O8 Newbie questions

[West]

Hey Buzz. I've reading some of these threads about concepts such as "flop odds" that you have described so well, and which I think can lead to improvements in my game, and so I want to get a handle on the math regarding possible hands and flops and so forth. I see something in your calculation here for AsAXX that doesn't make sense to me, and I think I figured out why.

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24*23/2 is the number of ways to choose two cards with both of them the same suit as one of the aces.

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Ok, I get where this comes from the combinations formula C(n,r) = n! / r!*(n-r)!

C (24,2) = 24! / 2!*22! = 24*23 / 2

If you want two cards, both of them the same suit as one of your aces, you're starting with a pool of 24 cards (if you have As Ah, there are 24 hearts and spades left.) But isn't the above formula also counting instances where you are first picking a heart, then a spade, and vice versa? Because C (24,2) is the formula for the number of combinations for picking two cards out of a group of 24.

Building on your method, I substituted:

C (12,2) = number of possible ways to pick two cards suited to one of your specific aces = 12*11 / 2 = 66. Multiply this by 2 (since you want to count the possible ways to have two cards suited to either one of your aces) for 132. Multiply this by the 6 possible combinations for aces, and you have 792 combinations of aces along with two cards suited to either one of the aces. Add this to the 6*24*24 = 3456 combinations in which you can select two aces, then one card suited to one of the aces (out of 24 possibilities) and one non ace unsuited to either ace (out of 24 possibilities) and you get a total of 4248.


Actually at first I mistakenly tried to calculate the number of combinations where you could select a card suited to one of your aces, and then a second card of the same suit in this way: 24 possible cards of the same suit as one of your aces, times 11 cards of the same suit as the card you just selected, equals 264 combinations, times the number of combinations of aces (6) = 1584. Add this to 3456 and you get a total of 5040. But then, I tried to get the same total combinations number in the way AKQJ10 was:

4 possible aces
x
3 remaining aces
x
24 cards suited to either ace
x
35 cards that are not an ace, and will not give you a double suited hand. 4x3x24x35=10080. I instantly realized this was twice 5040. I then realized that this method was double counting the number of combinations of aces. I THEN eventually realized that the reason for this is that you can only calculate combinations like this if each group that you are multiplying is mutually exclusive to the others. So for example, Buzz's calculation of the number of ways in which you can choose a pair of aces, and then one card suited to an ace, and one card unsuited to either ace (6*24*24) works because the first group of 24 cards is a completely different set of cards then the second group of 24 cards, as well as the aces. If you try and start by saying:

4 possible aces
x
3 remaining aces

...you are double counting combinations. Ah, then As is counted, but so is As, then Ah. I was making the same mistake in calculating the total of 5040. I *think* the total of 4248 may actually be right. Thoughts anyone?