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Chapter 7: Problem 1

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.

Short Answer

Expert verified
The expected winnings for the player when throwing a fair die and flipping a fair coin simultaneously is approximately 2.0833 units.

Step by step solution

01

List all possible outcomes and their respective winnings

First, we need to list all possible outcomes of rolling a fair die and flipping a fair coin and determine the respective winnings based on the given rules: 1. Die-roll: 1, Coin-flip: Heads - Winnings: 2(1) = 2 2. Die-roll: 1, Coin-flip: Tails - Winnings: 1/2(1) = 0.5 3. Die-roll: 2, Coin-flip: Heads - Winnings: 2(2) = 4 4. Die-roll: 2, Coin-flip: Tails - Winnings: 1/2(2) = 1 5. Die-roll: 3, Coin-flip: Heads - Winnings: 2(3) = 6 6. Die-roll: 3, Coin-flip: Tails - Winnings: 1/2(3) = 1.5 7. Die-roll: 4, Coin-flip: Heads - Winnings: 2(4) = 8 8. Die-roll: 4, Coin-flip: Tails - Winnings: 1/2(4) = 2 9. Die-roll: 5, Coin-flip: Heads - Winnings: 2(5) = 10 10. Die-roll: 5, Coin-flip: Tails - Winnings: 1/2(5) = 2.5 11. Die-roll: 6, Coin-flip: Heads - Winnings: 2(6) = 12 12. Die-roll: 6, Coin-flip: Tails - Winnings: 1/2(6) = 3
02

Calculate the probability of each outcome and the respective expected value

Next, we need to calculate the probability of each of the listed outcomes. Since there are 6 sides on a fair die and 2 sides on a fair coin, the total number of possible outcomes is 6 x 2 = 12. Therefore, the probability of one specific outcome is 1/12. Now, we can calculate the expected value of the winnings for each of these outcomes by multiplying the probability of each outcome by the respective winnings: 1. \(E(1, H) = \frac{1}{12} * 2 = \frac{1}{6}\) 2. \(E(1, T) = \frac{1}{12} * 0.5 = \frac{1}{24}\) 3. \(E(2, H) = \frac{1}{12} * 4 = \frac{1}{3}\) 4. \(E(2, T) = \frac{1}{12} * 1 = \frac{1}{12}\) 5. \(E(3, H) = \frac{1}{12} * 6 = \frac{1}{2}\) 6. \(E(3, T) = \frac{1}{12} * 1.5 = \frac{1}{8}\) 7. \(E(4, H) = \frac{1}{12} * 8 = \frac{2}{3}\) 8. \(E(4, T) = \frac{1}{12} * 2 = \frac{1}{6}\) 9. \(E(5, H) = \frac{1}{12} * 10 = \frac{5}{6}\) 10. \(E(5, T) = \frac{1}{12} * 2.5 = \frac{5}{24}\) 11. \(E(6, H) = \frac{1}{12} * 12 = 1\) 12. \(E(6, T) = \frac{1}{12} * 3 = \frac{1}{4}\)
03

Calculate the total expected value

Finally, we need to add up the expected values from each outcome to find the overall expected winnings for the player: \(E(total) = E(1, H) + E(1, T) + E(2, H) + E(2, T) + E(3, H) + E(3, T) + E(4, H) + E(4, T) + E(5, H) + E(5, T) + E(6, H) + E(6, T)\) \(E(total) = \frac{1}{6} + \frac{1}{24} + \frac{1}{3} + \frac{1}{12} + \frac{1}{2} + \frac{1}{8} + \frac{2}{3} + \frac{1}{6} + \frac{5}{6} + \frac{5}{24} + 1 + \frac{1}{4}\) \(E(total) = \frac{25}{12} = 2.0833\) The expected winnings for the player when throwing a fair die and flipping a fair coin simultaneously is approximately 2.0833 units.

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Most popular questions from this chapter

An urn contains 4 white and 6 black balls. Two successive random samples of sizes 3 and 5 , respectively, are drawn from the urn without replacement. Let \(X\) and \(Y\) denote the number of white balls in the two samples, and compute, \(E[X \mid Y=i]\), for \(i=1,2,3,4\)

Each of \(m+2\) players pays 1 unit to a kitty in order to play the following game. A fair coin is to be flipped successively \(n\) times, where \(n\) is an odd number, and the successive outcomes noted. Each player writes down, before the flips, a prediction of the outcomes. For instance, if \(n=3\), then a player might write down \((H, H, T)\), which means that he or she predicts that the first flip will land heads, the second heads, and the third tails. After the coins are flipped, the players count their total number of correct predictions. Thus, if the actual outcomes are all heads, then the player who wrote \((H, H, T)\) would have 2 correct predictions. The total kitty of \(m+2\) is then evenly - split up among those players having the largest number of correct predictions. Since each of the coin flips is equally likely to land on either heads or tails, \(m\) of the players have decided to make their predictions in a totally random fashion. Specifically, they will each flip one of their own fair coins \(n\) times and then use the result as their prediction. However, the final 2 of the players have formed a syndicate and will use the following strategy. One of them will make predictions in the same random fashion as the other \(m\).

Verify the formula for the moment generating function of a uniform random variable that is given in Table 7.2. Also, differentiate to verify the formulas for the mean and variance.

Suppose that balls are randomly removed from an um initially containing \(n\) white and \(m\) black balls. It was shown in Example \(2 \mathrm{~m}\) that \(E[X]=\) \(1+m /(n+1)\), when \(X\) is the number of draws needed to obtain a white ball. (a) Compute \(\operatorname{Var}(X)\). (b) Show that the expected number of balls that need be drawn to amass a total of \(k\) white balls is \(k[1+m /(n+1)]\). HINT: Let \(Y_{h}, i=1, \ldots, n+1\), denote the number of black balls withdrawn after the \((i-1)\) st white ball and before the \(i\) th white ball. Argue that the \(Y_{i}, i=1, \ldots, n+1\), are identically distributed.

For a standard normal random variable \(Z\), let \(\mu_{n}=E\left[Z^{\prime \prime}\right]\). Show that $$ \mu_{n}= \begin{cases}0 & \text { when } n \text { is odd } \\ \frac{(2 j) !}{2 j} j & \text { when } n=2 j\end{cases} $$ HINT: Start by expanding the moment generating function of \(Z\) into a Taylor series about 0 to obtain $$ \begin{aligned} E\left[e^{t Z}\right] &=e^{t^{2} / 2} \\ &=\sum_{j=0}^{\infty} \frac{\left(t^{2} / 2\right)^{j}}{j !} \end{aligned} $$
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