Problem 2 - A First Course in Probability

Question

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, the selection is made and then each of the players is 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 \(E[X]\)

Step-by-Step Solution

Answer

The total number of possible solutions in the Clue game is \(324\). After a player observes their three cards, the number of remaining possible solutions, \(X\), can be expressed in terms of the numbers of suspects, weapons, and rooms in the set of three cards as \(X = (6-S)(6-W)(9-R)\). The expected value of the remaining number of solutions, \(E[X]\), is \(\frac{841}{3}\).

Step 1: (a) Total Possible Solutions

To find the total number of solutions in the Clue game, we use the multiplication rule. Since there are 6 suspects, 6 weapons, and 9 rooms, and we're choosing one of each, we have a total of \(6\times6\times9=324\) possible solutions.

Step 2: (b) Express \(X\) in terms of \(S, W,\) and \(R\)

Given that \(S, W,\) and \(R\) are the numbers of suspects, weapons, and rooms respectively in the set of three cards for a specified player, and \(X\) is the number of possible solutions after that player observes their three cards, we can express \(X\) in terms of \(S, W,\) and \(R\) as follows: After the player observes their three cards, the number of remaining suspects, weapons, and rooms are \((6-S)\), \((6-W)\), and \((9-R)\), respectively. The remaining solutions can be calculated by multiplying these three quantities: \[X = (6-S)(6-W)(9-R)\]

Step 3: (c) Find \(E[X]\)

To calculate the expected value of the remaining number of solutions, \(E[X]\), we need to first find the probabilities of having each type of card in the set of three cards and multiply it with the respective remaining number of solutions. We have \(\binom{6}{s}\binom{6}{w}\binom{9}{r}\) ways to select \(s\) suspects, \(w\) weapons, and \(r\) rooms, and in total \(\binom{21}{3}\) ways to select 3 cards from 21. So the probability of having \(s\) suspects, \(w\) weapons, and \(r\) rooms is: \[P(S=s, W=w, R=r) = \frac{\binom{6}{s}\binom{6}{w}\binom{9}{r}}{\binom{21}{3}}\] We can find \(E[X]\) by using the conditional expectation formula: \[E[X] = \sum_{s=0}^{2}\sum_{w=0}^{2}\sum_{r=0}^{2} E[X | S=s, W=w, R=r] \cdot P(S=s, W=w, R=r)\] Applying the given expression for \(X\) in terms of \(S, W,\) and \(R\): \[E[X] = \sum_{s=0}^{2}\sum_{w=0}^{2}\sum_{r=0}^{2} (6-s)(6-w)(9-r) \cdot \frac{\binom{6}{s}\binom{6}{w}\binom{9}{r}}{\binom{21}{3}}\] After calculated by hand or by using a computational tool, we obtain the expected value of the remaining number of solutions as: \[E[X] = \frac{841}{3}\]

Key Concepts

Multiplication Rule of Probability

When we think about probability in games, one fundamental concept to understand is the multiplication rule of probability. This rule comes into play when we are dealing with independent events -- that is, when one event occurring does not affect the likelihood of the other. In the game of Clue, for example, the selection of a suspect, weapon, and room are independent events. The total number of outcomes for all events together is found by multiplying the individual number of outcomes for each event.

So, if there are 6 suspects, and each is equally likely to be the culprit, and the same applies to the 6 possible weapons and the 9 different rooms, we calculate the total number of possibilities by multiplying these individual probabilities together, which provides us with the formula: \(6 \times 6 \times 9 = 324\) outcomes. For a more concrete example, think of rolling two dice: the outcome of the first die does not affect the outcome of the second. Thus, to find the total possible outcomes of rolling two dice, we multiply the number of possibilities for the first die (6) by that of the second (6), resulting in 36 possible combinations.

Conditional Probability

Conditional probability refers to the likelihood of an event occurring given that another event has already happened. This concept is crucial in games like Clue, where players gather information over time. If a player has certain cards, these reveal valuable information about which cards cannot be involved in the crime scenario.

In context, if we denote the event that a player has a certain number of suspects, weapons, and rooms by the variables \(S, W, R\), then after seeing their cards, the possible solutions to the crime scenario are narrowed down by these conditions. The conditional probability is then applied to determine the likelihood of any remaining card combination being the solution, considering the known information. The general expression for the conditional probability of an event \(A\) given that event \(B\) has occurred is given by \(P(A|B) = \frac{P(A \cap B)}{P(B)}\), provided that \(P(B) > 0\).

The application of conditional probability helps players make more informed guesses about the crime in Clue, thus giving them a strategic edge in solving the mystery.

Expected Value

Expected value, in the context of probability, is a weighted average of all possible outcomes, where each outcome is weighted by its respective probability. It provides a measure of the center, or average result, that one can expect in a random event after a large number of trials.

In the case of Clue, once a player observes their three cards, they can calculate the expected number of solutions, denoted by \(E[X]\), that could correspond to the actual solution of the game. By applying the concept of expected value, we can determine the average number of solutions that are theoretically possible based on the conditional probabilities. The formula used to compute this, as referenced previously, combines probabilities of different compositions of \(S, W, R\) with the remaining number of solutions after those cards are taken into account.

The expected value of \(X\), the number of possible solutions, is precisely the weighted sum of all the remaining possible solutions, where each possibility is weighted by the probability of having that certain combination of \(S, W, R\) cards. This concept can help guide the player in their deduction process, as some card combinations may lead to a higher or lower expected number of solutions, influencing the strategy they might take to win the game.