Stud starting hand combinations?

[Ralph Wiggum]

How many different hand combinations are possible in Stud where you get 3 cards. I don't care which are door or hole cards, so A[img]/images/graemlins/spade.gif[/img] K[img]/images/graemlins/spade.gif[/img] Q[img]/images/graemlins/spade.gif[/img] is the same as K[img]/images/graemlins/spade.gif[/img] Q[img]/images/graemlins/spade.gif[/img] A[img]/images/graemlins/spade.gif[/img].

[Tom1975]

C(52,3)=22100

[Voltaire]

The way 52 cards can permutate is 52! ("fifty-two factorial") which equals 52*51*50...*2*1--a huge number with 68 digits. If you cared about the order that your starting three cards came in, the formula would be n!/(n-r)! equals 52*51*50*49...*2*1/49*48*47...*2*1, which would factor out to 52*51*50/1 which equals 132,600.

However you don't care about the order that your cards come in (and in general when figuring poker probabilities we do not care about the order). So what you want are three-card "combinations." The formula for figuring combinations is n!/r!(n-r)! Tom1975 is giving you the shorthand "52c3" version which is second nature to mathematicians. (The version for permutations is "52p3".) What the formula means is n things taken r at a time, where in your question the n is 52 and the r is 3.

So we have 52!/3!(52-3)! which factors out to 52*51*50/3*2*1 = 132,600/6 = 22,100 which is the number Tom1975 gave you.

I am (obviously!) not a mathematician and had to learn some stats and probability years ago on my own. So that is why I take the trouble to spell this out for non-mathematicians.

[JayTee]

Also, in case some of you guys didn't know, Google Calculator will solve these problems. Enter "52 choose 3" in the search box.