91Ó°ÊÓ

A player throws a fair die and simultaneously flips a fair coin. If the coin lands heads, then she wins twice, and if tails, then one-half of the value that appears on the die. Determine her expected winnings.

The game of Clue involves 6 suspects, 6 weapons, and 9 rooms. One of each is randomly chosen and the object of the game is to guess the chosen three. (a) How many solutions are possible? In one version of the game, after the selection is made each of the players is then randomly given three of the remaining cards. Let \(S, W\), and \(R\) be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player. Also, let \(X\) denote the number of solutions that are possible after that player observes his or her three cards. (b) Express \(X\) in terms of \(S, W\), and \(R\). (c) Find \(E[X]\).

An urn has \(n\) white and \(m\) black balls which are removed one at a time in a randomly chosen order. Find the expected number of instances in which a white ball is immediately followed by a black one.

If a die is to be rolled until all sides have appeared at least once, find the expected number of times that outcome 1 appears.

Let \(X\) be a random variable having finite expectation \(\mu\) and variance \(\sigma^{2}\), and let \(g(\cdot)\) be a twice differentiable function. Show that $$ E[g(X)] \approx g(\mu)+\frac{g^{\prime \prime}(\mu)}{2} \sigma^{2} $$ Hint: Expand \(g(\cdot)\) in a Taylor series about \(\mu\). Use the first three terms and ignore the remainder.

Suppose that \(A\) and \(B\) each randomly, and independently, choose 3 of 10 objects. Find the expected number of objects (a) chosen by both \(A\) and \(B\); (b) not chosen by either \(A\) or \(B\); (c) chosen by exactly one of \(A\) and \(B\).

A deck of \(n\) cards, numbered 1 through \(n\), is thoroughly shuffled so that all possible \(n !\) orderings can be assumed to be equally likely. Suppose you are to make \(n\) guesses sequentially, where the \(i\) th one is a guess of the card in position \(i\). Let \(N\) denote the number of correct guesses. (a) If you are not given any information about your earlier guesses show that, for any strategy, \(E[N]=1\). (b) Suppose that after each guess you are shown the card that was in the position in question. What do you think is the best strategy? Show that under this strategy $$ \begin{aligned} E[N] &=\frac{1}{n}+\frac{1}{n-1}+\cdots+1 \\ & \approx \int_{1}^{n} \frac{1}{x} d x=\log n \end{aligned} $$ (c) Suppose that you are told after each guess whether you are right or wrong. In this case it can be shown that the strategy that maximizes \(E[N]\) is one which keeps on guessing the same card until you are told you are correct and then changes to a new card. For this strategy show that $$ \begin{aligned} E[N] &=1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{n !} \\ &=e-1 \end{aligned} $$

Suppose that \(X_{1}\) and \(X_{2}\) are independent random variables having a common mean \(\mu\). Suppose also that \(\operatorname{Var}\left(X_{1}\right)=\sigma_{1}^{2}\) and \(\operatorname{Var}\left(X_{2}\right)=\sigma_{2}^{2} .\) The value of \(\mu\) is unknown and it is proposed to estimate \(\mu\) by a weighted average of \(X_{1}\) and \(X_{2}\). That is, \(\lambda X_{1}+(1-\lambda) X_{2}\) will be used as an estimate of \(\mu\), for some appropriate value of \(\lambda\). Which value of \(\lambda\) yields the estimate having the lowest possible variance? Explain why it is desirable to use this value of \(\lambda\).

For a group of 100 people compute (a) the expected number of days of the year that are birthdays of exactly 3 people; (b) the expected number of distinct birthdays.

How many times would you expect to roll a fair die before all 6 sides appeared at least once?