math quiz
 

As we all know AA and AK and KK is the most raised hands . So how many combinations are there using aces and kings?

I don't understand the question.

Do you mean how many different ways are they to be dealt AA, KK and AK?

Counting all the different suit combinations?

combinations using all aces and kings in the deck.;)

I was surprised to read in A Beautiful Mind (the original biography, that the movie was loosely based on) that probability was invented to analyze poker and other "parlour" games. I had just assumed that the theory came first, I guess!

There are 12 ways of getting AA (four possibilities out of the deck for the first card, times three possibilities for the second card). Same for KK.

For AK, there are 32 ways (eight possibilities for the first card, and four for the second).

So, overall there are only 56 combinations out of a possible 2652, meaning on any given hand we're 46:1 underdogs to get dealt one of these premium hands!

or, there are 220 hands possible, AA, AK, and KK are three of there, so 3 in 220 is easier for me:D

But there isn't a 1 in 220 chance of getting each of them - AK is more likely than AA or KK.

Actually... why 220 hands?

Holy Crap!!! i pulled the question from bluff magazine for feb.
They have the answer of 28.

Mathbabe has a far too superior mind:eek:

More isn't necessarily better. :-) I must have double-counted compared to what they were doing.

I will have to get back to you... it is past midnight here and my brain has turned into a pumpkin!

EDIT: Yeah, I double-counted - I took AsAc to be different than AcAs, for instance. If you don't care which card you get first, then divide all my numbers by 2. If they said 28 out of 1326, that's the same as my 56 out of 2652.

Bingo. It doesn't matter which card is on the left or right (online) or on top or on the bottom (live). It's the two card combo that counts.

Well done, MB.

Um, there is 220 possible starting hands according to what ive been told, but my answer sucks a bit as ive just realised it lists AK and AK suited as only 2 hands, so thats where my 3 comes from, a bit confusing but the 220 system works for me i think. Sorry to confuse people byt his, ignore me from now on, my odds are all fucked up.:D

There are 169 unique starting hands if you don't care about suits (meaning AK=1, AKs=1, AKo=1).

The chances of being dealt AA (or any given PP) are 220:1. Maybe that's what you are thinking of.

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

No prob.

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

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.

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

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