There are 20 cards (4 colors with 5 cards types each, e.g. ace).
Each of the 4 players randomly receives 5 cards.

What is the probability that one player has three aces after the drawing?

I came up with a solution, but it produces the wrong result (according to the end results our prof gave us.)


  • There are 4 aces in the deck
  • There are 20 cards in the deck
  • Each of the 4 players gets 5 cards

(# means amount)

#(possible cases) = #(draw 5 cards out of 20)

#(favorable cases) = #(draw 3 aces out of 4 aces) $\cdot$ #(draw 2 other cards out of remaining 16 cards)

Use the hypergeometric distribution formula:

$$P=\frac{\text{#favorable cases}}{\text{#possible cases}} = \frac{\binom{4}{3}\binom{16}{2}}{\binom{20}{5}} = 0.0396...$$

(Wolfram Alpha Link)

The solution our professor gave us says the end result is $0.1238$ though.

What did I do wrong? Did I forget to take anything into account?

I chose to do this exercise first, since the other probabilities to calculate seem even more complicated:

a) Each player has one ace.
b) Exactly one player has exactly two aces.
c) At least one player has exactly two aces.
d) (This one) One player has three aces.


Hint: Each individual player has the probability you calculated for getting three aces. You did that correctly. But the question stated that out of four players any one of them could get three aces. The probability of that is a bit higher than for the individual player. Does that get you on the right track?

  • $\begingroup$ Oh, if I just multiply the individual probability times 4 (players), I get the desired end result. I was kind of aware that I never took the four players into account but I couldn't figure out how to properly do that. Thanks! $\endgroup$ – LoLei Oct 23 '15 at 13:52
  • $\begingroup$ Could you give me a tip on how to do "exactly" one player has one ace? This means none of the other 3 remaining players has one ace. But how to use that... I need this for b). $\endgroup$ – LoLei Oct 23 '15 at 15:32
  • 1
    $\begingroup$ If I understand your question, you want to know how to calculate the probability of three players not having aces. The total probability for each player having no aces is 20 choose 5. The probability of no aces is 16 choose 5. And you have three players with no aces, so raise that to the power 3. $\endgroup$ – kleineg Oct 23 '15 at 16:02
  • $\begingroup$ I want to calculate the probability of three players not having exactly one ace in this example. But I think I already found a solution, thanks for your help! $\endgroup$ – LoLei Oct 23 '15 at 16:41
  • $\begingroup$ Glad I could help. $\endgroup$ – kleineg Oct 23 '15 at 17:20

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