Question 199709
<font face="Garamond" size="+2">


For any 5 card poker hand, the probability is the number of different ways that hand can be dealt divided by the number of possible 5 card hands in a 52 card deck.  Let's deal with the denominator first.


The number of possible 5 card hands dealt from a 52 card deck is given by the number of combinations of 52 things taken 5 at a time.  The number of combinations of *[tex \Large n] things taken *[tex \Large r] at a time is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ _nC_r = \frac{n!}{r!(n - r)!]


So (and you should verify my arithmetic):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ _{52}C_5 = \frac{52!}{5!(52 - 5)!} = 2,598,960]


Now, let's deal with the numerator:


First, the number of ways you can choose which rank your 4 of a kind will be, namely combinations of 13 things taken 1 at a time.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ _{13}C_{1} = \frac{13!}{1!(13 - 1)!} = 13]


(not surprisingly)


Then, the number of ways to draw the other three cards of that rank, namely combinations of 4 things taken 4 at a time.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ _{4}C_4 = \frac{4!}{4!(4 - 4)!} = 1]


Once you have decided on the rank, there's only one way to get the other three cards of that rank.


Finally, the number of ways to draw the last card.  Once the 4 of a kind are selected, there are 48 other cards to choose, so the number of combinations of 48 things taken 1 at a time:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ _{48}C_1 = \frac{48!}{1!(48 - 1)!} = 48]


And the total number of ways is the product of each of these counts, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ _{13}C_1\cdot_4C_4\cdot_{48}C_1 = 13\cdot1\cdot48 = 624]


Giving us the numerator.


Your probability then is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{_{13}C_1\cdot_4C_4\cdot_{48}C_1}{_{52}C_5} = \frac{624}{2,598,960} \approx 0.000240 = 0.0240%]


To put this in perspective, let's say you sat in on a poker game once per week and played 50 hands each time.  You might expect to be dealt 2 or 3 hands with 4 of a kind in 4 years.  Also, if one of the other guys seems to win a big pot with 4 of a kind every week, then someone is dealing off the bottom of the deck.


By the way, if you wanted the probability of getting exactly 4 Aces, for example, then reduce the numerator of the above calculation by a factor of 13.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>