# Probability that one player has three aces? (Solution check)

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.)

Facts:

• 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:

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

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.