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?

[Buzz]

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

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

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West - That formula works, and sometimes I use it. but I wasn’t even thinking of it here. I didn’t explain myself very well. Let me try to make it clear with an example - and I'll do it a different way. If you hold the ace of spades and the aces of hearts, there are 24 spades and hearts left in the deck. You could have AsAhXhXh, AsAhXhXs, or AsAhXsXs - and they all would be playable. In addition AsAhXhXn and AsAhXsXn would also be playable, where Xn means non-suited to either ace, in other words a club or diamond.

So what I want to do is add together
AsAhXhXh + AsAhXhXs + AsAhXsXs + AsAhXhXn + AsAhXsXn.

There’s only one way to get those two particular aces, but there’s also AsAd, AsAc, AhAd, AhAc, and AdAc, six different ways to pair two aces. So I’m going to substitute 6 in place of AsAh in each term. The result is:
6*XhXh + 6*XhXs + 6*XsXs + 6*XhXn + 6*XsXn.

Now XhXh = 12*11/2 = 66. That’s the number of two-card combinations of the other twelve hearts.

XhXs = 12*12 = 144.

XsXs = 66 (same as for hearts)

XhXn = 12*24 = 288 (I’m excluding another ace)

and XsXn = 288 (same as for a heart plus a club or diamond)

Substituting, and putting it all together
6*XhXh + 6*XhXs + 6*XsXs + 6*XhXn + 6*XsXn (copied from above).
Substituting:
6*66 + 6*144 + 6*66 + 6*288 + 6*288 =
6*(66+144+66+288+288) = 6*852 = 5112.

[ QUOTE ]
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.

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Yes!! That’s my two terms 6*66 and 6*66.

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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.

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Yes. You have done it correctly.

You left out the 6*XhXs = 6*12*12 = 864 term, but that’s my fault for not making it clear that if AAXX single suited is playable, then AAXX double suited is also playable.

Entirely my fault for not listing AAXX single suited and double suited separately. By lumping them together, I simply wasn't very careful about the way I presented the math.

In retrospect, I should have presented the math more carefully.

But when I originally did the math, I didn’t expect anybody would ever look at it. I was simply trying to quickly come up with the approximate percentage of starting hands a beginner might play in Omaha-8 as a response to an r.g.p. poster. As I set up the math, in my mind, if a hand with a suited ace was playable, then so was a hand that was double suited. I don't know how long it took me to write the twenty lines or so of math set-ups - probably less than five minutes. As I went down the list, what I had already included was very fresh in my mind, so that I was conscious of trying to write set ups so as not to have duplications.

But I’m not guaranteeing I didn’t make a mistake. Never a guarantee.

This is the third different way I’ve done that same basic problem and gotten the same answer, 5112, each time. If I did it next month, after having forgotten this, I might well do it yet another different way.

I don’t mind doing it. Kind of fun for me.

Years ago, I used to work out with a guy, lifting weights in his garage. He had somehow gotten a book of math thought problems that had answers in the back. After he had finished a set and while I was lifting and he was (in theory) spotting me, he would read a problem aloud and I would solve the problem in my head while I was lifting and tell him the answer. I don’t think I ever got one wrong. They were pretty easy problems from my perspective, but impossibly difficult from his. I think he wanted to stump me, but was also somehow amazed and delighted that I always got the right answer.

At any rate, sorry if I confused you (or anybody else). Might have been worthwhile for you to have gone through the math anyhow - and if you’re a bit like me, maybe it was kind of fun for you too.

Buzz

[West]

Ahhh, ok. If I'd have thought more carefully about the point of the whole calculation, I should have realized that the double suited hands needed to be counted here too.

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At any rate, sorry if I confused you (or anybody else). Might have been worthwhile for you to have gone through the math anyhow - and if you’re a bit like me, maybe it was kind of fun for you too.

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Definitely worthwhile.

[ejp1]

I think that 30 BB is enough for a session.At loose lower limit games if you just stick to strong starting hands that are working and pay attention to position you should be able to show a profit. [img]/images/graemlins/spade.gif[/img]

[yimmy9]

The best game I have found is P/L 25 on Party. very soft easy game.

[Mike Gallo]

I think 30 BB is a good starting bankroll. When I play a session now, I try to leave if I've lost 20 BB, b/c that may be a sign I'm on tilt.

Do you mean buy in? If you mean bankroll you left of a 0.

[Mike Gallo]

Excellent post.

Omaha takes a lof of playing to learn the nuances. And nut hands are completely destroyed on each of the turn and river. And counterfeiting will kill many low hands. I would start at a low stakes just to learn some of these but good play takes time. Good luck and enjoy but beware of the traps.

Great point. I played in a $5-$10 kill game the other night at the Borgata and watched players bleed money away due to these "traps".

I fell into one myself. I got counterfeited on the river and lost the hand. What makes the fact that I lost so bad was the fact that I did not realize I had the losing hand.

[Nick_Foxx]

i agree, i don't care how tight you are, if you have never played omaha before, prepare to take some hellacious beatings

i first learned omaha in a dealer's choice home game about 6-7 yrs ago and let's just say i ended up paying a lot for lessons - i got smoked... and if i recall, we were only playing 1/2

mike

[djcolts]

[ QUOTE ]
I think that 30 BB is enough for a session.At loose lower limit games if you just stick to strong starting hands that are working and pay attention to position you should be able to show a profit. [img]/images/graemlins/spade.gif[/img]

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Not right away. This game is tricky to learn - I've lost 36 BB over the weekend at Party .5/1. Playing "tight" pre-flop isn't enough. Just a warning to others learning to play - it doesn't happen right away - expect to lose some when you first play. Hopefully I'll be able to make a profit on this game someday...

[sammy_g]

[ QUOTE ]
Not right away. This game is tricky to learn - I've lost 36 BB over the weekend at Party .5/1. Playing "tight" pre-flop isn't enough. Just a warning to others learning to play - it doesn't happen right away - expect to lose some when you first play. Hopefully I'll be able to make a profit on this game someday...

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Yeah, I have to agree. When I first started playing, even though I was experienced at other games, I was hemorrhaging money. And I played tighter than most of my opponents before the flop. (I've made some other posts on the forum about this. [img]/images/graemlins/smirk.gif[/img])

Playing tight after the flop is also critical. Draw to the nuts. Today when I play, I'm really looking for a draw to the nut low, a draw to the nut flush, an open-ended or wraparound straight draw, or top set. Most other hands go into the muck.