math quiz

[eejit101]

Im digging through poker books i own right now, and have discovered your right. Until now ive been basing some of my calculations on adds that arent even right. How the hell am i winning?????

I have wrote a big post-It with the number "169" on it now, THANX TP

[Talking Poker]

No prob.

That number doesnt' really come into play much while playing. But that is the number.

[kmb13592]

It is a matter of permutations and combinations.

Formula: n!/ r!(n-r)!, using this there comes out to be 6 ways to get AA, 6 ways to get KK, and 28 ways to get AK.

This is what I learned in statistics class but in might be wrong.

[MathBabe]

Yes indeed. (I'll ramble on for a while here, since I already spent three hours tutoring statistics today, and it's sometimes hard to stop.) Although I'm not sure how many people on the board know what your n and r were, and how to figure out that ! thingy. (The ! thingy means multiply the number by all the numbers smaller than it - so 3! = 3*2*1 = 6. Numbers get big fast when you do that.)

The formula is for "how many ways can you choose r of n things". So, if there are four kinds of beer on tap, and you're getting one for your wife as well, how many choices of two of the four are there? 4!/2!(4-2)!, that's how many, which is 4*3*2*1/2*1*2*1, which is 6. It's important to note that this formula assumes you can't choose the same thing twice - wrong for my beer example (sure, you can both have Bud Light if you want), but good in Hold'em where you don't want to consider the same card coming up twice.

The most direct way to get to the original question of how many combinations of AA, KK, AK there are with the formula is to consider that you have 8 cards (four Aces and four Kings), and that you need to know how many ways there are to choose two of them. That's 8 choose 2, or 8!/2!(8-2)!, or 8*7/2 = 28. Voila.

MathBabe

[Talking Poker]

Nice post - especially for a "MathBabe." Statistics are fun.