1. ## cards

4 players are dealt 13 cards from a 52 card deck. What is the probability that one player gets exactly 3 aces?

2. Originally Posted by billym
4 players are dealt 13 cards from a 52 card deck. What is the probability that one player gets exactly 3 aces?
There are $\frac{52!}{(13!)^4}$ ways deal 13 cards to each of 4 players.

There are $\binom{4}{3}\binom{48}{10}$ to deal one hand with exactly 3 aces.

Can you finish? (Don’t forget the other three players.)

3. I know the answer is 0.041 but i just cant get it...

4. Shouldn't the total number of hands be $\binom{52}{13}$ ?

In which case it would be:

(number of 3 ace hands) / (total number of hands):

$
\binom{4}{3}\binom{48}{10}/\binom{52}{13} = 0.041
$

5. Originally Posted by billym
I know the answer is 0.041 but i just cant get it...
I completely disagree with that answer.
That is the probability that a particular player gets exactly three aces.
The probability that some player gets exactly three aces is four times that number.
You need to show this to whoever set this problem.

6. I guess I worded it wrong. The full question is:

"Consider a single hand of bridge. What is the probability for the hand to have exactly 3 aces?"

Oops.

What would the equation be if the question was the probability of some player getting 3 aces?

7. Originally Posted by billym
I guess I worded it wrong. The full question is:
"Consider a single hand of bridge. What is the probability for the hand to have exactly 3 aces?"
Worded like that, then the text's given answer is correct.