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Chapter 18: Problem 25

In the French Royal Lottery of the late eighteenth century, five numbered balls were drawn at random from a set of 90 balls. Originally, a player could buy a ticket on any one number or on a pair or on a triple. Later on, one was permitted to bet on a set of four or five as well as on a set given in the order drawn. Show that the odds against winning with a bet on a single number, a pair, and a triple are \(17: 1,399.5: 1\), and \(11,747: 1\), respectively. The payoffs on these bets are 15,270 , and 5,500 .

Short Answer

Expert verified
The odds against winning with a bet on a single number, a pair, and a triple are 17:1, 1,399.5:1, and 11,747:1, respectively.

Step by step solution

01

Find the total number of ways to choose 5 balls from 90 balls

We use the combination formula "n choose r", written as C(n, r) = n! / (r!(n-r)!) where n is the total number of balls, r is the number of balls chosen, and ! denotes the factorial of a number. Total number of ways to choose 5 balls from 90 balls is: C(90, 5)=\frac{90!}{5!(90-5)!}
02

Calculate the probability of winning with a bet on a single number

To win with a single number bet, the chosen number must be among the 5 drawn balls. There are 5 ways to choose 1 winning number from the 5 drawn balls and 85 ways to choose the other 4 non-winning numbers from the remaining 90-5=85 balls. So we have: Number of ways to win with a single number bet = C(5, 1) * C(85, 4) Probability of winning with a single number bet = \frac{C(5, 1) * C(85, 4)} {C(90, 5)}
03

Calculate the probability of winning with a bet on a pair

To win with a pair bet, the two chosen numbers must be among the 5 drawn balls. There are C(5, 2) ways to choose the 2 winning numbers from the 5 drawn balls and C(88, 3) ways to choose the other 3 non-winning numbers from the remaining 90-2=88 balls. So we have: Number of ways to win with a pair bet = C(5, 2) * C(88, 3) Probability of winning with a pair bet = \frac{C(5, 2) * C(88, 3)}{C(90, 5)}
04

Calculate the probability of winning with a bet on a triple

To win with a triple bet, the three chosen numbers must be among the 5 drawn balls. There are C(5, 3) ways to choose the 3 winning numbers from the 5 drawn balls and C(87, 2) ways to choose the other 2 non-winning numbers from the remaining 90-3=87 balls. So we have: Number of ways to win with a triple bet = C(5, 3) * C(87, 2) Probability of winning with a triple bet = \frac{C(5, 3) * C(87, 2)}{C(90, 5)}
05

Calculate the odds against winning

Odds against winning are calculated as (1 - probability of winning) / probability of winning. Odds against winning with a single number bet = \frac{1 - P(winning\_single\_number)}{P(winning\_single\_number)} = \frac{1-\frac{C(5,1)*C(85,4)}{C(90,5)}}{\frac{C(5,1)*C(85,4)}{C(90,5)}} = 17:1 Odds against winning with a pair bet = \frac{1 - P(winning\_pair)}{P(winning\_pair)} = \frac{1-\frac{C(5,2)*C(88,3)}{C(90,5)}}{\frac{C(5,2)*C(88,3)}{C(90,5)}} = 1,399.5:1 Odds against winning with a triple bet = \frac{1 - P(winning\_triple)}{P(winning\_triple)} = \frac{1-\frac{C(5,3)*C(87,2)}{C(90,5)}}{\frac{C(5,3)*C(87,2)}{C(90,5)}} = 11,747:1 Thus, the odds against winning with a bet on a single number, a pair, and a triple are \(17:1\), \(1,399.5:1\), and \(11,747:1\), respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a field of mathematics concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It is foundational for understanding probability in contexts like lotteries. In the French Royal Lottery example, the combinatorial task was to count the different ways five numbered balls could be drawn from a set of 90.

Combinatorics includes the use of 'combinations,' a key concept when discussing 'choosing' r items from a set of n without regard to order. The combination formula is expressed as \( C(n, r) = \frac{n!}{r!(n-r)!} \). Factorials are crucial in this formula—represented by the exclamation mark (!)—which means the product of all positive integers up to that number.

This branch of mathematics aids in understanding how many different groups of numbers can be formed, which is essential for calculating probabilities in lottery-style games where order does not matter. In the lottery example, we are using combinations to determine the probability of various winning bets.
Odds Calculation
The odds calculation is crucial in the world of gambling and lotteries. Odds are a way of expressing the likelihood of an event to occur versus it not occurring. In the lottery problem, after calculating the probability of winning, odds can be determined by the ratio \(\frac{1 - P(event)}{P(event)}\), where \(P(event)\) represents the probability of the event happening.

For a more intuitive understanding, if an event is certain not to happen, the odds are 0:1; whereas, if it's certain to happen, the odds are 1:0. In the lottery scenario, odds give a clearer picture of a player's chances of winning. For example, '17:1' indicates that for every 18 times the lottery is played (17 times not winning plus 1 time winning), you would expect to win only once, on average.
Factorial Notation
Factorial notation is a mathematical concept to succinctly represent the multiplication of a sequence of descending natural numbers. The symbol for factorial is an exclamation point (!). For instance, the factorial of 5, denoted as \(5!\), is equal to \(5 \times 4 \times 3 \times 2 \times 1\).

This notation is incredibly useful when dealing with large sequences in combinatorial problems because it compactly expresses complex multiplications. When calculating combinations, like in the lottery exercise, factorials handle the multitude of possibilities involved in drawing balls. It's important to understand that factorials grow extremely fast, which is why the numbers become very large in lottery probabilities, reflecting the long odds against winning in these scenarios.

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Most popular questions from this chapter

Imagine an urn with two balls, each of which may be either white or black. One of these balls is drawn and is put back before a new one is drawn. Suppose that in the first two draws white balls have been drawn. What is the probability of drawing a white ball on the third draw?

The so-called St. Petersburg Paradox was a topic of debate among those mathematicians involved in probability theory in the eighteenth century. The paradox involves the following game between two players. Player \(A\) flips a coin until a tail appears. If it appears on his first flip, player \(B\) pays him 1 ruble. If it appears on the second flip, \(B\) pays 2 rubles, on the third, 4 rubles, \(\ldots\), on the \(n\)th flip, \(2^{n-1}\) rubles. What amount should \(A\) be willing to pay \(B\) for the privilege of playing? Show first that \(A\) 's expectation, namely, the sum of the probabilities for each possible outcome of the game multiplied by the payoff for each outcome, is $$ \sum_{i=0}^{\infty} \frac{1}{2^{i}} 2^{i-1} $$ and then that this sum is infinite. Next, play the game 10 times and calculate the average payoff. What would you be willing to pay to play? Why does the concept of expectation seem to break down in this instance?

Show that if one defines the Bernoulli numbers \(B_{i}\) by setting $$ \frac{x}{e^{x}-1}=\sum_{i=0}^{\infty} \frac{B_{i}}{i !} x^{i} $$ then the values of \(B_{i}\) for \(i=2,4,6,8,10,12\) are the same as those calculated in the text and in Exercise 1 .

Use Bernoulli's formula to show that if greater certainty is wanted in the problem of Exercise 7 , say, \(c=10,000\), then the number of trials necessary is \(N=31,258\).

Add the highest-degree terms of the columns from Exercise 15 to get $$ s\left(\frac{s}{m}+\frac{1}{2 \cdot 3} \frac{s^{3}}{m^{3}}+\frac{1}{3 \cdot 5} \frac{s^{5}}{m^{5}}+\frac{1}{4 \cdot 7} \frac{s^{7}}{m^{7}}+\cdots\right) $$ which, setting \(x=s / m\), is equal to $$ s\left(\frac{2 x}{1 \cdot 2}+\frac{2 x^{3}}{3 \cdot 4}+\frac{2 x^{5}}{5 \cdot 6}+\frac{2 x^{7}}{7 \cdot 8}+\cdots\right) $$ Show that the series in the parenthesis can be expressed in finite terms as $$ \log \left(\frac{1+x}{1-x}\right)+\frac{1}{x} \log \left(1-x^{2}\right) $$ and therefore that the original series is $$ m x \log \left(\frac{1+x}{1-x}\right)+m \log \left(1-x^{2}\right) $$ Since \(s=m-1\) (or \(m x=m-1\) ), show therefore that the sum of the highest- degree terms of the columns of Exercise 15 is equal to $$ \begin{aligned} &(m-1) \log \left(\frac{1+\frac{m-1}{m}}{1-\frac{m-1}{m}}\right) \\ &\quad+m \log \left[\left(1+\frac{m-1}{m}\right)\left(1-\frac{m-1}{m}\right)\right] \end{aligned} $$ which in turn is equal to \((2 m-1) \log (2 m-1)-2 m \log m\).
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