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Frederick Mosteller, 'Fifty Challenging problems in probability', Q44:

A game consists of a sequence of plays, on each play either you or your opponent scores a point, you with probability $p$, your opponent with probability $1-p$. The total number of plays is even, and to win you need to get more than half the points. You get to choose the number of plays. What number do you choose?

I do not understand the answer:

Your probability of winning a game of $2n$ trials is your probability of getting $n+1$,$n+2$,...$2n$ points ie: $$P_{2n}=\sum_{x=n+1}^{2n} \binom{2n}{x}p^xq^{2n-x} \tag{1}$$ and in a game of $2n+2$ trials the probability of getting $n+2$ points or more : $$P_{2n+2}=\sum_{x=n+2}^{2n+2} \binom{2n+2}{x}p^xq^{2n+2-x} \tag{2}$$ A game of $2n+2$ plays can be regarded as a game of $2n$ plays with 2 trials added on. Unless player $A$ has won either $n$ or $n+1$ times in the $2n$ game, his status will not change in the $2n+2$ game.

Except for these 2 possibilities $P_{2n}$ and $P_{2n+2}$ would be identical. These exceptions are:

(a) having won $n+1$ times in $2n$ plays, $A$ loses the next 2, thus reducing his probability of winning in the $2n+2$ game by $$q^2 \binom{2n}{n+1}p^{n+1}q^{n+1} \tag{3}$$ (b) having won $n$ times in $2n$ plays, $A$ wins the next 2, thus increasing his probability of winning in the $2n+2$ game by $$p^2 \binom{2n}{n}p^{n}q^{n} \tag{4}$$

The optimum game length is such that $P_{2n-2} \lt P_{2n}$ and $P_{2n} \gt P_{2n+2}$ which given the above implies the inequalities : $$ q^2 \binom{2n-2}{n}p^{n}q^{n-2} \lt p^2 \binom{2n-2}{n-1}p^{n-1}q^{n-1} \tag{5} $$ $$ q^2 \binom{2n}{n+1}p^{n+1}q^{n-1} \gt p^2 \binom{2n}{n}p^{n}q^{n} \tag{6} $$ which reduces to $$(n-1)q \lt np \tag{7}$$ $$nq \gt (n+1)p \tag{8}$$ yielding $$N=2n=\frac{1}{1-2p} \tag{9}$$

I have typed the answer in full for those interested, however my question is just this:

what is the relationship between (1) ,(2), (3) and (4) ?

is it really just (2) = (1) + (3) + (4) ? the text does not make it clear ...

1 Answers 1

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(1) and (2) are really the same. (1) is the chance that A wins a majority of 2n games. (2) is the chance that A wins a majority of 2n+2 games. If we let L be the number of games the range of the summation in both is $\frac L2+1$ to $L$, which are the number of game wins that gives A the match. The two expressions are separately calculated using the same logic. The next paragraph explains where (3) and (4) come from. If A has won $n+2$ or more of the first $2n$, he can play the next two with no worry of converting a win into a loss. Similarly if he has won $n-1$ or less he cannot convert a loss into a win in two games. (3) and (4) compute the chance that the win/loss status of the match changes in the last two games.

Note that if $p \gt \frac 12$ the answer tells us to request a negative number of games if we are A. This reflects that the player with the advantage wants to play as many games as possible to improve his chances. In fact you will never satisfy (8) in that case.

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    So it is NOT true that (2) = (1) + (3) + (4) , am I right? I ask because of the statement 'a game of 2n+2 plays can be regarded as a game of 2n plays with 2 trials added on' which would imply that $P_{2n+2}$ is $P_{2n}$ + something, and (3) + (4) seem plausible candidates for that 'something' – 2017-01-30
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    No, (2)=(1)+(4)-(3) because (3) converts a win at 2n games into a loss at 2n+2 – 2017-01-30
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    it is very clear now , thank you very much Sir. – 2017-01-30
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    The basic idea is that even though playing more games is a disadvantage to A if $p \lt 1/2$ you decrease the chance that you tie, so the chance that you win can increase. We can look at choosing between $2$ and $4$ games. With two games, A wins with probability $p^2$. With four, A wins with probability $p^4+4p^3(1-p)$ If $p$ is very small, you want as few powers of $p$ possible so should just play two games. But $p^4+4p^3(1-p)=p^2(4p-3p^2)$ The last factor is greater than $1$ when $p \gt 1/3$ – 2017-01-30