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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 1 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 to estimate \(\mathrm{P}(\mathrm{A}\) wins the championship). d. If neither player earns any points from a draw, would the simulation in Part (c) take longer to perform? Explain your reasoning.

Short Answer

Expert verified
The probability that A wins the championship in just five games is \(0.3^5\). The probability that it takes just five games to obtain a champion is \(0.3^5 + 0.2^5\). To simulate the conditions for A winning the championship with half-points for a draw, assign a range of values to represent each outcome and use a random number generator. The simulation would likely take longer if draws did not contribute to the total score.

Step by step solution

01

Probability that A wins in five games

For A to win the championship in just five games, A must win every single game. Because the games are independent, we simply multiply the probabilities. Thus, the probability is \(0.3^5\).
02

Probability that it takes just five games to obtain a champion

Either A or B could be the champion in five games. So, it could be that A wins all five games or B wins all five games. The probability for this is the sum of the probabilities of these two scenarios, which is \(0.3^5 + 0.2^5\).
03

Performing a simulation for A as champion

To simulate the conditions, you can represent each outcome (A wins, B wins and draw) with numbers (for example 1-3 for A, 4-6 for B and 7-10 for draw) and use a random number generator to simulate games. Whenever there's a draw, you add half a point to each player. This continues until one player reaches or exceeds a total of 5 points, then the simulation ends.
04

Discussing impact of rule change on simulation

If neither player earns any points from a draw, this would likely make the simulation take longer as it would require more games for a player to reach the required 5 points. Draws in this case do not contribute to the total score, hence the condition of reaching 5 points is met in a greater number of games.

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Key Concepts

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

Independent Events
In probability theory, independent events are those whose outcomes do not affect each other. This means the outcome of one event gives no information about the outcome of another. When dealing with successive games in a championship, understanding independence is crucial.
For example, in a series of chess games between players A and B, each game is considered independent if the probability of A winning, B winning, or drawing remains constant, regardless of the outcomes of previous games.
In our chess championship scenario, since the games are independent, we can calculate probabilities of sequences of wins based on multiplication of the individual probabilities.
  • Player A winning: 0.3
  • Player B winning: 0.2
  • Draw: 0.5
The probability of winning or drawing any specific game remains unaffected by past game results, allowing us to apply straightforward calculations to identify possible outcomes.
Simulation
Simulation in probability theory refers to using random sampling to mimic the behavior of a complex system. Simulations can be a helpful tool in estimating probabilities that might be difficult to calculate analytically.
For the given chess championship problem, you can use a simulation to estimate the probability of player A winning through repeated trials of the sequence of games.
Here's a way to set it up:
  • Assign outcomes: Numbers 1 to 3 can indicate A wins, 4 to 6 for B, and 7 to 10 for a draw.
  • Use a random number generator to simulate game outcomes.
  • Track the score—A earns 1 point for a win, B the same, and each gets a half point for a draw (in one scenario).
  • Continue the simulation until one player accumulates the necessary points to win (5 points).
This simulation provides an effective approach to understand outcome distributions and estimate overall probabilities.
Game Theory
Game theory explores the strategic interactions in competitive situations. In situations like a chess championship, understanding these interactions can guide players in making optimal decisions.
In the context of the exercise, game theory would encourage players to evaluate strategies considering all possible outcomes of the games: wins, losses, and draws.
Games between A and B can be thought of as a series of decision nodes where each player can potentially benefit by adapting their strategy according to previous results. However, since the games are independent, past outcomes should not affect future decisions.
  • Players should rely on probabilities to optimize strategies.
  • Staying informed about opponents’ strategies and adapting when needed is key, though here probabilities remain static due to independence.
By using a strategic perspective, players might not increase their chances due to independence, but they ensure consistent play.
Probability Calculation
Probability calculation involves determining the likelihood of various outcomes in a given scenario. It's a fundamental concept in evaluating conditions in a probability theory framework.
In our exercise, calculating the probability that player A wins the championship in exactly five games means A needs to win all five games.
  • This probability is calculated by multiplying the probability of A winning one game (\(0.3\)) by itself five times: \(0.3^5\).
Additionally, finding the probability that a champion emerges in exactly five games (whether A or B) involves summing the probabilities of either scenario:
  • A wins all five games: \(0.3^5\)
  • B wins all five games: \(0.2^5\)
The sum of these probabilities \(0.3^5 + 0.2^5\) gives us the chance of a five-game decision for a champion.
Understanding these probability calculations ensures clear insight into likely outcomes and guides planning or strategy in competitive settings.

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Most popular questions from this chapter

Consider a Venn diagram picturing two events \(A\) and \(B\) that are not disjoint. a. Shade the event \((A \cup B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cap B^{C} .\) How are these two events related? b. Shade the event \((A \cap B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cup B^{C} .\) How are these two events related? (Note: These two relationships together are called DeMorgan's laws.)

A medical research team wishes to evaluate two different treatments for a disease. Subjects are selected two at a time, and then one of the pair is assigned to each of the two treatments. The treatments are applied, and each is either a success (S) or a failure (F). The researchers keep track of the total number of successes for each treatment. They plan to continue the chance experiment until the number of successes for one treatment exceeds the number of successes for the other treatment by 2 . For example, they might observe the results in the table for Exercise \(6.82\) given below. The chance experiment would stop after the sixth pair, because Treatment 1 has 2 more successes than Treatment \(2 .\) The researchers would conclude that Treatment 1 is preferable to Treatment \(2 .\) Suppose that Treatment 1 has a success rate of \(.7\) (that is, \(P(\) success \()=.7\) for Treatment 1\()\) and that Treatment 2 has a success rate of \(.4 .\) Use simulation to estimate the probabilities in Parts (a) and (b). (Hint: Use a pair of random digits to simulate one pair of subjects. Let the first digit represent Treatment 1 and use \(1-7\) as an indication of a success and 8,9 , and 0 to indicate a failure. Let the second digit represent Treatment 2, with 1-4 representing a success. For example, if the two digits selected to represent a pair were 8 and 3 , you would record failure for Treatment 1 and success for Treatment 2\. Continue to select pairs, keeping track of the total number of successes for each treatment. Stop the trial as soon as the number of successes for one treatment exceeds that for the other by \(2 .\) This would complete one trial. Now repeat this whole process until you have results for at least 20 trials [more is better]. Finally, use the simulation results to estimate the desired probabilities.) a. Estimate the probability that more than five pairs must be treated before a conclusion can be reached. (Hint: \(P(\) more than 5\()=1-P(5\) or fewer \() .\) ) b. Estimate the probability that the researchers will incorrectly conclude that Treatment 2 is the better treatment.

A library has five copies of a certain textbook on reserve of which two copies ( 1 and 2 ) are first printings and the other three \((3,4\), and 5\()\) are second printings. \(\mathrm{A}\) student examines these books in random order, stopping only when a second printing has been selected. a. Display the possible outcomes in a tree diagram. b. What outcomes are contained in the event \(A\), that exactly one book is examined before the chance experiment terminates? c. What outcomes are contained in the event \(C\), that the chance experiment terminates with the examination of book 5 ?

Suppose that a six-sided die is "loaded" so that any particular even-numbered face is twice as likely to land face up as any particular odd-numbered face. Consider the chance experiment that consists of rolling this die. a. What are the probabilities of the six simple events? (Hint: Denote these events by \(O_{1}, \ldots, O_{6}\). Then \(P\left(O_{1}\right)=p, P\left(O_{2}\right)=2 p, P\left(O_{3}\right)=p, \ldots, P\left(O_{6}\right)=2 p\) Now use a condition on the sum of these probabilities to determine \(p\).) b. What is the probability that the number showing is an odd number? at most three? c. Now suppose that the die is loaded so that the probability of any particular simple event is proportional to the number showing on the corresponding upturned face; that is, \(P\left(O_{1}\right)=c, P\left(O_{2}\right)=2 c, \ldots\), \(P\left(O_{6}\right)=6 c\). What are the probabilities of the six simple events? Calculate the probabilities of Part (b) for this die.

A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let \(n o t E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(E_{1}\) )? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?

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