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 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

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 solution | Comment 7 of 8 |
The first part is similar to where an ounce of water is taken from a quart of water and put into a quart of milk, then an ounce of the mixture is put back into the milk. Is there more water in the milk or vice versa. And of course it's the same.

So part 1: There's 100% probability that there are the same number of red cards in the first half as there are black cards in the second half.

Which leads to the solution of the second part. If each joker is on a separate side, then the jokers don't matter and the reds on one side match in number the blacks on the other side. But if both jokers are on the same side the two numbers don't match; there are more black cards on the side without the jokers than there are red cards on the side with the jokers.

What's the probability that each joker is on a different side? It's 27^2/C(54,2) = 27*27 / (54*53/2) = 27/53, and so that's also the probabilty that there are the same number of black cards on side 2 as red cards on side 1.

 Posted by Charlie on 2017-06-05 10:43:06

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