of 1/9. And so on.

Hence the game is fair.

Question: Is my reasoning correct?

Book: "Calculus: The Analysis of Functions" by Taylor

Answer 1

We need to sum up all the "infinite many turns", not just consider any single turn.

The Annie probability would be ${1\over 3} + {2\over 3}(1-{1\over 2}){1\over3}+{2\over 3}^2(1-{1\over 2})^2{1\over3}+{2\over 3}^3(1-{1\over 2})^3{1\over3}+...$ which is a geometric series which sums up to ${1\over 3}({1\over1-{2\over 3}(1-{1\over 2})})={1\over 2}$. Therefore Annie has exactly $1\over 2$ chance of winning overall and the game is fair.

Answer 2

Your reasoning is correct. The probability that each wins is the same: $1/3 + 1/9 + 1/27 + \cdots$. Because these two probabilities must sum to $1$, they must each be $1/2$, which you can also see by summing the series.

Another way to see this game is fair is to let $v$ be the amount of money Annie nets if both players stake \$$1$. There is probability $1/3$ of her netting \$$1$ right away. But with probability $2/3$ there is a $1/2$ chance of her losing \$$1$, and a $1/2$ chance of her netting the value $v$: $$v = \frac{1}{3}(1) + \frac{2}{3}\left(\frac{1}{2}(-1) + \frac{1}{2} v\right) = \frac{1}{3} v. $$ The solution of this is $v=0$, so the game is fair.

Answer 3

Your explaination is definitely right.
But we can go further. Let's call a 'round' when Annie and Bill both throws the die once. So, the same as your thought, the probability of Annie winning in the i-th round is$(\frac{4}{6})^{i-1}(\frac{3}{6})^{i-1}(\frac{2}{6})=(\frac{1}{3})^{i}$.
In fact, the probability of Annie winning is$\sum_{i=1}^{\infty}(\frac{1}{3})^{i}=\lim_{x\rightarrow \infty}\frac{1-(\frac{1}{3})^x}{3-1}=\frac{1}{2}$, natually the addtion of the probability of winning in each round.

Answer 4

Alternative approach.

Regardless of how many turns it takes, the game must eventually be resolved by the existence of some final round, where:

• Neither player wins in any previous round.

• Exactly one player wins in the final round.

However, just before this final round, the situation is exactly the same as it was at the start of the first round, except that you are guaranteed, that the game will be resolved in this final round.

Therefore, the problem can be attacked without resorting to the summation of an infinite series, and can instead be resolved by Bayes Theorem.

It is as if you are guaranteed that the game will be resolved in the very first round.

The probability of Anne winning in the very first round is $~1/3.$

The probability of Bill winning in the very first round is $~2/3 \times 1/2 = 1/3.$

Therefore, if you are given that the game lasted exactly 1 round, then Anne's probability of winning is

$$\frac{1/3}{1/3 + 1/3} = 1/2.$$