Just wondering how Pascal used the triangle in the problem where they wanted to determine how to divide the pot when the game ended early. It was the first major probability problem conceived in history. I couldn't really find a simple enough explanation.


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This is called the problem of points. An accessible explanation of Pascal's solution and how binomial coefficients (and hence the triangle) come up in it is given on Probability and Statistics blog, but it is a bit long to be reproduced here, see also Edwards, Pascal and the Problem of Points.

Coughlin and Kerwin in Mathematical Induction and Pascal's Problem of the Points give a simplified explanation for a special case, and show how to perform calculations directly on the triangle.

Suppose we have two players (teams), $A$ and $B$, with equal probabilities of winning each round. The game series was supposed to last until one of them has $5$ wins (points), and whoever does gets the whole pot $M$. Sadly, the game was interrupted when $A$ had only $3$ wins and $B$ had only $2$ wins. At that point, their fair shares of the pot are their probabilities of winning the series, if it continued, multiplied by $M$. To win the series, $A$ needs $2$ more wins and $B$ needs $3$ more wins:

"Pascal may have reasoned as follows. In this case the series would be decided in at most four games. A four-game series has $2^4$ possible outcomes: $AAAA,AAAB,AABB,...$. Team $A$ can win two games in $\binom42$ ways, three games in $\binom43$ ways, and all four games in $\binom44$ ways. Thus, team $A$ can win two or more games in $$\binom42+\binom43+\binom44$$ ways. These results represent theoretical possibilities even though it may not be necessary to play all four games. Consequently, the probability that team A will win the series is given by $$\frac1{2^4}\left[\binom42+\binom43+\binom44\right].$$ Using Pascal's triangle, we find $$\binom42=6,\ \ \binom43=4,\ \ \binom44=1;$$ and hence the probability that $A$ will win is $(6 + 4 + 1)/16 = 11/16$. Multiplying by $M$, we get $11/16\,M$..."

The share of $B$, accordingly, will be the complementary $5/16\,M$.

In general, $A$ and $B$ may have unequal probabilities of winning rounds, say $p$ and $1-p$ (no ties), so the sum of binomial coefficients has to be replaced by a weighted sum, with powers of $p$ and $1-p$ as weights. Pascal set up a recursion similar to the one used to construct the triangle of binomial coefficients, but involving $p$ and $1-p$. The idea of calculating the shares remains the same.

  • $\begingroup$ is there a rigorous proof for this? $\endgroup$
    – Markvz
    Aug 19, 2022 at 13:39
  • $\begingroup$ @Markvz Yes, see the linked sources. $\endgroup$
    – Conifold
    Aug 19, 2022 at 13:43

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