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Chapter 6: Problem 33

Two individuals, \(A\) and \(B\), are finalists for a chess championship. They will play a sequence of games, each of which can result in a win for A, a win for \(\mathrm{B}\), or a draw. Suppose that the outcomes of successive games are independent, with \(P(\) A wins game \()=.3, P(B\) wins game \()\) \(=.2\), and \(P(\) draw \()=.5\). Each time a player wins a game, he earns one point and his opponent earns no points. The first player to win 5 points wins the championship. For the sake of simplicity, assume that the championship will end in a draw if both players obtain 5 points at the same time. a. What is the probability that A wins the championship in just five games? b. What is the probability that it takes just five games to obtain a champion? c. If a draw earns a half-point for each player, describe how you would perform a simulation experiment to estimate \(P\) (A wins the championship). d. If neither player earns any points from a draw, would the simulation requested in Part (c) take longer to perform? Explain your reasoning.

Short Answer

Expert verified
a) The probability that A wins the championship in just five games is .00243. b) The probability that it takes just five games to obtain a champion is .00275. c) To simulate the game we can generate a random number for each game and the outcome is decided based on the range where this number falls into and a point is awarded accordingly. Repeat this until one gets 5 points and then perform such simulations for numerous times to calculate the probability. d) Simulations would take less time if no points are awarded for draws because less games are needed on average to determine a champion.

Step by step solution

01

Probability of A Winning

Since A wins the championship in just five games, it means A wins all the games. Since the outcomes of each game are independent, the combined probability can be determined by multiplying the individual probabilities. Therefore, the probability is \(.3^5 = .00243\).
02

Probability of Determining a Champion in Five Games

The event that it takes just five games to determine a champion means either A wins in five games or B wins in five games. Calculate probability of these two events and add them up which is \(.3^5 + .2^5 = .00243 + .00032 = .00275\).
03

Design a Simulated Experiment

To simulate this experiment, first, generate a random number. Then, if this number is less than .3, add one point to A; if it is between .3 and .5, add one point to B; and if it鈥檚 greater than .5, add a half point to both of A and B. Repeat these steps until one player gets 5 points. Repeat this simulation for a good amount of rounds and then calculate the probability A wins the championship from the data collected from these simulations. That probability could be the number of times A wins divided by the total number of simulations.
04

Effect of Not Awarding Points for Draws on the Simulation

If no points are awarded for draws, the probability of each game resulting in either a win for A or a win for B becomes larger. This would make the championship end sooner because a 'pointless' outcome has been eliminated. Therefore, the simulation would take less time to perform because less games are needed on average to determine a champion.

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Key Concepts

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

Probability Theory
Probability theory is the mathematical framework for analyzing uncertain events and outcomes. It is the backbone of various games of chance, statistical methods, and risk assessment procedures. When we talk about the probability of an event, we're referring to the likelihood of that event occurring, usually expressed as a value between 0 (impossible event) and 1 (certain event). In our textbook exercise dealing with a chess championship, we apply probability theory to find out the chances of player A winning in just five games. To calculate this, we use the rule of multiplication for independent events, where the occurrence of one event does not affect the others. Thus, the probability of A winning five games in a row is given by \(0.3^5 = 0.00243\), a rather low chance indicative of the difficulty of such a streak.
Statistical Simulation
Statistical simulation is a powerful tool to approximate probabilities and understand complex systems by using random sampling techniques. By simulating an experiment numerous times, we can estimate probabilities by observing the frequency of outcomes. In our chess scenario, simulating the tournament lets us answer queries about the probability of A's victory under different point-scoring rules. To construct a simulation, we could use a programming language or even spreadsheet software to generate random outcomes based on the given probabilities of A winning, B winning, or a draw. We then tally points according to these outcomes, repeating the process a substantial number of times to ensure accuracy. The ratio of simulations where A wins the championship to the total number of simulations gives us an empirical probability of A's victory鈥攁n invaluable tactic when analytical solutions are complex or unavailable.
Independent Events
In the context of probability, independent events are those whose outcomes do not influence one another. This concept is crucial for calculating the combined probability of sequential events occurring. In our chess finals example, the problem states that successive games' results are independent, meaning the outcome of one game has no bearing on the next. This is an essential assumption for both probabilistic calculations and for designing our simulations accurately. Without this assumption, calculating the probability of A or B winning the championship would require a more complicated model accounting for the relationship between games. The assumption of independence streamlines the complexity, making it feasible to use the multiplicative rule (\(P(A \text{ and } B) = P(A) \times P(B)\) when events A and B are independent) to find the probability of an event, such as A winning all five games.

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Most popular questions from this chapter

Four students must work together on a group project. They decide that each will take responsibility for a particular part of the project, as follows: $$\begin{array}{lllll} \hline \text { Person } & \text { Maria } & \text { Alex } & \text { Juan } & \text { Jacob } \\ \text { Task } & \begin{array}{l} \text { Survey } \\ \text { design } \end{array} & \begin{array}{l} \text { Data } \\ \text { collection } \end{array} & \text { Analysis } & \begin{array}{l} \text { Report } \\ \text { writing } \end{array} \end{array}$$ Because of the way the tasks have been divided, one student must finish before the next student can begin work. To ensure that the project is completed on time, a timeline is established, with a deadline for each team member. If any one of the team members is late, the timely completion of the project is jeopardized. Assume the following probabilities: 1\. The probability that Maria completes her part on time is 8 . 2\. If Maria completes her part on time, the probability that Alex completes on time is \(.9\), but if Maria is late, the probability that Alex completes on time is only .6. 3\. If Alex completes his part on time, the probability that Juan completes on time is \(.8\), but if Alex is late, the probability that Juan completes on time is only \(.5\). 4\. If Juan completes his part on time, the probability that Jacob completes on time is \(.9\), but if Juan is late, the probability that Jacob completes on time is only .7. Use simulation (with at least 20 trials) to estimate the probability that the project is completed on time. Think carefully about this one. For example, you might use a random digit to represent each part of the project (four in all). For the first digit (Maria's part), \(1-8\) could represent on time and 9 and 0 could represent late. Depending on what happened with Maria (late or on time), you would then look at the digit representing Alex's part. If Maria was on time, \(1-9\) would represent on time for Alex, but if Maria was late, only 1-6 would represent on time. The parts for Juan and Jacob could be handled similarly.

An article in the New York Times (March 2,1994 ) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P\) (survival) \(=.01\) for people who suffer a cardiac arrest in New York City (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings. Similar studies in smaller cities showed higher survival rates.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York

Three friends \((\mathrm{A}, \mathrm{B}\), and \(\mathrm{C})\) will participate in a round-robin tournament in which each one plays both of the others. Suppose that \(P(\mathrm{~A}\) beats \(\mathrm{B})=.7, P(\mathrm{~A}\) beats \(\mathrm{C})\) \(=.8\), and \(P(\) B beats \(C\) ) \(=.6\) and that the outcomes of the three matches are independent of one another. a. What is the probability that \(\mathrm{A}\) wins both her matches and that B beats \(C\) ? b. What is the probability that A wins both her matches? c. What is the probability that A loses both her matches? d. What is the probability that each person wins one match? (Hint: There are two different ways for this to happen. Calculate the probability of each separately, and then add. )

Approximately \(30 \%\) of the calls to an airline reservation phone line result in a reservation being made. a. Suppose that an operator handles 10 calls. What is the probability that none of the 10 calls results in a reservation? b. What assumption did you make in order to calculate the probability in Part (a)? c. What is the probability that at least one call results in a reservation being made?

A single-elimination tournament with four players is to be held. A total of three games will be played. In Game 1, the players seeded (rated) first and fourth play. In Game 2, the players seeded second and third play. In Game 3 , the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are given: $$\begin{aligned} &P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 4)=.8 \\ &P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 2)=.6 \\ &P(\text { Seed } 1 \text { defeats Seed } 3)=.7 \\ &P(\text { Seed } 2 \text { defeats } \text { Seed } 3)=.6 \\ &P(\text { Seed } 2 \text { defeats Seed } 4)=.7 \\ &P(\text { Seed } 3 \text { defeats Seed } 4)=.6 \end{aligned}$$ a. Describe how you would use a selection of random digits to simulate Game 1 of this tournament. b. Describe how you would use a selection of random digits to simulate Game 2 of this tournament. c. How would you use a selection of random digits to simulate the third game in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\) ) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the true probability? Explain.
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