All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Red vs Black Cuts (Posted on 2017-06-05)
A deck of 52 cards is shuffled and cut into two halves of 26 cards each. What is the probability that the number of red cards (hearts and diamonds) in the first half is equal to the number black cards (clubs and spades) in the second half?

Two jokers are added to the deck. The deck of 54 cards is shuffled and cut into two halves of 27 cards each. Now what is the probability that the number of red cards in the first half is equal to the number black cards in the second half?

 See The Solution Submitted by Brian Smith No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): Solution | Comment 5 of 8 |
(In reply to re: Solution by Daniel)

First joker can be anywhere.  Second is in one of the 53 possible places of which 27 are in the other half.

Suppose a deck of only 4 cards (r,b,j,j)
By my reasoning jokers opposing p=2/3 (3 possible places, 2 in the other half)
The 4!/2! = 12 possible shuffles
rb/jj
rj/bj*
rj/jb*
br/jj
jr/bj*
jr/jb*
bj/rj*
jb/rj*
jj/rb
bj/jr*
jb/jr*
jj/br
With the 8 outcomes of jokers on opposite sides asterisked.  8/12=2/3.

 Posted by Jer on 2017-06-05 09:51:37

 Search: Search body:
Forums (0)