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Chapter 3: Problem 64

\(A\) and \(B\) roll a pair of dice in turn, with \(A\) rolling first. A's objective is to obtain a sum of 6 , and \(B\) 's is to obtain a sum of 7 . The game ends when either player reaches his or her objective, and that player is declared the winner. (a) Find the probability that \(A\) is the winner. (b) Find the expected number of rolls of the dice. (c) Find the variance of the number of rolls of the dice.

Short Answer

Expert verified
The probability that A is the winner is approximately 0.4545. The expected number of rolls of the dice is approximately 2.2174, and the variance of the number of rolls of the dice is approximately 1.8036.

Step by step solution

01

Calculate the probability of A winning on the first turn

To win on the first turn, A needs to roll a sum of 6. There are 5 outcomes that result in a sum of 6: (1, 5), (2, 4), (3, 3), (4, 2), and (5, 1). Since a pair of dice have 36 possible outcomes, the probability is given by: \(P(A \ wins \ on \ first \ turn) = \frac{5}{36}\)
02

Calculate the probability of A winning on a subsequent turn

Let \(P_n\) be the probability that A wins on the nth turn. We know that \(P_1 = \frac{5}{36}\). Now, let's calculate \(P_2\). For this to happen, A must not win on the first turn, B must not win on their first turn, and A must win on the second turn: \(P(A \ wins \ on \ second \ turn) = P_2 = (1 - P_1) * (1 - \frac{6}{36}) * P_1\) Similarly, the probability of A winning on the 3rd, 4th, ... turns can be calculated iteratively: \(P_3 = (1 - P_1) * (1 - \frac{6}{36}) * P_2\) \(P_4 = (1 - P_1) * (1 - \frac{6}{36}) * P_3\) To find the probability of A winning the game, we sum up the probabilities of A winning on each turn: \(P(A \ wins) = \sum_{n=1}^{\infty} P_n\) This is a geometric series, with the sum given by: \(P(A \ wins) = \frac{P_1}{1 - (1 - P_1) * (1 - \frac{6}{36})}\)
03

Calculate the probability

Plug in the values for \(P_1\) and simplify: \(P(A \ wins) = \frac{\frac{5}{36}}{1 - (1 - \frac{5}{36}) * (1 - \frac{6}{36})} \approx 0.4545\) So, the probability that A is the winner is approximately 0.4545. #b) Find the expected number of rolls of the dice. # Let's use the fact that the expected number of rolls (E) can be found using the sum of the products of the probability and the number of rolls for each possible number of rolls: \(E = \sum_{n=1}^{\infty} n * P_n\) Since we already have the probabilities, plug them in and use the fact that it's a geometric series: \(E = \frac{1}{1 - (1 - P_1) * (1 - \frac{6}{36})}\)
04

Calculate the expected number of rolls

Plug in the values for \(P_1\) and simplify: \(E = \frac{1}{1 - (1 - \frac{5}{36}) * (1 - \frac{6}{36})} \approx 2.2174\) So, the expected number of rolls of the dice is approximately 2.2174. #c) Find the variance of the number of rolls of the dice.# Let's denote the variance of the number of rolls as V. Using the fact that variance can be found using the sum of the products of the squared difference between the number of rolls and the expected number of rolls, and the probability for each possible number of rolls: \(V = \sum_{n=1}^{\infty} (n - E)^2 * P_n\) Since we already have the probabilities, plug them in and use the fact that it's a geometric series: \(V = \frac{(1 - E)^2}{1 - (1 - P_1) * (1 - \frac{6}{36})}\)
05

Calculate the variance

Plug in the values for \(P_1\) and \(E\) and simplify: \(V = \frac{(1 - 2.2174)^2}{1 - (1 - \frac{5}{36}) * (1 - \frac{6}{36})} \approx 1.8036\) So, the variance of the number of rolls of the dice is approximately 1.8036.

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Key Concepts

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

Geometric Series in Probability
Understanding the role of a geometric series in probability is crucial when dealing with repetitive trials, such as rolling dice. In our dice game, we analyze the chances of a player winning over multiple rounds. The probability that player A wins on any given turn is a recursive event: each turn's outcome depends on the previous turns. A geometric series is used to sum the probabilities of A winning on the first, second, third turn, and so on.

This can be visualized as an infinite sequence of binary outcomes where each round can be a win for A or not, with the game continuing until A wins. To calculate the total probability of A winning, we acknowledge that this series converges, as the probability of A winning in later turns diminishes exponentially. Consequently, we apply the formula for the sum of an infinite geometric series, which requires the first term and the common ratio鈥攂oth derived from the probabilities of A winning and not winning in each round.

Students may find it challenging to understand this concept because it involves not just the immediate outcome of each dice roll, but the compounding effect of each subsequent roll. Highlighting the repetitive nature of the game and the decreasing likelihood of successive attempts can help clarify how a geometric series encapsulates this phenomenon.
Expected Number of Trials
The expected number of trials, or in our case, rolls of dice, is the average number it would take for a certain event to occur, like player A winning the game. It's found by combining each potential outcome's probability with the number of trials it takes to get there, a concept often overshadowed by the immediate probability of a single event.

In a repetitive game such as ours, we must consider every possible path to victory for player A and calculate the expected value by summing over all these paths. It is a mean weighted by probability, and we can think of it as what a player might anticipate as the 'length' of a game in number of dice rolls.

Explaining that the expected number of trials gives insight into the 'long-run' average experience can be helpful for students. They can understand it as judging the game not by one or a few rolls, but by imagining the game being played over and over again infinitely and averaging out all those experiences.
Variance of Number of Rolls
Variance in the number of rolls provides a measure of how much the number of rolls can differ from the expected number. It reflects the spread or dispersion of possible outcomes. High variance means the number of rolls could be very unpredictable, with a broad spread around the mean, while low variance indicates more consistent gameplay near the expected value.

Calculating variance involves squaring the difference between each possible number of rolls and the expected number, weighting those squared differences by the probability of each roll number, and then summing those values. It can be a challenging concept because it's not as intuitive as the expected number; variance is about the 'spread' rather than the 'middle' of all outcomes.

Students may benefit from understanding that while the expected number is an average, variance tells us about the reliability of that average 鈥� a game with low variance means you can be more confident the game will likely finish close to the expected number of rolls.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots\) be independent continuous random variables with a common distribution function \(F\) and density \(f=F^{\prime}\), and for \(k \geqslant 1\) let $$ N_{k}=\min \left\\{n \geqslant k: X_{n}=k \text { th largest of } X_{1}, \ldots, X_{n}\right\\} $$ (a) Show that \(P\left\\{N_{k}=n\right\\}=\frac{k-1}{n(n-1)}, n \geqslant k\). (b) Argue that $$ f_{X_{N_{k}}}(x)=f(x)(\bar{F}(x))^{k-1} \sum_{i=0}^{\infty}\left(\begin{array}{c} i+k-2 \\ i \end{array}\right)(F(x))^{i} $$ (c) Prove the following identity: $$ a^{1-k}=\sum_{i=0}^{\infty}\left(\begin{array}{c} i+k-2 \\ i \end{array}\right)(1-a)^{i}, \quad 0

An individual traveling on the real line is trying to reach the origin. However, the larger the desired step, the greater is the variance in the result of that step. Specifically, whenever the person is at location \(x\), he next moves to a location having mean 0 and variance \(\beta x^{2}\). Let \(X_{n}\) denote the position of the individual after having taken \(n\) steps. Supposing that \(X_{0}=x_{0}\), find (a) \(E\left[X_{n}\right]\); (b) \(\operatorname{Var}\left(X_{n}\right)\)

If \(X\) and \(Y\) are both discrete, show that \(\sum_{x} p_{X \mid Y}(x \mid y)=1\) for all \(y\) such that \(p_{Y}(y)>0\)

In a knockout tennis tournament of \(2^{n}\) contestants, the players are paired and play a match. The losers depart, the remaining \(2^{n-1}\) players are paired, and they play a match. This continues for \(n\) rounds, after which a single player remains unbeaten and is declared the winner. Suppose that the contestants are numbered 1 through \(2^{n}\), and that whenever two players contest a match, the lower numbered one wins with probability \(p\). Also suppose that the pairings of the remaining players are always done at random so that all possible pairings for that round are equally likely. (a) What is the probability that player 1 wins the tournament? (b) What is the probability that player 2 wins the tournament? Hint: Imagine that the random pairings are done in advance of the tournament. That is, the first-round pairings are randomly determined; the \(2^{n-1}\) first-round pairs are then themselves randomly paired, with the winners of each pair to play in round 2; these \(2^{n-2}\) groupings (of four players each) are then randomly paired, with the winners of each grouping to play in round 3, and so on. Say that players \(i\) and \(j\) are scheduled to meet in round \(k\) if, provided they both win their first \(k-1\) matches, they will meet in round \(k\). Now condition on the round in which players 1 and 2 are scheduled to meet.

Let \(N\) be a hypergeometric random variable having the distribution of the number of white balls in a random sample of size \(r\) from a set of \(w\) white and \(b\) blue balls. That is, $$ P\\{N=n\\}=\frac{\left(\begin{array}{c} w \\ n \end{array}\right)\left(\begin{array}{c} b \\ r-n \end{array}\right)}{\left(\begin{array}{c} w+b \\ r \end{array}\right)} $$ where we use the convention that \(\left(\begin{array}{c}m \\\ j\end{array}\right)=0\) if either \(j<0\) or \(j>m\). Now, consider a compound random variable \(S_{N}=\sum_{i=1}^{N} X_{i}\), where the \(X_{i}\) are positive integer valued random variables with \(\alpha_{j}=P\left\\{X_{i}=j\right\\}\) (a) With \(M\) as defined as in Section \(3.7\), find the distribution of \(M-1\). (b) Suppressing its dependence on \(b\), let \(P_{w, r}(k)=P\left\\{S_{N}=k\right\\}\), and derive a recursion equation for \(P_{w, r}(k)\). (c) Use the recursion of (b) to find \(P_{w, r}(2)\).
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