"choose"- another question from a math illiterate

[LuckYou777]

pretty much any cards mathematics question lists "choose" as an operation... (i.e., 52choose2 or something?)

my question is, how do you do the choose operation? not knowing this seems like a pretty serious mathematical flaw in my card game.

walk me through the steps seen here, taken from a forum about Kings Vs. Aces... what to do on the calculator when you see the "C"?

[ QUOTE ]
There are 6 ways to have AA, so the probability of 1 particular player having it is 6/C(50,2).
Multiply this by 8 players to get 8*6/C(50,2). This alone gets you very close, but it will double count all cases where 2 players have AA, so for the exact answer we have to subtract this off. The probability that 2 specific players have AA is 1/C(50,4) since there is only one way to deal 4 aces to 2 players. Since no more than 2 players can have AA, we can just multiply this by the number of 2 player pairs C(8,2). So the exact answer for the whole thing is:

8*6/C(50,2) - C(8,2)/C(50,4) = 1 in 25.6 = 24.6-to-1.

[/ QUOTE ]

all help appreciated.

monte /images/graemlins/club.gif

[ir0nphist]

Easiest way to do it is to use Excel. "=COMBIN(n,m)"
The actual formula is : C= n! / (n-m)! m!

ie. C(52,5) = 2,598,960

Hope that helps /images/graemlins/smile.gif

[Cobra]

You can get an inexpensive calculator that has the choose function on it. Look for a statistical calculator that has an nCr function key. The formula given in the other post was correct N!/((N-r)!*r!) were ! means factorial, this can also be found on many calculators.

Let me give you two examples and you will see how to figure the answer out if you do not have a calculator with these functions on them.

(52 choose 3) the number of unique flops=
52!/((52-3)!*3!) this simplifies to 52*51*50/1*2*3=22100
You start at 52 and go backwards a total of three terms then divide by 1 and go forward a total of three terms.

(52 choose 5)=52!/(47!*5!) this simplifies to 52*51*50*49*48/1*2*3*4*5 = 2598960 you notice that the top has five terms starting at 52 and the bottom has five terms starting at 1.

Cobra

[dkernler]

Just to add one more comment: the 'C' actually stands for 'combination'. It's like you wrote - you're choosing a number of cards. The key is that order doesn't matter. (Since AK is the same as KA, for example.) If we were picking 5 cards in order, there would be 52*51*50*49*48 ways. (One less card as you go.) Since order doesn't matter, we have to divide by the number of ways of ordering those 5 cards: 5*4*3*2*1. (5 choices for the first card, 4 for the 2nd, etc.)

Hence C(52,5) = (52*51*50*49*48)/(5*4*3*2*1) as others have already mentioned.

Nothing new to add for the calculation - just wanted to maybe give some context for it.