What is the probability that Cathy wins?

Question

Alice, Bob and Cathy take turns (in that order) in rolling a six sided die. If Alice ever rolls a 1, 2 or 3 she wins. If Bob rolls a 4 or a 5 he wins, and Cathy wins if she rolls a 6. They continue playing until a player wins. What is the probability (as a fraction) that Cathy wins?

This was I question I got when competing at a Mathematics competition earlier this week. I thought the answer was $\frac{1}{18}$ but it turns out the actual answer is $\frac{1}{13}$...

This is my working out of why I thought it was $\frac{1}{18}$:

For Cathy to win:

• Alice needs to get a 4, 5 or 6 and there is a $\frac{3}{6}$ chance of that happening when she rolls the die
• Bob needs to get a 1, 2, 3 or 6 and there is a $\frac{4}{6}$ probability of that happening when he rolls the die
• Cathy needs to get 6 and there is a $\frac{1}{6}$ probability of that happening when she rolls the die

Those three need to happen for Cathy to win so:

$$\frac{3}{6} \times \frac{4}{6} \times \frac{1}{6} = \frac{1}{18}$$

However, the answer was $\frac{1}{13}$ so something must be wrong! Could someone please demonstrate why it was $\frac{1}{13}$?

Answer 1

C wins when:

1) A loses, B loses, C wins

or

2) A loses, B loses, C loses, A loses, B loses, C wins

or

3) A loses, B loses, C loses, A loses, B loses, C loses, A loses, B loses, C wins

or

..........

There are infinitely many possibilities, what leads to infinite series $$ \sum_{i\ge 0}\left(\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{5}{6}\right)^i\cdot\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{1}{6} $$ where $\frac{3}{6}=P(A\ loses)$, $\frac{4}{6}=P(B\ loses)$, $\frac{5}{6}=P(C\ loses)$, $\frac{1}{6}=P(C\ wins)$.

The final answer is $\frac{18}{13}\cdot\frac{1}{18}=\frac{1}{13}$.

Answer 2

You are wrong because that is not the only way in which Cathy can win! Note that if Cathy rolls anything other than a 6, the game repeats itself until someone wins. Using very similar reasoning to yours, show that the probability of her winning is the infinite sum $$ \sum_{n=0}^\infty\Big(\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{5}{6}\Big)^n\Big(\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{1}{6}\Big) $$ Can you compute this sum now?

Edit: Here is a slicker way to compute the probability without an infinite sum. Either Cathy wins on the first round, which as you correctly computed, has probability $\frac{1}{18}$. Otherwise, with probability $\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{5}{6}$, nobody wins. In this case, the game continues in exactly the same way as if the first round had not happened. Thus $$ P=\frac{1}{18}+\frac{3}{6}\cdot\frac{4}{6}\cdot\frac{5}{6}P, $$ i.e. $P=\frac{1}{13}$.

Answer 3

If Cathy win at her first toss: $$\frac12 \times \frac23 \times \frac16$$

However, she might not be that lucky, she might make it in the second toss:

$$\frac12 \times \frac23 \times \frac56 \times \frac12 \times \frac23 \times \frac16$$

and so on.

Hence overall, the required probability is

\begin{align}\sum_{i=0}^\infty \left(\frac12 \times \frac23 \times \frac56\right)^i\left(\frac12 \times \frac23 \times \frac16\right)&=\sum_{i=0}^\infty \left(\frac5{18}\right)^i\left( \frac1{18}\right) \\ &=\frac1{18} \frac{1}{1-\frac{5}{18}}\\ &= \frac1{13}\end{align}

Answer 4

What you are leaving out is the possibility that nobody wins the first round. There is a $\frac{1}{18}$ chance that Cathy wins AND does so in the first round. Her overall odds will be $\sum_{n=1}^{\infty}(\frac{1}{2})^n(\frac{2}{3})^n(\frac{5}{6})^{n-1}(\frac{1}{6})=\frac{6}{5}\frac{1}{6}\sum_{n=1}^{\infty}(\frac{5}{18})^n=\frac{1}{5}\frac{5}{13}=\frac{1}{13}$