#1
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Red / Black Cards. Puzzle
This surprised me.
A single red card is removed from a standard 52 card deck. Then 13 cards are drawn and found to be the same color. What is the probability that they are black? |
#2
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Re: Red / Black Cards. Puzzle
C(26,13) / ( C(26,13) + C(25,13) ) = 66.66%
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#3
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Re: Red / Black Cards. Puzzle
Well, I was surprised. I know it's not a puzzle, I should have written puzzled ... then surprised. [img]/images/graemlins/ooo.gif[/img]
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#4
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Re: Red / Black Cards. Puzzle
[ QUOTE ]
C(26,13) / ( C(26,13) + C(25,13) ) = 66.66% [/ QUOTE ] I believe that is correct |
#5
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Re: Red / Black Cards. Puzzle
[ QUOTE ]
A single red card is removed from a standard 52 card deck. Then 13 cards are drawn and found to be the same color. What is the probability that they are black? [/ QUOTE ] Zero, you just told me they were red. |
#6
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Re: Red / Black Cards. Puzzle
[ QUOTE ]
[ QUOTE ] A single red card is removed from a standard 52 card deck. Then 13 cards are drawn and found to be the same color. What is the probability that they are black? [/ QUOTE ] Zero, you just told me they were red. [/ QUOTE ] I think he means the same color as each other, not necessarily the same color as the first card that was removed. |
#7
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Re: Red / Black Cards. Puzzle
Allright then, what was the probability they would be red.
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#8
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Re: Red / Black Cards. Puzzle
answer: 26c13/(26c13 + 25c13)
Now, 26c13 = 25c13 + 25c12 = 25c13 + 25c13 = 2 * 25c13 since n choose k = (n-1) choose k + (n-1) choose (k-1) and n choose k = n choose (n-k). Thus 26c13/(26c13 + 25c13) = (2 * 25c13)/(3 * 25c13) = 2/3. Therefore the answer is 2/3. This calculation also shows why the answer isn't surprising. There are twice as many ways to choose 13 cards from 26 as there are to choose 13 from 25. |
#9
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Re: Red / Black Cards. Puzzle
I found it surprising -- in the sense that it's counter-intuitive. When I read over it, I had to stop and pause --then look at it again.
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#10
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Re: Red / Black Cards. Puzzle
[ QUOTE ]
This surprised me. A single red card is removed from a standard 52 card deck. Then 13 cards are drawn and found to be the same color. What is the probability that they are black? [/ QUOTE ] <font color="white">I should be able to do this one! Number of ways to get 13 consecutive red = C(25,13) = 5200300 Number of ways to get 13 consecutive black = C(26,13) = 10400600 P(black) = 10400600 / (5200300 + 10400600) = 2/3</font> |
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