Probability of two specific cards to be in your hand in a game of bridge

Question

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?

Answer 1

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

$$\frac{13 \times 12 \times 50!}{52!} = \frac{1}{17}$$

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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,

$$\frac{1}{4} \times \frac{4}{17} = \frac{1}{17}$$

Answer 2

If you want to use the "slot" method ($4$ groups of $13$ slots each), since the question says my hand (of $13$),

$Pr = \dfrac{13}{52}\dfrac{12}{51} = \dfrac{1}{17}$

[In your attempt, $\Large\frac14$ was right, just writing it as $\Large\frac{13}{52}$ would have shown where you were going wrong]

And then, of course, there is the "standard" method of just looking at my hand which you already know, only I have written it in the full form as being more transparent

$\dfrac{\binom22\binom{50}{11}}{\binom{52}{13}} =\dfrac1{17}$