In the Clue game, A,B,C,D,E,F have to guess how Mr Bobby was murdered by drawing cards all different.
There is
- 9 "pieces";
- 6 "characters";
- 6 "weapons".
We randomly draw a card from each pile, then place them in an envelope. The object of the game is to determine which combination of cards is in the envelope.
18 remaining cards are divided equally between the 6 players; each player: 3 cards
- We haven't divided the cards yet.
(a) Suppose A had to guess the crime now. What would be the probability that his choice be the right one?
I think it'll be 1/(9*6*6) = 1/324 because we have 3 piles: Pile 1 with 9 cards, pile 2 and pile 3 with 6 cards for each and they are all different.
(b) Suppose A will have to guess the crime immediately after seeing his three cards. Will it increase the likelihood of guessing the crime correctly? Explain why or why not.
Yes, now the probability is 1/(8*5*5) = 1/200 and there are less options for the combination in the envelope
- We shuffled the cards, we distributed them
(a) What is the probability that A has Miss Scarlett in his game?
Miss Scarlett is one of the character cards, so the probability is 1/6 because we have 6 characters?
(b) What is the probability that Miss Scarlett is the murder knowing that A has not Miss Scarlett in his game?
Set E1 the event: A doesn't have Miss Scarlett in his game
S the event: Miss Scarlett is the murder. We have to find then P(S|E1) but I don't know where to begin. Please give me some help. Really appreciated!
1a)
You are correct. The probability of guessing the murder correctly before being given any information (beyond the rules of the game) will be
$$\frac{1}{9\times 6\times 6}$$
1b)
Clearly, once a player has looked at his/her cards the player has gained some knowledge as to which of the original possibilities for the murder are no longer possible. For instance, if the player holds the candlestick card the player knows that the murder could not have been done using the candlestick. Having ruled out whatever possibilities that involve the cards the player holds and guessing from the remaining possibilities the player will guess correctly with greater probability than had the player guessed before looking at their cards.
The probability that they guess correctly however will depend on the specific cards drawn and might not necessarily be $\frac{1}{8\times 5\times 5}$ since it is not guaranteed that the player will have drawn exactly one card of each type. Still, even without calculating explicitly we know it will be greater than the probability in the previous answer.
2a)
This is routine conditional probability calculation.
For player A to have drawn Mrs. Scarlet, first, the suspect that actually committed the murder must not have been Mrs. Scarlet. Next, from among the three cards the player draws from those not used in the murder, one of them must be Mrs. Scarlet.
$$\frac{5}{6}\times\frac{3}{18}$$.
2b)
It is possible for the player to hold three murder suspects in his/her hand. If that were the case, we could guess Mrs. Scarlet was the murderer with probability $\frac{1}{3}$. Otherwise with holding zero murder suspects, we could guess Mrs. Scarlet was the murderer with probability $\frac{1}{6}$.
Since we have no information given in the problem beyond knowing that Mrs. Scarlet isn't in our hand, and suspending the disbelief of having this knowledge and this knowledge alone... (e.g. you asked someone else to look at your hand and truthfully tell you whether or not they see Mrs. Scarlet in your hand)... we can approach with Bayes' Theorem.
Letting $E$ be the event that we don't have Mrs. Scarlet in our hand and $S$ the event that Mrs. Scarlet is the murderer, we ask to find $Pr(S\mid E)$. Breaking down by Bayes' Theorem:
$$Pr(S\mid E) = \dfrac{Pr(S\cap E)}{Pr(E)}=\dfrac{Pr(E\mid S)Pr(S)}{Pr(E)}$$
We calculated the opposite of $Pr(E)$ in the previous problem, $Pr(S)$ is obviously $\frac{1}{6}$, and$Pr(E\mid S)$ the probability that we don't hold Mrs. Scarlet in our hand given that Mrs. Scarlet is the true murderer is obviously equal to $1$.
$$\dfrac{\frac{1}{6}}{1-\frac{5}{6}\times\frac{3}{18}} = \dfrac{6}{31}$$