# A card game

1. Sep 22, 2004

I got into a debate with some people I know here at university and no consensus was reached.

For this hypothetical situation, there are four players playing cards.

So there are 52 cards, and 4 hands. All the cards are dealt out into four equal piles of 13.

What are the chances, that the SAME PERSON will have both the King of Diamonds, and the Ace of Diamonds. It doesn't have to be one specific person, just what are the chances of any of the four people getting both these two cards in their hand of thirteen.

2. Sep 23, 2004

### poolwin2001

For me the question looks trivial(must be my foolishness)

How can you select 13 cards out of 52?(too easy)
How can you select two specific cards out of possible 52 cards.
dividing both you get the answer(I think)

3. Sep 23, 2004

### mee

1 out of twentysix?

4. Sep 23, 2004

### HallsofIvy

Staff Emeritus
Since all 52 cards are dealt one person must get the ace of diamonds. Since you are only asking the probability that some person gets both the ace and king of diamonds, you are asking for the probability that that same person will get the king of diamonds.

It also doesn't matter in what order the cards were dealt so we may assume that the first card is the ace of diamonds and ask "what is the probability that the king of diamonds is one of the other 12 cards".

It is simple to calculate the probability that the king is not one of those 12 cards. Since there are 51 cards it might possibly be, the probability that the second card dealt is not a king is 50/51, if not the probability that the third card dealt is not a king is 49/51, etc to 39/40. The probability that none of the next 12 cards dealt to the person with the ace of diamonds is the king of diamonds is
(50*49*...*39)/(51*50*...*40)= 39/51. The probability that the same person will be dealt both the ace and king of diamonds is 1- 39/51= 12/51.