Mr. Brown always bets a dollar on the number 13 at roulette against the advice of Kind Friend. To help cure Mr. Brown of playing roulette, Kind Friend always bets Brown $20 at even money that Brown will be behind at the end of 36 plays. How is the cure working?

(Most American roulette wheels have 38 equally likely numbers. If the player’s number comes up, he is paid 35 times his stake and gets his original stake back; otherwise, he loses his stake)

Solution:

If Mr. Brown wins once in 36 turns, he is even with the casino. His probability of losing all 36 times is \displaystyle \left( \frac{37}{38} \right)^{36} \approx 0.383 . In a single turn, his expectation is

35\left( \frac{1}{38} \right)-1\left( \frac{37}{38} \right) = - \frac{2}{38}\ \text{dollars}

and in 36 turns

-\frac{2}{38}\left( 36 \right) = -1.89 \ \text{dollars}

Against Kind Friend, Mr. Brown has an expectation of

+20\left( 0.617 \right)-20\left( 0.383 \right)\approx +4.68 \ \text{dollars}

And so, all told, Mr. Brown gains +4.68 – 1.89 = + 2.79 dollars per 36 trials; he is finally making money at roulette. Possibly Kind Friend will be cured first. Of course, when Brown loses all 36, he is out $56, which may jolt him a bit.

Chuck-a-Luck is a gambling game often played at carnivals and gambling houses. A player may bet on anyone of the numbers 1, 2, 3, 4, 5, 6. Three dice are rolled. If the player’s number appears on one, two, or three of the dice, he receives respectively one, two, or three times his original stake plus his own money back; otherwise, he loses his stake. What is the player’s expected loss per unit stake? (Actually, the player may distribute stakes on several numbers, but each such stake can be regarded as a separate bet.)

Solution:

Let us compute the losses incurred (a) when the numbers on the three dice are different, (b) when exactly two are alike, and (c) when all three are alike. An easy attack is to suppose that you place a unit stake on each of the six numbers, thus betting six units in all. Suppose the roll produces three different numbers, say 1, 2, 3. Then the house takes the three unit stakes on the losing numbers 4, 5, 6 and pays off the three winning numbers 1, 2, 3. The house won nothing, and you won nothing. That result would be the same for any roll of three different numbers.

Next suppose the roll of the dice results in two of one number and one of a second, say 1, 1, 2. Then the house can use the stakes on numbers 3 and 4 to payoff the stake on number 1, and the stake on number 5 to payoff that on number 2. This leaves the stake on number 6 for the house. The house won one unit, you lost one unit, or per unit stake you lost 1/6.

Suppose the three dice roll the same number, for example, 1, 1, 1. Then the house can pay the triple odds from the stakes placed on 2, 3, 4 leaving those on 5 and 6 as house winnings. The loss per unit stake then is 2/6. Note that when a roll produces a multiple payoff the players are losing the most on the average.

To find the expected loss per unit stake in the whole game, we need to weight the three kinds of outcomes by their probabilities. If we regard the three dice as distinguishable –say red, green, and blue — there are 6 \times 6 \times 6= 216 ways for them to fall.

In how many ways do we get three different numbers? If we take them in order, 6 possibilities for the red, then for each of these, 5 for the green since it must not match the red, and for each red-green pair, 4 ways for the blue since it must not match either of the others, we get 6 \times 5 \times 4 = 120 ways.

For a moment skip the case where exactly two dice are alike and go on to three alike. There are just 6 ways because there are 6 ways for the red to fall and only 1 way for each of the others since they must match the red.

This means that there are 216 - 126 = 90 ways for them to fall two alike and one different. Let us check that directly. There are three main patterns that give two alike: red-green alike, red-blue alike, or green-blue alike. Count the number of ways for one of these, say red-green alike, and then multiply by three. The red can be thrown 6 ways, then the green 1 way to match, and the blue 5 ways to fail to match, or 30 ways. All told then we have 3 \times 30 = 90 ways, checking the result we got by subtraction.

We get the expected loss by weighting each loss by its probability and summing as follows:

\underbrace{\frac{120}{216}\times 0}_\text{none alike} + \underbrace{\frac{90}{216}\times \frac{1}{6}}_\text{2 alike}+\underbrace{\frac{6}{216}\times \frac{2}{6}}_\text{3 alike} = \frac{17}{216} \approx 0.079

Thus, you lose about 8% per play. Considering that a play might take half a minute and that government bonds pay you less than 4% interest for a year, the attrition can be regarded as fierce.

This calculation is for regular dice. Sometimes a spinning wheel with a pointer is used with sets of three numbers painted in segments around the edge of the wheel. The sets do not correspond perfectly to the frequencies given by the dice. In such wheels I have observed that the multiple payoffs are more frequent than for the dice, and therefore the expected loss to the bettor greater.

In a common carnival game, a player tosses a penny from a distance of about 5 feet onto the surface of a table ruled in 1-inch squares. If the penny (3/4 inch in diameter) falls entirely inside a square, the player receives 5 cents but does not get his penny back; otherwise, he loses his penny. If the penny lands on the table, what is his chance to win?

Solution:

When we toss the coin onto the table, some positions for the center of the coin are more likely than others, but over a very small square we can regard the probability distribution as uniform. This means that the proba­bility that the center falls into any
region of a square is proportional to the area of the region, indeed, is the area of the region divided by the area of the square. Since the coin is 3/8 inch in radius, its center must not land within 3/8 inch of any edge if the player is to win. This restriction generates a square of side 1/4 inch within which the center of the coin must lie for the coin to be in the square. Since the proba­bilities are proportional to areas, the probability of winning is \displaystyle \left( \frac{1}{4} \right)^2 = \frac{1}{16}. Of course, since there is a chance that the coin falls off the table altogether, the total probability of winning is smaller still. Also, the squares can be made smaller by merely thickening the lines. If the lines are 1/16 inch wide, the winning central area reduces the probability to \displaystyle \left( \frac{3}{16} \right)^{2} = \frac{9}{256} or less than \displaystyle \frac{1}{28}.

On the average, how many times must a die be thrown until one gets a 6?

Solution:

Let p be the probability of a 6 on a given trial. Then the probabilities of success for the first time on each trial are (let q = 1 - p):

The sum of the probabilities is

\begin{align*}
p+pq+pq^2+\ldots & = p\left( 1+q+q^2+\ldots \right) \\ \\
 & = \frac{p}{1-q} \\ \\
 & = \frac{p}{p} \\ \\
 & = 1
\end{align*}

The mean number of trials, m, is by definition,

m = p + 2pq + 3pq^2 + 4pq^3+ \ldots

Note that our usual trick for summing a geometric series works:

qm = pq + 2pq^2+3pq^3 + \ldots

Subtracting the second expression from the first gives

m-qm=p+pq+pq^2+\ldots

or

m\left( 1-q \right) = 1

Consequently,

mp=1

m=1/p

We see that p=1/6, and so m=6.

On the average, a die must be thrown 6 times until one gets a 6.