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There is a deck of 52 card.
1 card fall down. Then we take 2 cards from the rest. These 2 cards are spade.

What is the probablity of that fallen card is spade??

I think answer should be $\frac 14$ ... ??

Stefan4024
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T.J.
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    To see that $\frac14$ can't be the right answer, take the situation to an extreme: After the card falls down, 13 cards are taken form the rest and they all are spades. What is the probability that the fallen card is spade? – Hagen von Eitzen Sep 22 '13 at 15:04

2 Answers2

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We can exclude the two spade cards. Then we'll have $13-2 = 11$ spade cards in $52-2=50$ card deck. So the probability is:

$$\frac{11}{50}$$

Stefan4024
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  • but why ?? card fall down first . then we take the cards.?? – T.J. Sep 22 '13 at 15:04
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    Because we've already excluded 2 spade cards. We don't know the type of the fallen card.

    At the beginning the probability was $\frac 14$, but with the additional information the probability changes.

    Because the fallen card can be any card, think of the problem like this. Return the card back into the deck and the draw a random card. What's the probability on the random card to be spade?

     – Stefan4024 Sep 22 '13 at 15:09
  • Simular to the Monty Hall Problem. – Stefan4024 Sep 22 '13 at 15:10
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    @T.J. Suppose we pull out 12 spade cards. What is the probability then? (Certainly not $\frac{1}{4}$) – Don Larynx Sep 22 '13 at 15:12
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Informally, the fact that we picked $2$ spades makes it less likely that the missing card is a spade. Since it is easy to make a mistake, we do a formal conditional probability calculation.

Let $A$ be the event we picked $2$ spades, and $B$ the event the missing card is a spade. We find $\Pr(B|A)$. We have $$\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}.$$ We calculate the two probabilities on the right.

The event $A$ can happen in two disjoint ways: (i) the missing card is a spade and we drew $2$ spades or (ii) the missing card is a non-spade and we drew $2$ spades.

For (i), the probability the missing card is a spade is $\frac{1}{4}$. Given this has happened, the probability we drew $2$ spades is $\binom{12}{2}/\binom{51}{2}$. Thus the probability of (i) is $$\frac{1}{4}\frac{\binom{12}{2}}{\binom{51}{2}}.$$

The same sort of reasoning shows that the probability of (ii) is $$\frac{3}{4}\frac{\binom{13}{2}}{\binom{51}{2}}.$$

Add. We get $\Pr(A)=\frac{1}{4}(300)\frac{1}{\binom{51}{2}}$. We have already calculated $\Pr(A\cap B)$: it is the probability of (i).

Now we have all the information needed to calculate $\Pr(B|A)$. This simplifies to $\dfrac{11}{50}$.

André Nicolas
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