Excel:赢得决胜局游戏的条件概率 [英] Excel: Conditional probabilities of winning a tiebreaker game

问题描述

我一直在阅读以下论文( http://strategicgames.com.au/PhD.pdf-第21-22页)，而我试图在Excel中重现第23页的表2.5和2.3.

I have been reading the following paper (http://strategicgames.com.au/PhD.pdf - page 21-22) and I am attempting to reproduce table 2.5 and 2.3 on page 23 in Excel.

计算公式:

我的Excel输出:

我使用以下步骤来计算值.

I used the following steps to calculate the values.

1. 首先，我使用上述第三个公式-0.62 *(1-0.60)/(0.62 *(1-0.60)+(1-0.62)* 0.60)计算值0.52
2. 第二，我使用上面的第一个公式-=($ E $ 1 * C8)+($ G $ 1 * D7)计算值0.80，然后将该公式复制并粘贴到其他单元格中.

计算表值在哪里出错?

根据论文的表格，我应该收到以下答案.

I should be receiving the following answer based on the paper's table.

推荐答案

如果要使用这些公式，则如论文所述，您需要使用两个表P_A和P_B.

If you are going to use those formulas, then as the paper says, you need to use two tables, P_A and P_B.

您的计算假设玩家A一直待命，直到有人达到7或得分为(6,6).这错误地给了玩家A很大的优势.

Your calculation assumed that player A is on serve until someone reaches 7 or the score is (6,6). This incorrectly gives a big advantage to player A.

您可以使用一张既不是P_A也不是P_B的表格，而是一个隔行版本，我将其称为P_C，它给出了玩家A假设给定分数的前提下以给定分数获胜的概率.得分(a，b)，其中a + b mod 4为0或3.

You can use a single table that is neither P_A nor P_B in the paper, but instead an interlaced version that I will call P_C that gives the probability that player A wins at a given score assuming that player A serves first, and thus at scores (a,b) where a+b mod 4 is 0 or 3.

P_C(a,b) = p_A * P_C(a+1,b) + (1-p_A) * P_C(a,b+1) when a+b mod 4 is 0 or 3.
P_C(a,b) = (1-p_B) * P_C(a+1,b) + p_B * P_C(a,b+1) when a+b mod 4 is 1 or 2.
P_C(6,6) = p_A * (1-p_B)/(p_A * (1-p_B) + p_B * (1-p_A)).
P_C(7,_) = 1.
P_C(_,7) = 0.

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