Rolling a $4$ sided die until a $1$ shows up

Question

Annie, Bill, and Clara are playing a game. Annie goes first. She will roll a 4-sided die. If she rolls a 1 then she wins and the game ends. If she doesn't roll a 1 then Bill will roll the die. If he rolls a 1 then he wins and the game ends. Then Clara rolls the die, and the same conditions follow. They keep rolling in the order Annie, Bill, Clara, Annie, Bill, Clara, Annie... until someone rolls a 1. What is the probability Clara wins?

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I was thinking of using an infinite sum to try and figure this out, but I'm not sure how to calculate the sum. Can I have a hint please?

Answer 1

Usually in these cases you try to exploit the fact that after 3 turns the game repeats itself. That means that P(A) (annie wins) is basically $ P(A) = c_1 + c_2 \cdot P(A) $. Find the constants and solve for $ P(A) $.

Answer 2

Clara wins at her first roll when Annie and Bill don´t roll a 1 at their first roll and Clara roll a 1. The probability for that is $\left(\frac34\right)^2\cdot \frac14$. The probability that Clara win at her second roll is $\left(\frac34\right)^5\cdot \frac14$. Now you can use the infinite sum to calculate the probabilty that Clara wins the game.

$$P(\textrm{"Clara wins the game"})=\frac14\cdot \sum_{k=0}^{\infty} \left(\frac34\right)^{3k+2}$$

Answer 3

An alternate approach than the one suggested by the others is that you can think in terms of rounds and turns where each turn refers to each individual throw of the die by a player while a round is all three of your players having a turn.

Allow the game to continue until the completion of the current round so that all players have had the same number of turns even if a winner has already been determined. You will notice that it doesn't matter how many rounds have been played, that the overall winner of the game is dependent solely upon the outcomes of the final round.

Now, you can calculate the conditional probability that given a particular round is the final round that $C$ was the winner. For that, the first roll of the round must have not won, the second roll of the round must have not won, and the third roll of the round must have won while we are conditioning that this was the final round, i.e. it is not the case that noone won.

$$\dfrac{\frac{3}{4}\times\frac{3}{4}\times\frac{1}{4}}{1-\left(\frac{3}{4}\right)^3} = \frac{9}{37}$$

The related probabilities of $A$ winning and $B$ winning are $\frac{\frac{1}{4}}{1-\left(\frac{3}{4}\right)^3}=\frac{16}{37}$ and $\frac{\frac{3}{4}\times\frac{1}{4}}{1-\left(\frac{3}{4}\right)^3}=\frac{12}{37}$ respectively.

Answer 4

To clarify what's already been said, one approach is to think about each round as an independent experiment, and use the fact that if Clara doesn't win on the first round, the experiment resets. $$P(\text{Clara wins}) = P(\text{Clara wins on round 1}) + P(\text{Clara wins on round 2 or later})$$

Another way to phrase these events are: $$P(\text{Clara wins}) = P(\text{Clara wins on round 1}) + P(\text{Clara wins eventually but no one wins on round 1})$$

For Clara to win on round 1, Annie and Bill have to miss on the roll and Clara gets a 1, so this probability is $(3/4)(3/4)(1/4)$. The probability no one wins on round 1 is $(3/4)^3$. Now the experiment has 'reset' since in round 2 we have all the same probabilities as round 1.

With this in mind we can set up a relatively simple equation. Let $W$ be the event that Clara wins, and plugging into the above we have $$P(W) = (3/4)^2(1/4) + (3/4)^3 P(W)$$

To solve for $P(W)$ is a little bit of algebra: $$P(W) - (3/4)^3 P(W) = \left(\frac{3}{4}\right)^2 \frac{1}{4}$$ $$P(W)\left((1 - \left(\frac{3}{4}\right)^3\right) = \left(\frac{3}{4}\right)^2 \frac{1}{4}$$ $$P(W) = \frac{\left(\frac{3}{4}\right)^2 \frac{1}{4}}{1 - \left(\frac{3}{4}\right)^3} = 9/37$$

Epilogue: this is the same kind of work that proves infinite sums, so you were right to suppose there is a relationship between them and this problem. In this particular setting, I think the 'first-step analysis' framework is a more intuitive way to get the same result. You'll notice the final fraction, before simplification, looks a lot like the closed form of an infinite sum.