# Card Deck Question

• Sep 11th 2010, 08:28 AM
CuriosityCabinet
Card Deck Question
A pack of cards consists of 52 different cards. A malicious dealer changes one of the cards for a second copy of another card in the pack and he then deals the cards to four players, giving thirteen to each. What is the probability that one player has two identical cards?
• Sep 11th 2010, 09:13 AM
Wilmer
12/51
• Sep 11th 2010, 09:20 AM
Unknown008
I think it's like this:

The probability that a particular player gets a particular card is 1/4.

Say, Ace of spades. It can either be with player 1, or player 2, or player 3 or player 4. Hence, 1/4.

Now, Let's say that the identical cards are King of Hearts.

The probability that player 1 gets the card is 1/4.

When the second King of Hearts is dealt, the probability that it ends up with player 1 is again 1/4.

Hence, the probability that player 1 gets identical cards is 1/4 x 1/4 = 1/16

Now, that could happen to any other player, each with the equal chances, hence, the final probability becomes 1/16 x 4 = 1/4.

EDIT: Okay, seems like I was wrong. Sorry, I thought it could be solved like this. (Doh)
• Sep 11th 2010, 09:26 AM
CuriosityCabinet
Quote:

Originally Posted by Wilmer
12/51

Why?
• Sep 11th 2010, 09:41 AM
Wilmer
Somebody WILL get one of the 2 cards...
51 cards left: he gets 12 others; so 12/51

I also tested it running 100,000,000 deals; definitely 12/51.
• Sep 11th 2010, 01:13 PM
Soroban
Hello, CuriosityCabinet!

I agree with Wilmer . . .

Quote:

A pack of cards consists of 52 different cards. A malicious dealer changes
one of the cards for a second copy of another card in the pack
and he then deals the cards to four players, giving thirteen to each.
What is the probability that one player has the two identical cards?

Consider any player and the probability that he gets the two duplicate cards.

There are: . $\displaystyle {52\choose13}$ possible hands he could get.

But he gets the two duplicate cards (1 way) and 11 more cards: ${50\choose11}$ ways.

His probability is: . $\displaystyle \frac{{50\choose11}}{{52\choose13}} \;=\;\frac{\frac{50!}{11!\,39!}}{\frac{52!}{13!\,3 9!}} \;=\;\frac{50!\,13!}{52!\,11!} \;=\;\frac{13\cdot12}{52\cdot51} \;=\;\frac{1}{17}$

Since this could happen to any of the four players,

. . the probability is: . $4 \times \dfrac{1}{17} \;=\;\dfrac{4}{17}$