# Rolling Dice Game, Probability of Ending on an Even Roll

Mathematics Asked by IanFromWashington on January 3, 2022

The game is described as follows. $$A$$ and $$B$$ take turns rolling a fair six sided die. Say $$A$$ rolls first. Then if $$A$$ rolls {1,2} they win. If not, then $$B$$ rolls. If $$B$$ rolls {3,4,5,6} then they win. This process repeats until $$A$$ or $$B$$ wins, and the game stops.

What is the probability that the game ends on an even turn when $$A$$ rolls first?

Now the book gives the answer as $$frac{4}{7}$$, however, when try to calculate I end up with $$frac{2}{11}$$.

Below is my work:

To calculate this probability, we decompose the event into two disjoint events, (a) the event where $$A$$ wins on an even roll, and (b) the event where $$B$$ wins on an even roll.

(a) Now, the probability $$A$$ wins can be calculated as follows
begin{align*} biggr(frac{2}{3} cdot frac{1}{3}biggr)biggr(frac{1}{3}biggr) + biggr(frac{2}{3} cdot frac{1}{3}biggr)biggr(frac{2}{3} cdot frac{1}{3}biggr)biggr(frac{2}{3} cdot frac{1}{3}biggr)biggr(frac{1}{3}biggr) + dots = sum_{k=0}^infty biggr(frac{2}{9}biggr)^{2k+1}frac{1}{3}\ = sum_{k=0}^infty frac{2}{27}biggr(frac{2}{9}biggr)^{2k} = sum_{k=0}^infty frac{2}{27}biggr(frac{4}{81}biggr)^k = frac{2}{27}cdot frac{1}{1- frac{4}{81}} = frac{6}{77}. end{align*}

(b) Similarly we calculate the probability $$B$$ wins on an even roll as
begin{align*} biggr(frac{2}{3} cdot frac{1}{3}biggr)biggr(frac{2}{3}cdot frac{2}{3}biggr) + biggr(frac{2}{3} cdot frac{1}{3}biggr)biggr(frac{2}{3} cdot frac{1}{3}biggr)biggr(frac{2}{3} cdot frac{1}{3}biggr)biggr(frac{2}{3}cdotfrac{2}{3}biggr) + dots = sum_{k=0}^infty biggr(frac{2}{9}biggr)^{2k+1}frac{4}{9}\ = sum_{k=0}^infty frac{8}{81}biggr(frac{2}{9}biggr)^{2k} = sum_{k=0}^infty frac{8}{81}biggr(frac{4}{81}biggr)^k = frac{8}{81}cdot frac{1}{1- frac{4}{81}} = frac{8}{77}. end{align*}

Therefore, it follows that the probability of the game ending on an even number of rolls is
$$begin{equation*} frac{6}{77} + frac{8}{77} = frac{2}{11}. end{equation*}$$

Am I missing something?

Working under the assumption that the intended interpretation of the question was merely asking the probability that $$B$$ wins (i.e. distinguishing between the term "rounds" as iterating whenever A has a turn and "turns" iterating whenever either A or B has a turn) two other approaches have already been written. Here I will include yet another approach:

Consider the final round, that is a roll of $$A$$ followed by a roll of $$B$$, where we allow $$B$$ to roll even in the event that $$A$$ has already won despite the roll not influencing the final result of the game.

Ordinarily there are $$6times 6 = 36$$ equally likely results for a round. Here, we condition on the fact that it is the last round, implying that it was not the case that both players missed their respective targets. This gives $$6times 6 - 4times 2 = 28$$ equally likely possible final rounds.

Of these, $$4times 4 = 16$$ of them end with $$A$$ missing their target and $$B$$ hitting theirs.

The probability of $$B$$ winning the game is then: $$dfrac{16}{28} = dfrac{4}{7}$$

Answered by JMoravitz on January 3, 2022

Thanks to the comment of @JMoravitz I realized my mistake. I was interpreting turns as the rolls $$A$$ AND $$B$$, as in $${A_1,B_1}, {A_2,B_2}, dots$$. In reality the question is merely asking what the probability of $$B$$ winning if $$A$$ rolls first.

The work is as follows: We calculate the probability of $$B$$ winning. Denote the probability of $$B$$ winning on their $$i$$th roll as $$S_i$$. Now, the probabilities of $$B$$ winning on her first roll, second roll, third roll, etc., are as follows: $$begin{equation*} P(S_1) = biggr(frac{2}{3}biggr)biggr(frac{2}{3}biggr), quad P(S_2) = biggr(frac{2}{3}biggr)biggr(frac{1}{3}biggr)biggr(frac{2}{3}biggr)biggr(frac{2}{3}biggr), quad P(S_3) = biggr(biggr(frac{2}{3}biggr)biggr(frac{1}{3}biggr)biggr)^2biggr(frac{2}{3}biggr)biggr(frac{2}{3}biggr), dots end{equation*}$$ It then follows that in general that $$displaystyle P(S_i) = biggr(frac{2}{9}biggr)^{i-1} biggr(frac{4}{9}biggr).$$ Thus, it follows that the probability of $$B$$ winning is calculated as $$begin{equation*} P(S) = Pbiggr(bigcup_{i=1}^infty S_ibiggr) = sum_{i=1}^infty P(S_i) = sum_{i=1}^infty biggr(frac{2}{9}biggr)^{i-1} biggr(frac{4}{9}biggr) = frac{4}{9} sum_{i=1}^infty biggr(frac{2}{9}biggr)^{i-1} = frac{4}{9} cdot frac{9}{7} = frac{4}{7}. end{equation*}$$

Answered by IanFromWashington on January 3, 2022

The problem is not clear as stated.

Interpretation $$#1$$: If you interpret it as "find the probability that the game end in an evenly numbered round" you can reason recursively.

Let $$P$$ denote the answer. The probability that the game ends in the first round is $$frac 26+frac 46times frac 46=frac 79$$. If you don't end in the first round, the probability is now $$1-P$$. Thus $$P=frac 79times 0 +frac 29times (1-P)implies boxed{P=frac 2{11}}$$

Interpretation $$#2$$: If the problem meant "find the probability that $$B$$ wins given that $$A$$ starts" that too can be solved recursively. Let $$Psi$$ denote that answer and let $$Phi$$ be the probability that $$B$$ wins given that $$B$$ starts. Then $$Psi=frac 46times Phi$$ and $$Phi=frac 46 +frac 26times Psi$$ This system is easily solved and yields $$boxed {Psi=frac 47}$$ as desired.

Answered by lulu on January 3, 2022

The game has to end by either A winning or B winning

Let's say A wins. He is just as likely to roll a 1 or a 2 on the last roll. Therefore in a game that A wins, probability of an even roll ending the game is 1/2, as 1(odd) and 2(even) are equally likely.

Let's say B wins. He is just as likely to roll a 3/4/5/6 on the last roll. Therefore in a game that B wins, probability of an even roll ending the game is 1/2, as 4 and 6 are favourable outcomes.

P.S. I have assumed that "ending on an even roll" as written in the title means the die outputs an even number. I agree that while the body of the question seems to suggest an even turn, this seems like the correct interpretation to me.

Answered by user635640 on January 3, 2022