Assume that a 52-card deck is distributed among 4 players, each with 13.
What is the probability that two specific cards, say the Ace of Hearts and Ace of Diamonds to be in my hand?
I have two approach in solving this, each giving a different answer.
The first approach to consider the sample space to be the position of the two aces among the four players. There is a 1/4 chance that the Ace of hearts to be in your hand and a 1/4 chance that the Ace of diamonds to be in your hand, so the chance that both Aces to be in your hand is 1/16
The second approach is using a combination approach, the sample size being all possibilites of a 13 card hand from 52 cards.
Then, we can calculate as follows:
$\frac{{50}\choose{11}}{{52}\choose{13}} = \frac{1}{17} $
Which one is the correct answer and where did one of the working went wrong?
Best Answer
Here's an alternative approach.
We arrange the $52$ cards in a row. You get the first $13$ cards. The next player gets the next $13$ and so on.
The total number of arranging the cards is $52!$.
For you to get the Ace of Hearts ($A_H$) and the Ace of Diamonds ($A_D$), they must be placed somewhere in the first $13$ positions. That can be done in $13 \times 12$ ways. The remaining $50$ cards can then be arranged in $50!$ ways.
So the required probability would be
$$\frac{13 \times 12 \times 50!}{52!} = \frac{1}{17}$$
Problem with your first approach
While the probability of you getting one specific card is $1/4$, that of getting another specific card when you already have the first in hand isn't $1/4$. Let me explain.
As I just said, the probability of you getting the first card is indeed $1/4$. Each player has $13$ openings which will each be filled with one card. So there is symmetry. Hence the probability of a specific card going to any specific player is $1/4$, regardless of the player.
However, once you have the first specified Ace (say $A_H$), now the situation is no longer symmetrical. You have only $12$ openings left while others have all $13$ left. So the probability of you getting the second specified Ace is $< 1/4$. To be exact, the probability is
$$\frac{12}{51} = \frac{4}{17}$$
[$12+13 \times 3 = 51$]
Now you get the same result,
$$\frac{1}{4} \times \frac{4}{17} = \frac{1}{17}$$