# find the probability of a spade from a five-card deck [closed]

I am trying to solve a simple exercise from a probability book but I have no idea how. Given a five-card deck containing: ace of spades, king of spades, king of hearts, queen of spades, and queen of hearts, there are 2 players and each one has exactly one card. You observe evidence that one player has a picture card (king or queen). What’s the probability that the other player has a spade?

I think at conditional probabilities, but I don't know how to model this events. P(A|B) where A is the event of extracting a spade and B represents event of having a spade from a picture card. I know that if I observe picture card than the probability of being a spade is 1/2.

## closed as unclear what you're asking by JMoravitz, Ethan Bolker, Shailesh, Leucippus, Parcly TaxelMay 24 '18 at 14:25

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• How many cards does each player hold? Did we see the rank of all of the one players' cards, or just that one? – C Monsour May 20 '18 at 19:26
• It matters how the cards are split up among the players. Consider the extreme example of where the youngest player is always given all of the cards. In such a scenario, it is impossible for the older player to hold a spade (or in fact any card at all). Consider the other extreme where for each card a coin is flipped. If heads was flipped the first player gets the card. If tails, then the second player gets the card. Consider a third example where the first player is always dealt the top two cards and the second player gets the remaining three. These all have different final answers – JMoravitz May 20 '18 at 19:38
• each player has exact one card. All that I know is that one of them has a picture card. (it can be spade or not). I have to find probability that the other player has a spade. – user1979704 May 21 '18 at 10:27

either it is a spade face card (Pr = ?) in which case the other card has a Pr of $\frac14$ of being a spade,
or it is a non-spade face card (Pr = ?) in which case the other card has Pr of $\frac24$ of being a spade
• $$Pr(A) = Pr(A|B) * P(B) + Pr(A|C) * P(C)$$ where A is the event of having a spade, B if "open" card is a spade and C if "open" card is not a spade. $$Pr(B) = \frac{1}{2}, Pr(A|B) = \frac{2}{4}, Pr(C) = \frac{1}{2}, Pr(A|C) = \frac{3}{4}$$ thus resulting $$Pr(A) = \frac{2} {8} + \frac{3}{8} = \frac{5}{8}$$. – user1979704 May 22 '18 at 12:37