提前终止的博弈：公平分配赌注

(a) Peter 还需 2 胜、Paula 还需 4 胜。等价于最多再打 5 局：Peter 在 5 局内赢至少 2 局则先到达。\n\nPeter 获胜概率\n\n$$\n1-\\mathbb{P}(\\text{5 局内 Peter \\le 1 胜})=1-\\left[\\binom{5}{0}+\\binom{5}{1}\\right]\\left(\\frac12\\right)^5=\\boxed{0.8125}.\n$$\n\n因此 100 应按 81.25 与 18.75 分。\n\n(b) 一般：Peter 还需 $a$ 胜、Paula 还需 $b$ 胜，则\n\n$$\n\\boxed{\\mathbb{P}(\\text{Peter})=\\sum_{k=a}^{a+b-1}\\binom{a+b-1}{k}\\left(\\frac12\\right)^{a+b-1}}.\n$$

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英文解析

a. Peter and Paula seem not to have agreed on how to proceed when the game is interrupted before one of them has won 5 points. Clearly, then, one option is that both simply retain their original 50 because the game has not been completed according to

erableprogresstowardwinning5pointsandthereforeclaimmorethanhisoriginal50. A rational basis of this claim could be to consider in how many similar cases Peter would finally have won over Paula - if the game had con- tinued. More precisely, Peter needs 2 more points, whereas Paula still needs twice as many, namely 4. Thus we may legitimately ask: what is the probability that Peter would have won the 2 required points before Paula had won4 points?

The hard way to compute this probability is to enumerate all possible sequences that end favorably for Peter and to sum their probabilities. Note that Peter's victory necessarily ends with a point made by him and may be preceded by 1, 2,3, or at most 4 rounds of which Peter has won exactly one. The complete list contains 10 possible sequences of successive winners ( $\mathbf{A} =$ Peter wins a round, $\mathbf{B} =$ Paula wins a round): AA, ABA, BAA, ABBA, BABA, BBAA, ABBBA, BABBA, BBABA, BBAAA, and their probabilities are easily found to add up to $26 / 32 = 0.8125$ . This result suggests that Peter might claim 81.25, in whichcasePaulareceives18.75. Clearly, when thenumber of rounds that Peter and Paula would still need to win gets larger, then this procedure of explicit enumeration becomes fairly laborious.

A more elegant way to go about this problem is to realize that after at most 5 more rounds either Peter or Paula must have won: if Peter wins 2 or more of these 5 rounds, then Paula has won at most 3 rounds- not enoughd for her to achieve the required total of 5 points. Conversely, if Paula wins 4 or 5 rounds, Peter has won at most 1. Therefore, by an argument somewhat similar to the one we used to solve Problem 1.1, a procedure equivalent to the original rules is to pretend to play exactly 5 more rounds: if Peter wins 2or more of these, this is equivalent to him reaching a total of 5 points before Paula did. On the other hand, if Paula wins 4 or more of these 5 rounds, then she must have reached a total of 5 points before Peter has. The advantage of this reconceptualization is that for a fixed number of rounds (namely, 5) the required probabilities are easily computed from the binomial distribution with n $= 5$ $\mathfrak{p} =$ . For example, the probability that Peter wins no round or only 1 round out of 5 is

$${\binom{5}{0}}\left({\frac{1}{2}}\right)^5+\left({\frac{5}{1}}\right)\left({\frac{1}{2}}\right)^5={\frac{6}{32}}$$

in agreement with the result obtained by the lengthy explicit enumeration described earlier.

b. More generally, if Peter needs to win a more rounds and Paula needs to win b more rounds for an overall win, they could pretend that they play a fixed number of a $+b - 1$ further rounds. Exactly one of the following two mutually exclusive events will then obtain:

1. Peter wins a or more points, in which case Paula wins less than b, or

2. Paula wins b or more points, in which case Peter wins less than a.

The first event implies that Peter had won a points before Paula had won b; the second event means that Paula had won b points before Peter had won a. Therefore, with respect to the winning probabilities, the procedure is functionally equivalent to the rules originally set out, namely to stop when either Peter has collected a, or when Paula has collected b further points.

The probability, e.g., of the first event enumerated here is then easily found by summing the respective binomial terms:

$$\mathrm{P}(\mathrm{Peter~wins~overall}\mid a,b) = \frac{1}{2^{a + b - 1}}\cdot \sum_{i = a}^{a + b - 1}\left( \begin{array}{c}a + b - 1\\ i \end{array} \right)$$