# Probability problem involving combinations

• Oct 3rd 2010, 08:08 PM
tonyc4l
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 $\displaystyle \frac{4 \binom{13}{4}}{\binom{52}{4}}$ 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 $\displaystyle \frac{52!}{(13!)^4}$, the number of ways to deal 52 cards to 4 people in groups of 13. I don't get where $\displaystyle \binom{52}{4}$ comes from. If anyone has any advice or a hint or something I'd love to hear it.
• Oct 4th 2010, 07:49 AM
Pandevil1990
Your result isn't correct.Take a look here :Hypergeometric distribution - Wikipedia, the free encyclopedia
(from wikipedia)
• Oct 4th 2010, 12:09 PM
Soroban
Hello, tonyc4l!

Quote:

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 $\displaystyle \frac{4 \binom{13}{4}}{\binom{52}{4}}$
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 $\displaystyle \dfrac{52!}{(13!)^4}$
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 $\displaystyle \binom{52}{4}$ 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, $\displaystyle \text{Mr.}\,X.$

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

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

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

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

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

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

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

Since this could happen to any of the four players:

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

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

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

. . which happens to equal [1].

But I too can't see the reasoning behind that answer.
• Oct 4th 2010, 12:18 PM
undefined
Nevermind! I wrote something dumb.
• Oct 4th 2010, 01:01 PM
Plato
I too wish someone could explain the reasoning behind the given answer.

$\displaystyle \dfrac{\binom{49}{9}\frac{(39!)}{(13!)^3(3!)}}{\fr ac{52!}{(13!)^4(4!)}}$ also gives the correct answer.
• Oct 4th 2010, 04:00 PM
awkward
Quote:

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 $\displaystyle \frac{4 \binom{13}{4}}{\binom{52}{4}}$ 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 $\displaystyle \frac{52!}{(13!)^4}$, the number of ways to deal 52 cards to 4 people in groups of 13. I don't get where $\displaystyle \binom{52}{4}$ 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 $\displaystyle \binom{52}{4}$ possible locations of the aces in the deck, all of which we assume are equally likely. Of these, $\displaystyle \binom{13}{4}$ have all the aces among the first 13 cards. So Fred's probability of getting all the aces is
$\displaystyle \frac{\binom{13}{4}}{\binom{52}{4}}$.

Multiply by 4 to obtain the probability that some player, not necessarily Fred, gets all the aces.
• Oct 4th 2010, 04:05 PM
undefined
Quote:

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 $\displaystyle \binom{52}{4}$ possible locations of the aces in the deck, all of which we assume are equally likely. Of these, $\displaystyle \binom{13}{4}$ have all the aces among the first 13 cards. So Fred's probability of getting all the aces is
$\displaystyle \frac{\binom{13}{4}}{\binom{52}{4}}$.

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

Thanks! This problem had been bugging me.
• Oct 5th 2010, 03:21 AM
downthesun01
Can someone explain to me how you guys worked the problem? I used the hypergeometric distribtion and it was really simple.

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

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!
• Oct 5th 2010, 03:31 AM
Plato
Quote:

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