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\(A\) and \(B\) alternate rolling a pair of dice, stopping either when \(A\) rolls the sum 9 or when \(B\) rolls the sum 6\. Assuming that \(A\) rolls first, find the probability that the final roll is made by \(A .\)

Short Answer

Expert verified
The probability that A wins the game by rolling a sum of 9 is \(\frac{4}{9}\).

Step by step solution

01

Probability of rolling a 9 for A and 6 for B

To find the probability that A rolls a 9 and B rolls a 6 in a single turn, we will find the possible combinations of dice rolls and divide them by the total number of possible dice rolls (36). There are four combinations for A to roll a 9: (3,6),(4,5),(5,4),(6,3) There are five combinations for B to roll a 6: (1,5),(2,4),(3,3),(4,2),(5,1) So the probability of A rolling a 9 is \(\frac{4}{36} = \frac{1}{9} \), and the probability of B rolling a 6 is \(\frac{5}{36}\).
02

Probability of the game continuing

Now we will find the probability that the game continues after each turn. The probability that neither A nor B rolls their target number in a single turn is equal to one minus the probability that one of them does: \(P(\textrm{continue}) = 1 - [P(\textrm{A rolls 9}) + P(\textrm{B rolls 6})] = 1 - [\frac{1}{9} + \frac{5}{36}] = 1 - \frac{9}{36} = \frac{27}{36} = \frac{3}{4}\) The probability of the game continuing is \(\frac{3}{4}\).
03

Summing up the probability

Now we will create an infinite geometric series for the probability that A wins the game. The first term of this series is the probability that A wins on the first turn which is \(\frac{1}{9}\), and the common ratio is the probability of the game continuing (\(\frac{3}{4}\)), so each successive term corresponds to the probability of the game continuing for some number of turns before A wins. The infinite geometric series formula \(S = \frac{a}{1 - r}\) can be used to find the sum of this series where S is the sum, a is the initial term, and r is the common ratio. \(S = \frac{\frac{1}{9}}{1 - \frac{3}{4}} = \frac{\frac{1}{9}}{\frac{1}{4}} = \frac{4}{9}\)
04

Final answer

The probability that A wins the game by rolling a sum of 9 is \(\frac{4}{9}\).

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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 geometric series in probability can reveal insightful patterns in events with repeated trials. In the context of dice games, especially those with alternate turns, the geometric series is a powerful tool to calculate cumulative probabilities of an event occurring over several rounds.

Consider a situation where a player A rolls a pair of dice and wins if he gets a sum of 9. Let's say he is playing against player B, who wins if he gets a sum of 6. They continue to take turns rolling the dice until one of them gets their winning sum. The probability of A rolling a 9 in one throw is \frac{1}{9} and the probability of the game progressing to the next round after both have rolled is \frac{3}{4}. The probability that A eventually wins is not just \frac{1}{9}, but is the sum of probabilities that A wins on the first roll, or the game continues and he wins after B's first roll, and so forth.

This is where the geometric series comes into play. By using the formula for the sum of an infinite geometric series, given by \(S = \frac{a}{1 - r}\), where \(a\) is the first term and \(r\) is the common ratio, we can find the total probability of A winning the game. Here, \(a\) would be \(\frac{1}{9}\) and \(r\) would be \(\frac{3}{4}\). Thus, \(S = \frac{\frac{1}{9}}{1 - \frac{3}{4}} = \frac{4}{9}\), which computes A's overall winning probability.
Alternating Turns Probability
When analyzing games involving alternating turns, such as the dice game between A and B, understanding how to calculate the probability in such scenarios is crucial. Each roll is an independent event, but the overall outcome of the game depends on a sequence of these independent events.

In such games, it becomes important to consider the probabilities of both players not achieving their target on a single turn - this introduces the concept of the game continuing. If we consider that player A and B have respective probabilities of winning on a given turn, the probability that neither wins on that turn is crucial to calculate the chances of the game extending to additional rounds.

As established in the example problem, the probability of the game continuing after each turn is \(\frac{3}{4}\). The probability that A wins on his turn is independent of the turns that come before. Hence, when calculating A's probability of winning, we can regard A's chances in the upcoming round as a 'reset' each time. The repeated multiplication of the probability of the game continuing gives us the probability space for every turn that A has, a concept that closely ties with the geometric series in probability.
Dice Roll Combinations
Dice roll combinations play a central role in the calculation of probabilities in dice games. A standard six-sided die has 36 possible outcomes when rolled in pairs (6 sides on the first die multiplied by 6 sides on the second die). Each combination is equally likely, giving us a simple base probability of \(\frac{1}{36}\) for any specific outcome.

In our example, player A wins if any of four specific combinations occur, giving a probability of \(\frac{4}{36}\) or simplified to \(\frac{1}{9}\) of winning on a single roll. Similarly, player B has five specific combinations to win, which translates to a winning probability of \(\frac{5}{36}\) on a single roll.

Understanding dice roll combinations requires recognizing that each die roll is independent and the outcome of one doesn't affect the other. The total number of combinations is simply a matter of combinatorics, the branch of mathematics dealing with counting combinations and permutations, which gives a systematic way to count the possible outcomes and analyze their probabilities, enhancing strategies for games involving dice.

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Most popular questions from this chapter

\(A\) and \(B\) play a series of games. Each game is independently won by \(A\) with probability \(p\) and by \(B\) with probability \(1-p .\) They stop when the total number of wins of one of the players is two greater than that of the other player. The player with the greater number of total wins is declared the winner of the series. (a) Find the probability that a total of 4 games are played. (b) Find the probability that \(A\) is the winner of the series.

Barbara and Dianne go target shooting. Suppose that each of Barbara's shots hits a wooden duck target with probability \(p_{1},\) while each shot of Dianne's hits it with probability \(p_{2} .\) Suppose that they shoot simultaneously at the same target. If the wooden duck is knocked over (indicating that it was hit), what is the probability that (a) both shots hit the duck? (b) Barbara's shot hit the duck? What independence assumptions have you made?

An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the urn. What is the probability that all of the balls selected are white? What is the conditional probability that the die landed on 3 if all the balls selected are white?

Suppose that an insurance company classifies people into one of three classes: good risks, average risks, and bad risks. The company's records indicate that the probabilities that good-, average-, and bad-risk persons will be involved in an accident over a 1 -year span are, respectively, .05, .15 and \(30 .\) If 20 percent of the population is a good risk, 50 percent an average risk, and 30 percent a bad risk, what proportion of people have accidents in a fixed year? If policyholder \(A\) had no accidents in \(2012,\) what is the probability that he or she is a good risk? is an average risk?

Suppose that 5 percent of men and 0.25 percent of women are color blind. A color-blind person is chosen at random. What is the probability of this person being male? Assume that there are an equal number of males and females. What if the population consisted of twice as many males as females?

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