# Math Help - Probability problem involving combinations

1. ## Probability problem involving combinations

Here's the problem statement:

A deck of 52 cards contains 4 aces. If the cards are shuffled and distributed in a random manner to four players so that each player receives 13 cards, what is the probability that all 4 aces will be received by the same player?

The answer is $\frac{4 \binom{13}{4}}{\binom{52}{4}}$ but I'm really having trouble interpreting that answer and seeing how to get it. I understand that it's written as (# of favorable outcomes) / (total # of outcomes) but I would think that the total number of outcomes would be $\frac{52!}{(13!)^4}$, the number of ways to deal 52 cards to 4 people in groups of 13. I don't get where $\binom{52}{4}$ comes from. If anyone has any advice or a hint or something I'd love to hear it.

2. Your result isn't correct.Take a look here :Hypergeometric distribution - Wikipedia, the free encyclopedia
(from wikipedia)

3. Hello, tonyc4l!

A deck of 52 cards contains 4 Aces.
If the cards are delat randomly to four players so that each player receives 13 cards,
what is the probability that all 4 Aces will be received by the same player?

The answer is $\frac{4 \binom{13}{4}}{\binom{52}{4}}$
but I'm really having trouble interpreting that answer and seeing how to get it.

I understand that it's written as (# of favorable outcomes) / (total # of outcomes)
but I would think that the total number of outcomes would be $\dfrac{52!}{(13!)^4}$
the number of ways to deal 52 cards to 4 people in groups of 13.

This is correct, but unnecessary.

I don't get where $\binom{52}{4}$ comes from. .
Neither do I.
If anyone has any advice or a hint or something, I'd love to hear it.

Consider one of the four players, say, $\text{Mr.}\,X.$

What is the probability that he gets all four Aces?
(Note that we don't care what the other players get.)

There are: . ${52\choose13}$ possible hands he could get.

He must get the four Aces: . ${4\choose4} = 1$ way.

And he gets 9 of the other 48 cards: . ${48\choose9}$ ways.

Hence, there are: . $1\!\cdot\!{48\choose9}$ hands which contain the four Aces.

$\displaystyle \text{So: }\; P(\text{X gets 4 Aces}) \;=\;\frac{{48\choose9}}{{52\choose13}} \;=\;\frac{48!}{9!\,39!}\cdot\frac{13!\,39!}{52!} \;=\;\frac{13!\,48!}{9!\,52!}$ .[1]

. . . . . . $=\;\dfrac{13\cdot12\cdot11\cdot10}{52\cdot51\cdot5 0\cdot49} \;=\;\dfrac{11}{4165}$

Since this could happen to any of the four players:

. . $P(\text{someone gets 4 Aces}) \;=\;4\cdot\dfrac{11}{4165} \;=\;\boxed{\dfrac{44}{4165}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

The given answer has: . $\displaystyle \frac{{13\choose4}}{{52\choose4}} \;=\;\frac{13!}{4!\,9!}\cdot\frac{4!\,48!}{52!} \;=\;\frac{13!\,48!}{9!\,52!}$

. . which happens to equal [1].

But I too can't see the reasoning behind that answer.

4. Nevermind! I wrote something dumb.

5. I too wish someone could explain the reasoning behind the given answer.

$\dfrac{\binom{49}{9}\frac{(39!)}{(13!)^3(3!)}}{\fr ac{52!}{(13!)^4(4!)}}$ also gives the correct answer.

6. Originally Posted by tonyc4l
Here's the problem statement:

A deck of 52 cards contains 4 aces. If the cards are shuffled and distributed in a random manner to four players so that each player receives 13 cards, what is the probability that all 4 aces will be received by the same player?

The answer is $\frac{4 \binom{13}{4}}{\binom{52}{4}}$ but I'm really having trouble interpreting that answer and seeing how to get it. I understand that it's written as (# of favorable outcomes) / (total # of outcomes) but I would think that the total number of outcomes would be $\frac{52!}{(13!)^4}$, the number of ways to deal 52 cards to 4 people in groups of 13. I don't get where $\binom{52}{4}$ comes from. If anyone has any advice or a hint or something I'd love to hear it.
I had difficulty seeing the "book solution", but here is, I think, a plausible explanation.

Suppose one of the players is Fred, and let's see if we can find the probability that he gets all 4 aces. For convenience of analysis, let's say the first 13 cards in the deck are dealt to Fred. If we simply distinguish between aces and non-aces, and we consider the aces indistinguishable, then there are $\binom{52}{4}$ possible locations of the aces in the deck, all of which we assume are equally likely. Of these, $\binom{13}{4}$ have all the aces among the first 13 cards. So Fred's probability of getting all the aces is
$\frac{\binom{13}{4}}{\binom{52}{4}}$.

Multiply by 4 to obtain the probability that some player, not necessarily Fred, gets all the aces.

7. Originally Posted by awkward
I had difficulty seeing the "book solution", but here is, I think, a plausible explanation.

Suppose one of the players is Fred, and let's see if we can find the probability that he gets all 4 aces. For convenience of analysis, let's say the first 13 cards in the deck are dealt to Fred. If we simply distinguish between aces and non-aces, and we consider the aces indistinguishable, then there are $\binom{52}{4}$ possible locations of the aces in the deck, all of which we assume are equally likely. Of these, $\binom{13}{4}$ have all the aces among the first 13 cards. So Fred's probability of getting all the aces is
$\frac{\binom{13}{4}}{\binom{52}{4}}$.

Multiply by 4 to obtain the probability that some player, not necessarily Fred, gets all the aces.
Thanks! This problem had been bugging me.

8. Can someone explain to me how you guys worked the problem? I used the hypergeometric distribtion and it was really simple.

$4* \frac{{{4}\choose{4}}{{48}\choose{9}}}{{{52}\choos e{13}}}$

When you work it out into decimal form, gives the same solution as posted in the original post; and I think it's much easier to read and understand.

I'd love to understand how you guys came to your answer though.

Nevermimd. Awkward's post makes sense. I still like the hypergeometric method more though. Thanks!

9. Originally Posted by downthesun01
Can someone explain to me how you guys worked the problem? I used the hypergeometric distribtion and it was really simple.
$4* \frac{{{4}\choose{4}}{{48}\choose{9}}}{{{52}\choos e{13}}}$