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  #1  
Old 03-01-2007, 12:31 AM
Phat Mack Phat Mack is offline
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Default How many callers does Buzz need?

[ QUOTE ]
I just needed four opponents out of the five to chase in order to have favorable odds to bet. Here’s the reasoning:

Since a king flush is going to lose to an ace flush about one time out of three, let’s only count two thirds of the clubs as outs. That would have us counting 6 outs for the club flush. But, since one of those is the ten which pairs the board, let’s only count five club outs.

Note that 8*2/3 is actually 5.33, rather than 5. But 5 is close enough.

Then we also have 3 outs for the Broadway. But since we’ll have to split with another Broadway some of the time, those 3 outs are actually slightly less than 3. But 3 is close enough.

Thus we have a total of about 8 outs. And we have two chances.
Here’s the math set-up for the next step:
(8/44) + (36/44)*(8/43)

Then solving (8/44) + (6.7/44) = (14.7/44) = about one third.

Thus we expect to win about one third of the time and lose two thirds of the time. We lose one chip on each of the two times we lose. Therefore we need to win two chips each time we win.

With two callers, we should have roughly a break even proposition.


[/ QUOTE ]

Buzz is pondering a flop bet in his March article. He starts by saying we need 4 callers, but then ends by saying we need 2. This is what puzzles me: the math says he is 2:1 dog for hitting his hand, but the math is figured for the next two cards. Betting the flop only gets us one card. We may have to pay more to see the river. It's possible that the flop bet will provide us the right price for betting or calling the turn if we miss, but is the flop bet itself correct? We are getting 2:1 for an 8/44 shot. I think needing 4 callers is more correct.
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  #2  
Old 03-01-2007, 03:14 AM
Buzz Buzz is offline
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Default Re: How many callers does Buzz need?

[ QUOTE ]
Buzz is pondering a flop bet in his March article. He starts by saying we need 4 callers, but then ends by saying we need 2.

[/ QUOTE ]Hi Mack - It’s an obvious error. I wrote four and then went on to write two. In truth I was thinking two at the time I did the mental calculations as I played the hand.

I don't know what I was thinking when I typed "four." Just a mental lapse. I caught the error when I proof-read and I thought I had corrected it, but somehow it slipped through. Sorry.

Two. I needed two callers, not four.

[ QUOTE ]
We are getting 2:1 for an 8/44 shot. I think needing 4 callers is more correct.

[/ QUOTE ]But we will have two chances to make the draw. On the 36/44 times we miss, we’ll have an 8/43 shot on the second chance. Combined, it’s (8/44) +(36/44)*(8/43) = 14.7/44.

I didn’t do the math in my head in exactly that way. I thought of the 35/44 as about nine elevenths, which is about 0.8. Then I multiplied that .8 by the same number of outs we have the second time (the river) if we miss the first time (the turn). So the fraction became (8+6.4)/44, or roughly 1/3. (15/45 would be 1/3).

I believe two callers is the correct number (and, rightly or wrongly, that is also what I thought at the time) to justify putting fresh money into the pot. I thought if I made my hand and managed to collect more on the river, that implied amount was not assured, but would be frosting on the cake. (And it turned out that I got some frosting too, from one disgruntled opponent).

At any rate, thanks for the correction. I'll look it over and think about it some more when my head is clearer, but at this point I think two is correct.

Regards,

Buzz
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  #3  
Old 03-01-2007, 09:28 AM
duh duh is offline
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Default Re: How many callers does Buzz need?

[ QUOTE ]
But we will have two chances to make the draw.

[/ QUOTE ]

Yes, but you will have to pay for both of those cards, unless you can count on the turn being checked through. (you're 4:1 on each street)
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  #4  
Old 03-01-2007, 03:46 PM
C Dubya C Dubya is offline
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Default Re: How many callers does Buzz need?

Buzz is going to play the hand until the river anyway. It may cost him more chips but his decision is made.
Therefore, on this bet only, two callers are sufficient and three is a money maker.
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  #5  
Old 03-01-2007, 06:51 PM
Buzz Buzz is offline
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Default Re: How many callers does Buzz need?

[ QUOTE ]
Yes, but you will have to pay for both of those cards, unless you can count on the turn being checked through.

[/ QUOTE ]Duh - You’re right about that.

If two opponents call my one bet on the second betting round and if all three of us also each put in one (big) bet on the third betting round, I’m still getting 2 to 1 odds on the fresh money going into the pot on the next two betting rounds.

(But if two opponents call here and then one drops out on the next betting round, then I’m not). I guess I have to cross my fingers for this betting round and keep them crossed for the turn, the next betting round and the river. (That's how you play Omaha-8, at least figuratively. It's somehow kind of exciting.) [img]/images/graemlins/crazy.gif[/img]

If the board pairs on the turn, then it’s only costing me one bet. If the turn is one of three non-club nines, then it’s ugly. Otherwise I either make the club flush, a Broadway, or pick up a low draw. I was more or less aware of these possibilities, but didn’t completely think them through while I was playing the hand. But, yes, there are some possible disasters here. Omaha-8 is definitely a drawing game and there are always possible disasters when you’re drawing. For example, I could make the nut club flush with the ace of clubs on the turn and then the board could pair on the river with the result that I would lose it all. (Fortunately for me, that didn’t happen).

As I look at the hand and the flop now, I think Hero is probably ahead after the flop against five hands with random cards. Of course my opponents don’t have random cards, but with this particular flop,
J[img]/images/graemlins/club.gif[/img], T[img]/images/graemlins/diamond.gif[/img], 5[img]/images/graemlins/club.gif[/img], I think they might be better off with random cards. At any rate, I think Hero, with
A[img]/images/graemlins/heart.gif[/img], K[img]/images/graemlins/club.gif[/img], 4[img]/images/graemlins/spade.gif[/img], 3[img]/images/graemlins/club.gif[/img], probably has the best hand here, although he doesn’t actually have anything yet. If Hero’s ahead here, even though he doesn’t have anything yet, I think he wants fresh money going into the pot.

[ QUOTE ]
(you're 4:1 on each street).

[/ QUOTE ]I think something like that way when I’m calling, but I don’t think that way when I’m initiating fresh money into the pot. Assuming the board doesn’t pair on the turn, I’m seeing the river with this hand/flop.

Buzz
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  #6  
Old 03-01-2007, 06:59 PM
Buzz Buzz is offline
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Default Re: How many callers does Buzz need?

C Dubya - Thanks. That's pretty close to what I was thinking.

Buzz
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  #7  
Old 03-06-2007, 01:26 PM
Phat Mack Phat Mack is offline
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Default Re: How many callers does Buzz need?

Hey Buzz,

I'm still not comfortable here.

[ QUOTE ]
Thus we have a total of about 8 outs. And we have two chances.
Here’s the math set-up for the next step:
(8/44) + (36/44)*(8/43)

Then solving (8/44) + (6.7/44) = (14.7/44) = about one third.

Thus we expect to win about one third of the time and lose two thirds of the time. We lose one chip on each of the two times we lose. Therefore we need to win two chips each time we win.

[/ QUOTE ]

Let's look at the 36/44 number. The 36 represents the missed clubs. But what if the missed clubs pair the board? If we subtract the 9 non-club board pairers, we get 27/44, which lowers our chances.

On the other hand, 20 of the 36 provide the first leg of runner-runner low. This might be good and it might be bad: good if it gives us extra outs; bad if we are playing for half the pot. The 3 non-club 2s might be worth another 1.5 outs, giving us 9.5/44 + 27/44 * 8/43 = ~.3.

Strangely, however I massage these numbers, I end up ~.3. [img]/images/graemlins/smile.gif[/img]
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  #8  
Old 03-06-2007, 04:46 PM
bobbyi bobbyi is offline
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Default Re: How many callers does Buzz need?

Isn't the main reason to bet the flop so that we fold out the tight players behind us whom hero says are very likely to have A2 so that we win if the backdoor low comes in?
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  #9  
Old 03-06-2007, 08:01 PM
Buzz Buzz is offline
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Default Re: How many callers does Buzz need?

Hi Mack - [ QUOTE ]
Let's look at the 36/44 number. The 36 represents the missed clubs.

[/ QUOTE ]Hmm. Not exactly. Usually after the flop there are 45 cards whose whereabouts are unknown. (And that is actually the case here). Some of those 45 cards are favorable while others are unfavorable. As you know, we call the favorable ones “outs.” I didn’t want to count the ten of clubs (which makes the flush but also pairs the board) one way or the other. I wanted to just consider the other 44 cards and use 44 for the denominator rather than the normal 45. Any respectable mathematician or math teacher might cringe at the idea of taking such liberties with the math - but I was trying to make a fast approximation while playing a hand of poker, and that’s how I was thinking.

I know we can’t just discount the ten of clubs and not count it one way or the other. I know we really need to figure the percentage of time Hero will win with the ten of clubs and add that to the “outs” column, while adding the percentage of time the ten of clubs will lose to the “non-outs” column. These two columns are going to add up to 45 if done properly (because 45 is how many cards Hero cannot see after the flop).

But I have been through the rigorous math for other similar situations and realize that I can take this small libery with the math to get an approximation that is good enough for my purposes. To be more precise, I could have fractionated the ten of clubs, counting part of it as good and part of it as bad. Instead I ignored the ten of clubs and changed the denominator from 45 to 44. Can I do this and still be rigorous about the math? Of course not. Can I do this and get close enough? I think so.

At any rate, rightly or wrongly, I ignored the ten of clubs and changed the denominator from 45 to 44.

Then I had 11 outs. 1, the ace of clubs, for the nut flush, the other 7 clubs for the 2nd nut flush and 3 non-club queens for the Broadway.
44-11 = 33, so that I also had 33 non-outs.

But the 11 outs won’t win 100% of the time. I estimated that collectively they would win about 70% of the time (and would lose 30% of the time)
11*.70 = 8 (close enough)

So I mentally changed the 11 outs to 8 outs. And he added 3 to the 32 non-outs (to make 35 bricks).

So that is where the 8 came from. I didn’t do it exacly that way, but that is about what it amounts to and maybe is more understandable.

[ QUOTE ]
But what if the missed clubs pair the board?

[/ QUOTE ]Only one club pairs the board on the turn (the ten). However, now that you mention it, Hero could make the club flush or Broadway on the turn, then the board could subsequently pair on the river, and Hero could possibly lose to a full house or quads.

At the time, I estimated Hero needed two opponents to call the flop bet to justify initiating fresh money into the pot from an odds standpoint. In making this estimation, I neither directly considered implied pot odds nor reverse implied pot odds. In other words, if I made the flush on the turn, I would presumably be able to collect from somebody on the third and fourth betting rounds - but if I made the flush on the turn, the board could pair on the river, possibly making an opponent a full house or quads. Those two effects probably sort of, although not exactly, cancel each other.

You think Hero should have four opponents to call the bet. I certainly like four better than two, but four is not what I was thinking at the time. In retrospect and to be on the safe side, from the viewpoint of getting proper odds to initiate fresh money into the pot, maybe Hero needed three opponents instead of just two.

There were actually five opponents in the hand when Hero bet the flop. Hero thought he needed just two to call, but as I think more about it now, perhaps should have been thinking he needed three. (I still think four is more than needed). Actually three of the five opponents called the bet, which was fine.

There were also some other excellent reasons to bet the flop, including knocking out the very probable A2XY hand(s) behind me - and just the general good that comes from playing aggressively (but not over-aggressively).

When I wrote the article, I was merely trying to describe what (rightly or wrongly) I was thinking and how I was approximating the odds while playing the hand.

I hope this response makes the matter clearer for you. Keep after me if it doesn’t and maybe I can think of a better explanation.

Take care.

Buzz
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  #10  
Old 03-06-2007, 08:12 PM
Buzz Buzz is offline
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Default Re: How many callers does Buzz need?

Hi Bobby -[ QUOTE ]
Isn't the main reason to bet the flop so that we fold out the tight players behind us whom hero says are very likely to have A2 so that we win if the backdoor low comes in?

[/ QUOTE ]I don't know if I would call it the "main" reason, but it certainly is one of the important considerations.

"Reasons" to bet are complex in my own mind. I think the various pros outweigh the various cons in this particular case.

Buzz
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