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There's this game called Cluedo. I'm trying to calculate probability of what cards are hidden in envelope just by people showing someone else a card.

Simply There are 6 cards of characters, 6 cards of weapons and 9 cards of rooms. One card of each is hidden in envelope, the goal is to guess the cards. The other cards are split between all players. If 6 people are playing each player has a 3 cards, one of each kind. If less people are playing someone has a extra card. All players guess in their turn a combination of each kind if someone is holding one of the cards, will show it to the one that was asking.

At the beginning the probability to guess a character is 1/5 as i'm holding one. 1/5 for the weapons, 1/8 for the rooms.

If someone would ask a combination ( of cards i'm not holding ), someone else would show a card to that player. There's a 33.33% probability for each item in the combination that it's not in the envelope.

At this point how would i calculate the probability for of the items from that combination that they are in the envelope and probability of items that wasn't in the combination. The probability of these have to be higher now, right?

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  • $\begingroup$ an interesting question with this "game" is, for perfect players, the probability to win in relation with their index in the list ( of the players in order of play ). $\endgroup$ – user354674 Sep 13 '16 at 18:03
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This was an answer to the previous version of the question.

Hint: $P(A|B)=\frac{P(A\cap B)}{P(B)}$. This is called the conditional probability of $A$ given the condition $B$.

Define $A$ to be the event that you can guess correctly.

Define $B$ to be the event that the card with number 4 is not in the envelope.

[I have assumed you want to calculate $P(A|B)$. However, your English is not very clear. If this is not what you want, please edit your question or comment.]

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  • $\begingroup$ Thanks i guess that's what i was looking for. $\endgroup$ – Sugar Sep 13 '16 at 15:32
  • $\begingroup$ I couldn't make it work. I edited the initial question to give more information. $\endgroup$ – Sugar Sep 13 '16 at 17:40

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