Rolling Dice Game, Probability of Ending on an Even Roll

Question

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}\cdot\frac{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?

Answer 1

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)^2\biggr(\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) = P\biggr(\bigcup_{i=1}^\infty S_i\biggr) = \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*}

Answer 2

The answer = 1/2

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.

Answer 3

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

as in your solution.

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 $$\Psi=\frac 46\times \Phi$$ and $$\Phi=\frac 46 +\frac 26\times \Psi$$ This system is easily solved and yields $$\boxed {\Psi=\frac 47}$$ as desired.

Answer 4

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 $6\times 6 = 36$ 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 $6\times 6 - 4\times 2 = 28$ equally likely possible final rounds.

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

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