Chapter 18

Calculate the Bernoulli numbers \(B_{8}, B_{10}\), and \(B_{12} .\) The sequence of Bernoulli numbers is usually completed by setting \(B_{0}=1, B_{1}=-\frac{1}{2}\), and \(B_{k}=0\) for \(k\) odd and greater than 1 .

Write out explicitly, using Bernoulli's techniques, the formulas for the sums of the first \(n\) fourth, fifth, and tenth powers. Then show that the sum of the tenth powers of the first 1000 positive integers is $$ 91,409,924,241,424,243,424,241,924,242,500 $$ Bernoulli claimed that he calculated this value in "less than half of a quarter of an hour" (without a calculator).

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 .

Complete Bernoulli's calculation of his example for the Law of Large Numbers by showing that if \(r=30\) and \(s=20\) (so \(t=50\) ) and if \(c=1000\), then $$ n t+\frac{r t(n-1)}{s+1}>m t+\frac{s t(m-1)}{r+1} $$ where \(m, n\) are integers such that $$ m \geq \frac{\log c(s-1)}{\log (r+1)-\log r} $$ and $$ n \geq \frac{\log c(r-1)}{\log (s+1)-\log s}. $$ Conclude that in this case the necessary number of trials is \(N=25,550\).

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 \(N=31,258\).

In his Letter to a Friend on Sets in Court Tennis, written in 1687 but not published until 1713, Jakob Bernoulli analyzed the probabilities at any point in a game or set of court tennis, whose scoring rules are virtually identical with those of tennis today. He determined the odds both when the players were evenly matched and when one player was stronger than the other. If two players \(A\) and \(B\) are evenly matched in a tennis game with the score \(15: 30\), determine the probability of player \(A\) winning. (Remember that one must win by two points.)

Continuing from Exercise 9, suppose that player \(A\) is twice as strong as player \(B\). Suppose that the score is \(30: 30\). Determine the probability of player \(A\) winning. What is the probability of \(A\) winning if the score is \(15: 30\) ?

Suppose that the probability of success in an experiment is \(1 / 10\). How many trials of the experiment are necessary to ensure even odds on it happening at least once? Calculate this both by De Moivre's exact method and his approximation.

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 \((2 m-1) \log (2 m-1)-2 m \log m\).

Derive De Moivre's result \(\log \left(\frac{Q}{M}\right) \approx-\frac{2 t^{2}}{n} \quad\) or equivalently \(\quad \log \left(\frac{M}{Q}\right) \approx \frac{2 t^{2}}{n}\) (Hint: Divide the arguments of the first two logarithm terms in the expression in the text by \(m\). Then simplify and replace the remaining logarithm terms by the first two terms of their respective power series.)

De Moivre's result developing the normal curve implies that the probability \(P_{\epsilon}\) of an observed result lying between \(p-\epsilon\) and \(p+\epsilon\) in \(n\) trials is given by $$ P_{\epsilon}=\frac{1}{\sqrt{2 \pi n p(1-p)}} \int_{-n \epsilon}^{n \epsilon} e^{-\frac{t^{2}}{2 n p(1-p)}} d t $$ Change variables by setting \(u=t / \sqrt{n p(1-p)}\) and use symmetry to show that this integral may be rewritten as $$ P_{\epsilon}=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\frac{\sqrt{n} \epsilon}{\sqrt{p(1-p)}}} e^{-\frac{1}{2} u^{2}} d u $$ Calculate this integral for Bernoulli's example, using \(p=\) \(.6, \epsilon=.02\), and \(n=6498\), and show that in this case \(P_{\epsilon}=\) \(0.999\), a value giving moral certainty. (Use a graphing utility.) Find a value for \(n\) that gives \(P_{\epsilon}=0.99\).

Calculate \(P(r

Show that if an event of unknown probability happens \(n\) times in succession, the odds are \(2^{n+1}-1\) to 1 for more than an even chance of its happening again.

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?

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 .

In Euler's analysis of the lottery for the case \(k=2\), determine the general formulas for the "fair" prizes \(a\) and \(b\) for matching two numbers and for matching one number, respectively, in terms of \(n\) and \(t\), where \(t\) tokens are drawn out of a total of \(n\).

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?

Outline a lesson for a statistics course deriving Bayes's theorem and discussing its usefulness.