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logic question

 
 
lovejoy
 
Reply Mon 2 Nov, 2009 12:19 pm


A jokerless deck of 52 cards has been shuffled thoroughly. Take the cards and divide them into two stacks of 26 cards each, labelled A and B

If you divide the cards as described above and check the contents of each pile 1000 times, how many times could you expect the number of red cards in one pile to match the number of black cards in the other?
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Type: Question • Score: 0 • Views: 1,867 • Replies: 8
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engineer
 
  1  
Reply Mon 2 Nov, 2009 12:47 pm
@lovejoy,
The number of red cards in pile A will always match the number of black cards in pile B.

Let X equal the number of red cards in pile A
The number of black cards in pile A is (26-X)
The number of remaining red cards for pile B is (26-X)
The number of black cards in pile B is 26 - (26-X) = X
chai2
 
  1  
Reply Mon 2 Nov, 2009 12:53 pm
@engineer,
I'm glad the answer is the same as I got, as I didn't use any formula to figure it.

It just seemed immediately seemed the only way it would work

Criss cross.....Reds in one pile must mean equal black in the other.

pile # 1
1 red card, 25 black
pile #2 must have
25 red and 1 black.
engineer
 
  1  
Reply Mon 2 Nov, 2009 01:16 pm
@chai2,
OK, here is a more fun one. Given the same situation, what is the probability that both piles are 13 red, 13 black?
chai2
 
  1  
Reply Mon 2 Nov, 2009 02:05 pm
@engineer,
engineer wrote:

OK, here is a more fun one. Given the same situation, what is the probability that both piles are 13 red, 13 black?



really low?

Laughing


no really, I'll think about that and get back to you.
acutally, I'll not think about that. answers to stuff like this seem to come to me when I'm not really thinking about them.
0 Replies
 
joefromchicago
 
  1  
Reply Mon 2 Nov, 2009 02:15 pm
@engineer,
1/2^13 or 8192 to 1
engineer
 
  1  
Reply Mon 2 Nov, 2009 02:23 pm
@joefromchicago,
Nope, surprisingly high! So high, I had to write a quick program to test the results to convince myself my math is correct.

The number of possible half decks with 13 red, 13 black is calculated by finding the possible combinations of red cards times the possible combinations of black cards.

( 26! / 13!^2 ) ^2

The number of total possible half deck combinations is

52! / 26!^2

The resulting ratio is 21.8%. I ran a 10,000 run simulation and got 2149 actual results, so right in line.
0 Replies
 
ebrown p
 
  1  
Reply Mon 2 Nov, 2009 02:32 pm
@joefromchicago,
You are way off joe.

The answer is clearly less then 26 to 1.

Consider the probability there is zero red cards in my half (i.e. all are black)
Then consider the probability that there are one red card in my half (and the rest are black).
The consider three red cards... and four
Continue that sequence until there are 26 red cards (and no black cards).

Two things are clear

1) The sum of all of these 26 different probabilities is 1.
2) That there are 13 red cards in my half is the most likely outcome.
engineer
 
  1  
Reply Mon 2 Nov, 2009 02:56 pm
@ebrown p,
Here's the entire probability table

Red Cards Percentage
0 0.000%
1 0.000%
2 0.000%
3 0.000%
4 0.000%
5 0.001%
6 0.011%
7 0.087%
8 0.492%
9 1.969%
10 5.689%
11 12.037%
12 18.808%
13 21.813%
14 18.808%
15 12.037%
16 5.689%
17 1.969%
18 0.492%
19 0.087%
20 0.011%
21 0.001%
22 0.000%
23 0.000%
24 0.000%
25 0.000%
26 0.000%
0 Replies
 
 

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