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

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?

• Another approach: Find the probability that the ace of diamonds is in the same hand as the ace of hearts. Then divide by four to get the probability that that hand is your hand. Commented Feb 15 at 4:53
• The first approach wrongly assumes independence. Taken to an extreme, it could suggest the probability all $52$ cards were in your $13$ card hand would be $\left(\frac{1}{4}\right)^{52}>0$. Commented Feb 15 at 14:57

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

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

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,

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

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

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