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 $$P=frac 79times 0 +frac 29times (1-P)implies boxed{P=frac 2{11}}$$

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 $$Psi=frac 46times Phi$$ 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 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.

Answered by user635640 on January 3, 2022

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