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I'm having trouble with a probability problem and would love for an explanation. The problem is as follows:

Two cards from an ordinary deck of 52 cards are missing. What is the probability that a random card drawn from the deck is a spade?

For whatever reason I can't wrap my head around the change in probability from the random removal—best I've got is 52-choose-50.

mattpolicastro
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1 Answers1

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So, in this case there are three possibilities you need to consider:

  1. Both removed cards are spades
  2. One removed card is a spade
  3. Neither of the removed cards is a spade

Assuming there is an even chance that any card from the deck is missing, what is the probability of each of the above events occurring?

  1. The probability of both lost cards being spades is $\frac{13}{52}*\frac{12}{51}=\frac{1}{17}$
  2. The probability of one lost card being a spade is $\frac{13}{52}*\frac{39}{51} + \frac{39}{52}*\frac{13}{51} = \frac{13}{34}$
  3. The probability that neither is a spade is $\frac{39}{52}*\frac{38}{51}=\frac{19}{34}$

Now in each scenario, you have 50 cards remaining, but 11 spades remain in the first scenario, 12 in the second scenario and 13 in the third scenario. The probabilities of picking a spade in each scenario respectively is $\frac{11}{50}, \frac{12}{50},\frac{13}{50}$.

So, now all that remains is to apply the law of total probability to consider all 3 cases:

P(Pick a spade) = P(Event 1)*P(Pick spade given event 1) + P(Event 2)*P(Pick spade given event 2) + P(Event 3)*P(Pick spade given event 3)

P(Pick a spade) = $\frac{1}{17} * \frac{11}{50} + \frac{13}{34} * \frac{12}{50} + \frac{19}{34} * \frac{13}{50} = \frac{1}{4}$

Why does this answer make sense?

As David pointed out in his answer, card suits are symmetric which make them easy to deal with. This problem can really be simplified by just thinking of it as, if you were to draw cards from a deck, what is the probability that the 3rd card drawn is a spade? Obviously, since each is suit is identical, there is an equal chance that each suit would be the 3rd card picked. So, the probability is $\frac{1}{4}$

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