The wheel of fortune

Problem

A supermarket organizes a competition in order to raise money for the equipement of the local kindergarden. The five sectors of the wheel of fortune used for this purpose are numbered from 1 to 5. The size of the sectors is proportional to the value of the numbers, e.g., the sector with number 3 is three times as large as the sector with the number 1. After the player has paid six euros, the wheel of fortune will be turned once. Does the player obtain one of the numbers 1 to 4, he will receive the corresponding value in a corresponding amount of euros. If he obtains the number 5, he receives a ticket for a leisure park with a value of fifteen euros.

1. Determine the angle spanned by the sector with number 1 as well as the probability that the player will win the ticket in a single game.

   (Partial result: Size of angle: 24°)

2. Determine the expectation value for the payment per game if winning a ticket is equivalent to receiving a payment of fifteen euros. Interprete the result.

3. The supermarket needs to pay to the leisure park only ten euros per ticket. Therefore, as a result of the competition, one can expect a surplus to be donated to the local kindergarden. Determine the expected surplus provided that the game is played 6000 times.

Solution of part 1a

The angle spanned by sector 1 can be obtained as follows:

\[x+2x+3x+4x+5x =360°\quad\Leftrightarrow\quad x=24°\,.\]

The probability to win a ticket is given by

\[\frac{5}{1+2+3+4+5}=\frac{5}{15}=\frac{1}{3}.\]

Solution of part 1b

The expectation value is determined by means of the formula

\[E=\sum\limits_{X}P(X)\cdot X,\]

where \(X\) is the amount of the payment. Assuming that winning the ticket is equivalent to a payment of fifteen euros, we obtain

\[E=\frac{1}{15}+\frac{2}{15}\cdot2+\frac{3}{15}\cdot3+\frac{4}{15}\cdot4+\frac{5}{15}\cdot15=7\,.\]

A player on average receives seven euros per round.

A corresponding simulation can be carried out with Sage.

Solution of part 1c

In contrast to part b, the payment now has to be replaced by the yield for the supermarket, i.e., \(5\), \(4\), \(3\), \(2\) and \(-4\). We thus obtain for the expectation value

\[E=\frac{1}{15}\cdot5+\frac{2}{15}\cdot4+\frac{3}{15}\cdot3+\frac{4}{15}\cdot2+\frac{5}{15}\cdot(-4)=\frac{2}{3}.\]

The yield per game for the supermarket amounts to 67 cents. On the basis of 6000 games, a surplus of about

\[6000\cdot\frac{2}{3}\,\mathrm{euros}=4000\,\mathrm{euros}\]

should result. For such a large number of repetitions, the actual result should not deviate by too much from the expectation value.

We simulate the game with Sage as seen from the supermarket or the kindergarden.

An impression of the bandwidth of the results can be obtained by means of a simulation. A series of 6000 games is repeated many times and the frequency of surplusses is displayed in a histogram.