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#1
07-15-2005, 06:03 PM
 yellowjack Senior Member Join Date: Nov 2004 Location: Vancouver, Canada Posts: 263
An Example from my Probability Class I Need Help w/

A deck of 52 cards is divided into two equal sets randomly. What is the probability that each set contains 13 red cards?

Answer:
You choose 13 of 26 for the first set, and 13 of the remaining 13 for the second set.

26_C_13 * 13_C_13 = # of ways the deck can be split with 13 red cards for each set

52_C_26 = total # of ways that 52 cards can be split into 2 equal sets

26_C_13 * 13_C_13 = 26_C_13, therefore

(26_C_13) / (52_C_26) = answer

---

For the first part of this question, are we completely ignoring the black cards here and just finding out how many ways we can pick exactly 13 red cards from the 26 total red cards for one set?

Also for the 52 choose 26, this is total number of ways 26 cards (red and black) grouped together can be chosen for one set of the 52 cards in half. Is there I can do (i.e. double-checking w/ other possible calculations) to assure myself that this value will make sense with the 26_C_13 bit, or should it come as intuitional?

Thanks.
#2
07-15-2005, 06:36 PM
 llamaoo7 Junior Member Join Date: Nov 2004 Posts: 6
Re: An Example from my Probability Class I Need Help w/

[ QUOTE ]
A deck of 52 cards is divided into two equal sets randomly. What is the probability that each set contains 13 red cards?
26_C_13 * 13_C_13 = # of ways the deck can be split with 13 red cards for each set

[/ QUOTE ]

This is the probability if you only have a stack of 26 red cards to divde up. But what about the black cards? Your answer gives the number of ways for 13 red cards in each stack but you need to multiply by another 26_C_3, I believe. Seems like you might need to divide by 2 incase each stack isn't unique.
#3
07-16-2005, 02:25 AM
 emp1346 Junior Member Join Date: Mar 2004 Posts: 1
Re: An Example from my Probability Class I Need Help w/

I think that yellowjack's original post is close to correct...

Firstly, 52c26 should be the number of ways a deck can be cut in half, and doesn't need to be doubled as this would redundantly count combinations...

Secondly, 26c13 should be the number of ways that the red cards can be divided up... However, I think you also need to consider that the black cards aren't necessarily divided up the same every time, which would bring in another 26c13...

Ultimately, the answer would then be:

26c13*26c13
----------- = .2181
52c26
#4
07-16-2005, 09:52 AM
 irchans Senior Member Join Date: Sep 2002 Posts: 157
Comparison to flipping a coin 26 times

I agree with emp1346's answer. I thought it might be interesting to compare this question to a similar coin toss probability question.

The probability of getting n red cards when drawing 2 n cards from a size 4 n deck is

C(2*n, n)^2/C(4*n, 2*n) ~= Sqrt(2/Pi/n).

The probability of getting n heads when flipping a coin 2 n times is

C(2*n, n)/2^n ~= Sqrt(1/Pi/n).

So the probability of getting n heads is about the probability of getting n red cards divided by Sqrt(2). Hmmm.
#5
07-16-2005, 04:44 PM
 yellowjack Senior Member Join Date: Nov 2004 Location: Vancouver, Canada Posts: 263
Re: An Example from my Probability Class I Need Help w/

Thanks for the replies guys. I understand completely now, the prof messed it up.
#6
07-16-2005, 05:32 PM
 yellowjack Senior Member Join Date: Nov 2004 Location: Vancouver, Canada Posts: 263
Re: An Example from my Probability Class I Need Help w/

Oh [censored] I just emailed my probability prof from my poker email account.
#7
07-18-2005, 01:10 AM
 AaronBrown Senior Member Join Date: May 2005 Location: New York Posts: 505
Re: Comparison to flipping a coin 26 times

That makes some sense. Let's say you put the red cards in one pile and the black cards in another. You flip a coin 26 times, if it's heads you take a card from the red pile, if it's tails you take a card from the black pile. In that case, the chance of having 13 red cards is the same as flipping 13 heads in 26 attempts.

But when you shuffle, the probability of getting a red card is the number of remaining red cards divided by the total number of remaining cards. That makes it more likely to get exactly 13 because if you already have a lot of red cards the next card is more likely to be black; and if you don't have many red cards, the next card is likely to be red.

It's interesting that the difference approaches the square root of two. I'm sure there's a good reason for it, but I don't see it immediately.
#8
07-18-2005, 11:42 AM
 bfc Junior Member Join Date: May 2005 Location: Sweden Posts: 21
Re: An Example from my Probability Class I Need Help w/

emp1346's answer is correct. This is an application of the hypergeometric distribution; it's possible your prof may may have wanted you to state that although I would be more impressed by solutions argued from first principles.

One can calculate hypergeometric probabilities in MS Excel
=HYPGEOMDIST(13,26,26,52)

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