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We lost two cards from a deck of $52$ cards. If we extract a card of this deck, what is the probability we get a diamond?

I'm confused here.

Suppose we lost two diamonds, then

$$P(\text{we get a diamond})=\frac{11}{50}$$

but, if we lost two cards of the other suits, we have

$$P(\text{we get a diamond})=\frac{13}{50}$$

if we lost one of the diamonds and one card from the other suits, we have:

$$P(\text{we get a diamond})=\frac{12}{50}$$

but here I'm not sure, can someone help me?

N. F. Taussig
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rcoder
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    Good start. Now use the law of total probability for formal computation. Assuming cards are lost at random, you need probabilities of losing exactly 0, 1, or 2 diamonds. // Same as asking for probability that the third of three cards drawn without replacement is a diamond (where suits of first two draws not revealed). Answ 1/4, by symmetry. – BruceET Apr 23 '18 at 00:37

4 Answers4

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Another approach is to see that by symmetry, the answer must be the same for any of the four suits, since there was no information about the suit of either of the lost cards. Since the probabilities associated with each of the four suits must still sum to one, the probability of drawing a diamond is just $1/4$.

Now, if you had to determine the probability that, for instance, at least one of the first two draws was a diamond, well, that might get a little harder...

Brian Tung
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    I think Brian Tung's answer hit the nail on the head. Just to beat a dead horse, suppose that you lost 51 cards out of the deck, with no info on which cards were lost. What would then be the probability of the remaining card being a diamond. – user2661923 Apr 23 '18 at 01:27
  • succinctly: all suits are created equal :P – qwr Apr 23 '18 at 03:01
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Using the second method in my Comment, and using R as a calculator. There are d = 52*51*50 ways to choose three cards without replacement. The number of ways for the third card to be a Diamond is n = 39*38*13 + 39*13*12 + 13*39*12 + 13*12*11. The ratio n/d is the answer: 1/4.

d = 52*51*50
n = 39*38*13 + 39*13*12 + 13*39*12 + 13*12*11
n/d
## 0.25
BruceET
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Here is an equivalent way to understand Brian’s answer:

Suppose that you will draw three cards from the deck. What is the probability that the third card you draw from the deck will be a diamond?

Theoretical Economist
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Suppose you deal out all the cards, face down, in a line, and then turn over the leftmost card. What's the probability that it is a diamond? $\frac14$, of course.

Now deal out the cards as before, but this time, before you turn over the leftmost card, you take two cards off the right-hand end and, without looking at them, you throw them away. What is the probability now that the leftmost card is a diamond?

MJD
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