Random Game

Problem

In order to earn money for the equipment of the playing area in the children’s unit of the hospital a prize draw is offered. After the player paid two euros, three balls are drawn at random without replacement from a container in which there are three red, three green and three blue balls. If the three balls have the same colour, the player wins and receives a certain amount of money; otherwise he looses and obtains no money. Afterwards, the drawn balls are put back into the container.

1. Show that the probability for winning one game equals \(\frac{1}{28}\).

2. Compute which amount of money, in case of winning, has to be paid out so that an average gain of 1.25 euros per game for the equipment of the playing area can be expected.

Solution of part a

The probability that three balls with the same colour are drawn can be computed as follows. First, there are 9 balls in the container. Now, one ball is drawn at random. Consequently, eight balls remain in the container. Two of these balls have the same colour as the already drawn ball. The probability of drawing one of these two balls in the next turn thus is \(\frac{2}{8}\). Afterwards, the last ball with the same colour has to be drawn from the container with the remaining seven balls. The corresponding probability is \(\frac{1}{7}\).

The total probability of winning hence is:

\[\frac{2}{8} \cdot \frac{1}{7} = \frac{1}{28} \approx 0.0357\,.\]

This prize draw can be simulated with Sage. For that purpose, we shuffle the balls and consider the colour of the first three balls.

Solution of part b

The expected earnings \(E\) per game are computed for a stake of 2 euros per game and a prize money of \(x\) euro to be

\[E(x) = 2 - \frac{1}{28} \cdot x\,.\]

If the earnings per game are to be 1.25 euros, i.e. \(E(x)=1{,}25\), the result is \(x=21\). The prize money hence has to be 21 euros.

With Sage, we can test what impact a prize money of 21 euros has on the earnings per game.