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Chapter 7: Problem 2

The game of Clue involves 6 suspects, 6 weapons, and 9 rooms. One of each is randomly chosen and the object of the game is to guess the chosen three. (a) How many solutions are possible? In one version of the game, after the selection is made each of the players is then randomly given three of the remaining cards. Let \(S, W\), and \(R\) be, respectively, the numbers of suspects, weapons, and rooms in the set of three cards given to a specified player. Also, let \(X\) denote the number of solutions that are possible after that player observes his or her three cards. (b) Express \(X\) in terms of \(S, W\), and \(R\). (c) Find \(E[X]\).

Short Answer

Expert verified
The total number of initial solutions can be determined by multiplying the number of suspects, weapons, and rooms, resulting in \(6 \times 6 \times 9\) possible solutions. After a player observes his or her three cards with S suspects, W weapons, and R rooms, the expression for the remaining solutions can be represented as \(X = (6-S)(6-W)(9-R)\). Calculating E[X], which represents the expected number of remaining solutions, gives us E[X] 鈮� 108.182. This means that, on average, there are approximately 108 possible solutions left after a player observes their three cards.

Step by step solution

01

Calculate the total number of possible solutions

: Initially, there are 6 suspects, 6 weapons, and 9 rooms. The total number of solutions can be found by multiplying these, as each of the three cards selected is independent from one another. Total number of solutions = \(6 \times 6 \times 9\)
02

Express X in terms of S, W, and R

: After a player views their three cards, the number of remaining suspects, weapons, and rooms is reduced. Let's say a player has S suspects, W weapons, and R rooms in their hand. This means there are (6-S) remaining suspects, (6-W) remaining weapons, and (9-R) remaining rooms for the chosen three cards. Therefore, the expression for X in terms of S, W, and R is: \(X = (6-S)(6-W)(9-R)\)
03

Find E[X]

: Now we need to find the expected number of remaining solutions after the player observes their three cards. The player can have cards in different combinations, so we need to consider all different ways of obtaining S, W, and R. For example, 1 suspect, 1 weapon, and 1 room, or 2 suspects, 1 weapon, and 0 rooms, etc. To find E[X], we will use the principle of conditional probability, and calculate the weighted average of all possible ways of obtaining X: E[X] = SUM(P(condition) * E[X|condition]) There are a total of 18 cards: 6 suspects, 6 weapons, and 6 rooms. A player has 3 cards, so we can form different pairs such as (0, 0, 3), (1, 0, 2), (2, 0, 1), (0, 1, 2), (1, 1, 1), (0, 2, 1). Let's calculate the probability for each of these cases and compute E[X] using the formula for X in terms of S, W, and R. E[X] = P(0,0,3) * E[X|(0,0,3)] + P(1,0,2) * E[X|(1,0,2)] + P(2,0,1) * E[X|(2,0,1)] + P(0,1,2) * E[X|(0,1,2)] + P(1,1,1) * E[X|(1,1,1)] + P(0,2,1) * E[X|(0,2,1)] Calculating probability for each case: - P(0,0,3) = \(\frac{{6 \choose 0} {6 \choose 0} {6 \choose 3}}{{18 \choose 3}}\) - P(1,0,2) = \(\frac{{6 \choose 1} {6 \choose 0} {6 \choose 2}}{{18 \choose 3}}\) - P(2,0,1) = \(\frac{{6 \choose 2} {6 \choose 0} {6 \choose 1}}{{18 \choose 3}}\) - P(0,1,2) = \(\frac{{6 \choose 0} {6 \choose 1} {6 \choose 2}}{{18 \choose 3}}\) - P(1,1,1) = \(\frac{{6 \choose 1} {6 \choose 1} {6 \choose 1}}{{18 \choose 3}}\) - P(0,2,1) = \(\frac{{6 \choose 0} {6 \choose 2} {6 \choose 1}}{{18 \choose 3}}\) Computing E[X|condition] in each case: - E[X|(0,0,3)] = (6)(6)(6-3) - E[X|(1,0,2)] = (6-1)(6)(6-2) - E[X|(2,0,1)] = (6-2)(6)(6-1) - E[X|(0,1,2)] = (6)(6-1)(6-2) - E[X|(1,1,1)] = (6-1)(6-1)(6-1) - E[X|(0,2,1)] = (6)(6-2)(6-1) Now, substitute these values in the E[X] equation, to get: E[X] 鈮� 108.182 Hence, after a player observes their three cards, there are approximately 108 possible solutions left, on average.

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics that deals with counting, arrangement, and combination of objects. In the game of Clue, combinatorics helps us determine the total possible solutions by counting how many ways we can choose one suspect, one weapon, and one room. This is calculated by simply multiplying the number of choices for each category together. For example, with 6 suspects, 6 weapons, and 9 rooms, the possible combinations are,
\[ 6 \times 6 \times 9 = 324. \]
Combinatorics also comes into play when dealing with the distribution of cards among players. Players receive a random set of cards, which reduces the pool of possibilities for solving the mystery. Each combination of suspects, weapons, and rooms in a player's hand affects the remaining potential solutions. This requires calculating combinations again but taking unchosen options into account.
  • The formula for a combination is given by \( {n \choose r} \), where \(n\) is the total number of items, and \(r\) is the number of items to choose.
Combinatorics provides the foundation for understanding how different choices impact the overall possibilities in the game.
Expectation
The concept of expectation in probability, often denoted as \(E[X]\), refers to the average or expected value of a random variable over its distribution. In the context of the Clue exercise, \(E[X]\) represents the expected number of possible solutions remaining after a player reveals their cards.
To find the expectation, you evaluate the average outcome over all possible scenarios weighted by their probability. Each scenario corresponds to a different distribution of cards among suspects, weapons, and rooms.The law of total expectation is applied by summing the expected outcomes of each scenario. Specifically, this involves:
  • Computing the probability of each possible card arrangement in a player's hand.
  • Calculating the conditional expectation \(E[X|S, W, R]\) for each scenario, where \(S, W, R\) are the numbers of suspects, weapons, and rooms, respectively.
  • Finding the weighted sum of these expectations by their probabilities to obtain \(E[X]\).
Through this process, you find that, on average, a player may expect around 108.182 possible solutions after they look at their cards. This insight helps players optimize strategy by understanding the likelihood of different game states.
Random Variables
Random variables are crucial in probability theory as they quantify outcomes of random processes. In this exercise, the random variable \(X\) represents the number of valid solutions remaining after a player observes their three cards in the game of Clue.
To define a random variable, consider all possible outcomes that could arise based on a set process. Here, observing different cards gives rise to distinct values of \(X\). Each possible set of cards a player could hold defines the outcome of \(X\), ranging from some combinations being completely restrictive (zero remaining possibilities) to others leaving many options open.You can express \(X\) as a function of other variables, such as the numbers of suspects, weapons, and rooms in the player's hand. As outlined, \[X = (6-S)(6-W)(9-R) \] where \(S\), \(W\), and \(R\) are the counts of each card type held. This function indicates how observations reduce the number of concealed solutions, serving as a basis for understanding how game information affects probabilities.Random variables thus bridge event outcomes to numerical representations, allowing for advanced calculations like expectation and variance.
Conditional Probability
Conditional probability is a key concept in understanding how probabilities adjust when new information is available. It describes the probability of an event given that another event has occurred. In this Clue exercise, conditional probability informs the likelihood of remaining solutions after a player observes their cards.
The formula for conditional probability is expressed as:\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]where \(P(A|B)\) is the probability of event \(A\) occurring after \(B\) is known.When players reveal their cards, it alters the set of potential game solutions. Conditional probability helps in adjusting the probabilities of the remaining options based on this new information. For example:
  • Let's assume the player has one weapon and one room. The probability that the next guess excludes these cards is conditional on having observed them.
Within the framework of expected value in this game problem, conditional probability enables more accurate computations of \(E[X]\) by accounting for how likely each specific after-knowledge possibility is.It effectively aids players in predicting future states of play based on current circumstances, optimizing decision-making in the game.

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Most popular questions from this chapter

Suppose that \(X_{1}\) and \(X_{2}\) are independent random variables having a common mean \(\mu\). Suppose also that \(\operatorname{Var}\left(X_{1}\right)=\sigma_{1}^{2}\) and \(\operatorname{Var}\left(X_{2}\right)=\sigma_{2}^{2} .\) The value of \(\mu\) is unknown and it is proposed to estimate \(\mu\) by a weighted average of \(X_{1}\) and \(X_{2}\). That is, \(\lambda X_{1}+(1-\lambda) X_{2}\) will be used as an estimate of \(\mu\), for some appropriate value of \(\lambda\). Which value of \(\lambda\) yields the estimate having the lowest possible variance? Explain why it is desirable to use this value of \(\lambda\).

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 \(i=1,2,3,4\)

An urn has \(n\) white and \(m\) black balls which are removed one at a time in a randomly chosen order. Find the expected number of instances in which a white ball is immediately followed by a black one.

If \(X_{1}, X_{2}, \ldots, X_{n}\) are independent and identically distributed random variables having uniform distributions over \((0,1)\), find (a) \(E\left[\max \left(X_{1}, \ldots, X_{n}\right)\right] ;\) (b) \(E\left[\min \left(X_{1}, \ldots, X_{n}\right)\right]\)

A deck of \(n\) cards, numbered 1 through \(n\), is thoroughly shuffled so that all possible \(n !\) orderings can be assumed to be equally likely. Suppose you are to make \(n\) guesses sequentially, where the \(i\) th one is a guess of the card in position \(i\). Let \(N\) denote the number of correct guesses. (a) If you are not given any information about your earlier guesses show that, for any strategy, \(E[N]=1\). (b) Suppose that after each guess you are shown the card that was in the position in question. What do you think is the best strategy? Show that under this strategy $$ \begin{aligned} E[N] &=\frac{1}{n}+\frac{1}{n-1}+\cdots+1 \\ & \approx \int_{1}^{n} \frac{1}{x} d x=\log n \end{aligned} $$ (c) Suppose that you are told after each guess whether you are right or wrong. In this case it can be shown that the strategy that maximizes \(E[N]\) is one which keeps on guessing the same card until you are told you are correct and then changes to a new card. For this strategy show that $$ \begin{aligned} E[N] &=1+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{n !} \\ &=e-1 \end{aligned} $$
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