# Ross chapter 7 The game of clue

1. Mar 10, 2015

### mattclgn

1. The problem statement, all variables and given/known data.
The game of Clue involves six subjects, 6 weapons and 9 rooms. One of each is randomly chosen and the object of the game is to guess the 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. S, W, R represent suspects weapons, and rooms in the set of three cards given to a specified player. lex x denote the number of solutions that are possible after that player observes their three cards
b) Express X in terms of S,W,R
c) Find E[x]

2. Relevant equations

Not sure n * k / N . The mean of a hypergeometric maybe?

3. The attempt at a solution.
a) is 324 since 6*6*9
b) is X= (6-S, W-6, R-9)...I think.
c) is where it gets tricky. The back of the book said the answer was 199.6. I think it's Hypergeometric. How many players am I working with here? It doesn't say... I mean, if i have 3 of them, so....in the first there are 5*5*8 which is approx above, but that can't be it. is it there are 18 cards and you take three and then do a combination of what they could be and that is probability and you multiply by whatever remains by the combination of whatever it can be?

2. Mar 10, 2015

### RUber

After reading the question a few times, it seems like you are choosing 3 cards, so you have options of SSS, WWW, RRR, ..., SWR.
Each combination has a probability associated with it and a number of solutions based on the set of 3 cards.
Find the expected value as $\sum_{I=1}^n p(X)*X$.

3. Mar 11, 2015

### haruspex

I think the OP is using S, W, R to represent the number of cards of each type, so S+W+R=3.
Isn't X a count of possible solutions, not a vector? Also, I believe you mean 6-W etc.

4. Mar 11, 2015

### RUber

When I worked this out the first time, I failed to account for the prior draw of the real solution from each of the categories and got 198.75.
The total number ways to draw the 3 cards would be 18C3, and the number of ways to draw any combination would be (5CS)*(5CW)*(8CR). Using this, I was able to validate the book's solution you posted above.