Red / Black Cards. Puzzle

[SheetWise]

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?

[CallMeIshmael]

C(26,13) / ( C(26,13) + C(25,13) ) = 66.66%

[SheetWise]

Well, I was surprised. I know it's not a puzzle, I should have written puzzled ... then surprised. [img]/images/graemlins/ooo.gif[/img]

[slickpoppa]

[ QUOTE ]
C(26,13) / ( C(26,13) + C(25,13) ) = 66.66%

[/ QUOTE ]

I believe that is correct

[JoshuaMayes]

[ 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.

[slickpoppa]

[ 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.

[SheetWise]

Allright then, what was the probability they would be red.

[jason_t]

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.

[SheetWise]

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.

[Homer]

[ 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>