## Task 19/71

Two people A and B are passing the time with a game of chance. The stake should be received by the one who first reaches three winning points (in each game, a winning point is awarded; a draw in any game is impossible). Certain circumstances require the game to be interrupted at a score of $1: 0$ for A.

How should the stake be distributed between the two players?
It is about determining the probability with which each of the two players wins. To do this, the possible game outcomes need to be analyzed. A feasible approach is to apply a probability tree, which simplifies the necessary considerations and reduces them to a minimum (figure).

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-209.jpg?height=483&width=1602&top_left_y=227&top_left_x=227)

From this, it becomes clear: The chances for A to win after two more games are \(\frac{1}{4}\). The chances for A to win after three more games are \(\frac{2}{6}\) of the remaining \(\frac{3}{4}\), which is \(\frac{1}{4}\).

In \(\frac{3}{6} \cdot \frac{3}{4} = \frac{3}{8}\) of the cases, the game is still undecided after three more games, and half of these are favorable for A's win, so A has a winning chance of \(\frac{1}{2} \cdot \frac{3}{8} = \frac{3}{16}\) after the 4th game. Since the different possibilities are mutually exclusive, A's winning chance is:

\[
\frac{1}{4} + \frac{1}{4} + \frac{3}{16} = \frac{11}{16}
\]

Correspondingly, B's chance is calculated as:

\[
\frac{1}{4} \cdot \frac{3}{4} + \frac{2}{6} \cdot \frac{1}{2} \cdot \frac{3}{4} = \frac{5}{16}
\]

This value could also have been obtained from \(1 - \frac{11}{16}\). The win is therefore to be distributed in the ratio of 11 parts for A to 5 parts for B.