# Deck of Cards Probability Question

1. Jan 9, 2010

### mr.physics

1. The problem statement, all variables and given/known data

What is the probability of gettting four-of-a-kind in a thirteen card hand dealt from a standard fifty two card deck?

2. Relevant equations

3. The attempt at a solution

2. Jan 9, 2010

### pbandjay

Are you asking for the probably at least one four of a kind? Or is having multiple four of a kinds in the hand illegal?

3. Jan 10, 2010

### mr.physics

I am asking for at least one four of a kind.

4. Jan 10, 2010

### vela

Staff Emeritus
I got about 3.43%. What have you tried so far?

5. Jan 10, 2010

### pbandjay

I got this answer as well.

6. Jan 10, 2010

### mr.physics

I got .0753 = ((13)(52C9))/(52C13) although i suspect that there are some repetitious combinations ie. 4,4,4,4,5,6,7,7,7,7,5,6,2 and 7,7,7,7,5,6,4,4,4,4,5,6,2.

Could you explain how you arrived at your answer?

7. Jan 10, 2010

### vela

Staff Emeritus
Your mistake is in the numerator. After you draw the four of a kind, how many cards are left from which to draw the remaining 9 cards?

8. Jan 10, 2010

### mr.physics

ahh, i see
Thanks!
I'm still not sure how to eliminate the repetitious combinations though...

9. Jan 10, 2010

### vela

Staff Emeritus
You haven't eliminated repetitions like getting two or three sets of four of a kind, but that's okay because you said you wanted the probability of at least one four of a kind, not exactly one four of a kind.

10. Jan 10, 2010

### mr.physics

"You haven't eliminated repetitions like getting two or three sets of four of a kind"

In calculating the combinations for any particular four-of-a-kind, for example for the combinations of the four-of-a-kinds for cards 4 and 7, there will be one combination that is counted twice: 4,4,4,4,5,6,7,7,7,7,5,6,2 and 7,7,7,7,5,6,4,4,4,4,5,6,2.

11. Jan 10, 2010

### vela

Staff Emeritus
Yeah, you're right. I'll have to ponder this a bit.

12. Jan 10, 2010

### vela

Staff Emeritus
This might help:

$$\{\texttt{hands with three four-of-a-kinds}\} \subset \{\texttt{hands with at least two four-of-a-kinds}\}$$
$$\subset \{\texttt{hands with at least one four-of-a-kind}\}$$

Last edited: Jan 10, 2010
13. Jan 10, 2010

### LCKurtz

I agree, but remember probabilities are numbers between 0 and 1, not percentages.

So you mean .034 approximately.