Cards probability

Oct 2018
1
0
Israel
I draw 9 cards out of a deck of 52 cards. what's the probability that i draw one of them 4 times (different shape). no returns to the deck.

thanks!
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
I assume you mean what is the probability of getting at least one four-of-a-kind. Here goes:

Choose 4 cards out of the 9: $\dbinom{9}{4}$

Choose 1 card out of 13 (you want the card number): $\dbinom{13}{1}$

Now for the remaining 5 cards, we want five cards from the deck: $\dbinom{48}{5}$

So, if we multiply them out, we get: $\dbinom{9}{4}\dbinom{13}{1}\dbinom{48}{5}$.

This is overcounting. It is possible that in the other five cards, you will also have 4-of-a-kind (so you are counting those possibilities twice). So, you need to subtract off that possibility, and then divide by the total number of possible draws of 9 cards.

Let's count the number of ways that we can get two 4-of-a-kinds:

$$\dbinom{9}{8}\dbinom{13}{2}\dbinom{44}{1}$$

So, that means the probability we get at least one four-of-a-kind will be:

$$\dfrac{\dbinom{9}{4}\dbinom{13}{1}\dbinom{48}{5} - \dbinom{9}{8}\dbinom{13}{2}\dbinom{44}{1}}{\dbinom{52}{9}}$$
 
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SlipEternal

MHF Helper
Nov 2010
3,728
1,571
I assume you mean what is the probability of getting at least one four-of-a-kind. Here goes:

Choose 4 cards out of the 9: $\dbinom{9}{4}$

Choose 1 card out of 13 (you want the card number): $\dbinom{13}{1}$

Now for the remaining 5 cards, we want five cards from the deck: $\dbinom{48}{5}$

So, if we multiply them out, we get: $\dbinom{9}{4}\dbinom{13}{1}\dbinom{48}{5}$.

This is overcounting. It is possible that in the other five cards, you will also have 4-of-a-kind (so you are counting those possibilities twice). So, you need to subtract off that possibility, and then divide by the total number of possible draws of 9 cards.

Let's count the number of ways that we can get two 4-of-a-kinds:

$$\dbinom{9}{8}\dbinom{13}{2}\dbinom{44}{1}$$

So, that means the probability we get at least one four-of-a-kind will be:

$$\dfrac{\dbinom{9}{4}\dbinom{13}{1}\dbinom{48}{5} - \dbinom{9}{8}\dbinom{13}{2}\dbinom{44}{1}}{\dbinom{52}{9}}$$
Oops! This is wrong. Here is the correct answer:

If four cards are the same number, then you can have the following possible distributions.

6 numbers represented: $\dbinom{13}{6}\dbinom{6}{1}\dbinom{4}{1}^5$
(Choose the six numbers, choose one of the numbers to be the four-of-a-kind, for the remaining five cards, choose the suit)

5 numbers represented: $\dbinom{13}{5}\dbinom{5}{3}\dbinom{2}{1}\dbinom{4}{2}\dbinom{4}{1}^3$
(choose the five numbers, choose three of them to be singletons, of the remaining two numbers, choose one of them to be the four-of-a-kind and the other is a pair, choose the two suits of the pair, choose the suit of the three singletons.)

4 numbers represented: $\dbinom{13}{4}\dbinom{4}{2}\dbinom{2}{1}\left( \dbinom{4}{3} \dbinom{4}{1}^2+\dbinom{4}{1}\dbinom{4}{2}^2\right)$
(choose the four numbers, choose two of them. This choice will either be the 4-of-a-kind and 3-of-a-kind or the 4-of-a-kind and the singleton. Choose one of them to be the 4-of-a-kind. Now, in the case where you chose the 3-of-a-kind, choose 3 suits and for the other cards, choose a single suit each. For the case where you choose a 4-of-a-kind and a singleton, choose the suit of the singleton and for the two pairs, choose the suits.)

3 numbers represented: $\dbinom{13}{3}\dbinom{3}{1}\dbinom{4}{1}+\dbinom{13}{3}\dbinom{4}{3}\dbinom{4}{2}$
(choose the three numbers. Either you have 4-of-a-kind twice and a singleton or 4-of-a-kind, 3-of-a-kind, pair)

Then add it up and divide by $\dbinom{52}{9}$:

You get around 0.6%.

Note: If you have two numbers represented among the 9 cards, then you would need at least five cards of the same number. Since there are only four suits in the deck, this is not possible. So, you have at least three numbers represented. And if you have four of one number, and one of five distinct numbers, then you can have a maximum of 6 numbers represented.
 
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