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?
let there be r red cards in the first half,
then that means there are 26-r red cards in the other half,
then that menas there are 26-(26-r)=r black cards in the scond half
thus the two values always match and the probability is 100%
adding two jokers we can use the same reasoning except now there are instead
27-(26-r)=27-26+r=1+r non red cards (either joker or black). For the number of
black cards to be r that means there has to be exactly 1 joker in the second half
thus the other joker must be in the first half. So in summary, for the two values
to be equal we simply need for there to be a single joker in each pile.
Now there are a total of 54C27 ways of selecting the cards in the first half
52C26 of those result in a single joker in each half. Thus our probability is
52C26/54C27
[52!/(26!*(52-26)!)]/[54!/(27!*(54-27)!)]
(52!*27!*27!)/(54!*26!*26!)
(26*26)/(54*53)
338/1431
or approximately 23.62%
I find it quite interesting that the addition of just 2 cards reduces the odds so significantly
I went into this knowing that the first would be 100% and the second would be less but I was
guessing the second answer would be much higher.
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Posted by Daniel
on 2017-06-05 09:26:17 |
solution
