also let me give u the scenario so maybe u can get some info i might have missed like the crucial one above..

when sb donks small...I knew he didn't have a low draw, he loved playing his low draws fast, so I put him on a high hand, not a low one...any high hand leading here cannot be a 1 pair hand with no low draw...AKJT is c/folding...he has to have a high hand that connects two pair or better to lead...since his high hand is not a wheel he doesn't have 45...therefore two of his cards are accounted for and neither is a 4

then i realized that the short stack IP has to have a high hand or he raises the flop...again the only high hand that raises must not be 45 since it's the low...

on his stack he is also folding one pair hands...there is still very few hands he can continue with and if he doesn't have any low, his high hand must be two cards that connect to the flop...neither one can be a 4

the other scenario is they can both have have A4xx or A5xx, but since niether one can have A45xx, it's a wash...

so just to simplify, I'm going to give the cards they can have denoting N as any card not a 4

NNxx
A4xx
A5NN

those are the only hand combinations possible for each player...make sense?

now u see that x denotes any random cards...so a four being dealt to 1 is 4/13...the chance it is paired with exactly a 6 is 1/13, then there is a chance they both hold A5NN but I don't even have to worry about that hand because i'm protected from having my low counterfieted by making that tha high so we can ignore A5NN from our calcs we get that they will only have 46 2.3%

ok that's way higher then i calculated at first but ok that's i'm sure correct

i think u can also do this analysis with hand combo's....if we eliminate 45 we eliminate virtually all the possible hand combinations that make two pair or better the ones left are

A2xx
A3xx
AAxx
22xx
33xx
23xx

that is the key to this hand..we assign them both two pair or better high hands, therefore eliminating 45 eliminates the vast majority of high hand combos and subsequently almost all 46 combos

Really, Unreal, both your opponents could have 45 between them.

This hand is about as interesting as a banana.

but I'm voting 5 stars.

Heyy man I like bananas.

The chance that a player is dealt at least one 4 is 1-(48/52)(47/51)(46/50)(45/49)=
.281 not 4/13. The chance of being dealt at least one 4 and one 6 is even more complicated. Its been a while since I took stat so I could be wrong.

Inspired by your post, I went out to the kitchen and ate a third of a small banana.

Buzz

i'm in the camp that this thread is ridiculous however a suggestion

just use the propokertools sim. if you use the new generic syntax i believe the range you are looking for is (a2,a3,a4,a5,a7,23,27,37)!45
you can also use the count function to see that there are alot 41k hands in this range

I eat 1/2 of a banana Monday through Saturday without fail.

Aside from PF being a moldy piece of gouda, hero seems to be wondering how to get value out of the hand by repping it. Essentially it's almost never going to be a hand to get value from EP in a FR NLO8 game, so it's only "value' is going to be from turning it into a bluff, which is only going to happen starting on the flop.
Only hands like X45K/Qhh are going to be slowplaying the flop, so almost every other 45 hand is going to be raising to get value out of the xxxK/Qhh, A4xx,A5xx,25xx and 35xx that may draw to the wheel. If I somehow found myself in this awful situation, i'd think the only way to play it would be to raise/fold vs. SB and raise/call UTG+2 OTF and B/F any blank turn vs. the SB and bet/call any turn vs. UTG+2.

The importance of the % of times someone here has 46 is about as relevent as considering blockers. What range you get played back at with matters so much more and what % of that range consists of 46(and 45).

And as far as balance goes, I can't think of a viable arguement that would include this as a reasonable hand to include for balance in a FR NLO8 game UTG.

I don't know what math you want.

Are we wondering after the flop but before the turn? If so, if neither of them flopped a wheel, then if either the turn or river was a four or a five, they still could make wheels.

After the flop, but before the turn, there simply is not any way you can get around the possibility at least one of them either already already has a wheel or will make one on the turn or river.

And then again after the turn, but before the river, there simply is not any way you can get around the possibility at least one of them either already already has a wheel or will make one on the river.

Are we only concerned with the possibility one or both of them have a four plus a six or a five plus a six?

If so, if SB didn't flop a wheel, and if two of his cards are not fours, fives, or sixes, then how restrictive do you want his other two cards to be? If we give SB a set of aces, then the odds SB would have 46AA or 56AA, from Hero's perspective is rather low, I think in the neighborhood of 26 or 27 to 1 against, something like that.
(from 1/0.03743316)

Buzz

The flaw in your logic is (1) you cannot know for certain SB has not flopped a wheel and (2) you cannot know for certain SB is not doing about the same thing you, yourself, seem to be doing. SB could have 56**, the same as you, or SB could do you one better with 46**. And the same is true of UTG+2.

That's a possibility.

Same flaw.

That's about what I did myself. (I used "n").

It makes sense, but it's wrong. Either player could have 56NN, the same as you, or either player could do you one better with 46NN.

I don't follow that.

Did you see the way I would construct a chart for an opponent holding one or two fours and one or two sixes? It would be a four line chart.
4466
446*
466*
46**
are the possibilities for the chart. "*" is any card other than a four or a six. And then I'd calculate the number of ways an opponent could hold one of those.

I don't see any way around making a chart like that. I'm not saying there's no other way to do it, but if there is, I don't know what.

Do you want to insist your opponent has an ace? If so, the four-six-ace chart looks like this:
446A
466A
46AA
46A*

For this kind of straightforward probability problem, I always make a chart, covering the various possibilities.

We know where the three cards of the flop plus the four cards of Hero's hand are.

There are 45 missing cards, including four fours.
The probability SB has a four is:
P=1-C(41,4)/C(45,4).
And we solve that.
Turns out P=1-0.67968724=~0.320.
4/13=0.308 is your value.
Close.

• Short explanation
  C(41,4) means choose any four cards from 41 cards. There are different ways to calculate that. One of the easier ways is 41!/4!/37!. "!" means "factoral."
  4!=4*3*2*1=24.
  
  41!/4!/37! quickly simplifies to 41*40*39*38/24
  
  Thus C(45,4) = 41!/4!/37! = 41*40*39*38/24
  
  Similarly, C(45,4) = 45*44*43*42/24
  
  Then dividing the first by the second, 41*40*39*38/45/44/43/42 is the probability of no four. And 1 minus that is the probability Villain has a four.
  
  Anyhow, that's a way to do this type of probability problem. You could probably get a better explanation from a good math teacher.

If we didn't know the whereabouts of the seven cards we can see after the flop, the probability of a four being dealt to anyone would be
P=1-C(48,4)/C(52,4).
And we'd solve that to get P=1-0.71873673=~0.281.

I'm wondering if that is true, and if so, why it would be true. It does not seem the chance of a four being paired with a six would be 1/13, but I don't know what your parameters are. You seem to be grabbing numbers out of the air. It may seem that I'm doing the same. How would you ever know for sure? (I guess you wouldn't).

The probability one of your opponents has a four and a six depends on what your restrictions are for the other two cards. That's not clear to me. But if you want to insist that the other two cards be make a set with the board, considering the cards we can see, the following are the possibilities:

46AA	36	4*3*3
4622	36	4*3*3
4633	36	4*3*3
total	108	total

And if we insist the other two cards make a set with the board and the slots I show as taken by a four and a six can be anything (but something to make trips), then
3*3*44*43/2= 8514

Then 108/8514=0.0127, or about 1%. But if you don't make your requirements for your opponent's hands that restrictive, then the probability someone would incidentally have a four and a six would be higher.

But, my thinking is someone might not need to specifically have a high hand. One of your opponents could be pulling the same sort of stunt you yourself are pulling, and with an even better low hand. 46** (or 4466, or 446*, or 466*).

Brilliant for you to put both of your opponents on sets, since they both had sets this time and nobody slow played a flopped wheel. But I think usually both opponents wouldn't have sets and sometimes one of them will slow play a flopped wheel, or SB will bet half the pot on the second betting round hoping someone behind will figure him to not have the wheel and raise, allowing him (SB) to re-raise all-in.

Buzz

That's correct if you can't see your own hand or any board cards.

If you can see Hero's hand here and the board, I think P=~0.320.

That's still not 4/13... but 4/13=0.308 is kind of close, in the ball park.

I think it's about 6.4% if there are no restrictions on the other cards in the hand. (That's what I calculated).

Buzz

yeah and i forgot that i hold one of the 6s so there are really only 3 6s available

u guys are understanding now thanks..this came to me subconsciously on the spot, but they need 2 pair or better here to continue....so when u look for two pair or better hands, the first one that comes up is 45, but since they don't have the low, they can't have that hand....this restricts all two pair or better hands to having no 4s, so the 4s will be in their hand only randomly and since i have one of the sixes, they will have 46 randomly <2%

they will have a 4 2.81% of the time...that's 9 cards in the deck i can "see", only 3/43 are sixes

so .281x3/43 = ~2%

amirite?


so since they only have 46 randomly, I have the essential nut low right, since the only other low that can break me it's low gives me the high? now my nut outs counterbalance, so how often will i scoop?

ok now if they've connected two to the flop...2 pair or better, that restricts the number of flush draws they could have...but by how much?

The probability one of your opponents has a four and a six depends on what your restrictions are for the other two cards. That's not clear to me. But if you want to insist that the other two cards be make a set with the board, considering the cards we can see, the following are the possibilities:

46AA	36	4*3*3
4622	36	4*3*3
4633	36	4*3*3
total	108	total

And if we insist the other two cards make a set with the board and the slots I show as taken by a four and a six can be anything (but something to make trips), then
3*3*44*43/2= 8514

Then 108/8514=0.0127, or about 1%.'


my restrictions are exactly clear...both 4 card hands have to make 2 pair or better on this board, excluding the wheel, so u need to throw in all two pair combos

i'm pretty sure this is gonna work out to being scooped far less then 1% of the time which as a gambler and poker player is the nut low imho

so whatever the math is given my restrictions is what it is...we calculate a profite when we get all stacks in...i'm sure i'm losing pennies the times they show up with 46

now we calculate the times they have A4, which i think would be about 28%....and give it 3 clean outs against us...they will have A5 less freq then a4 but we are free rolling that hand, so we can add money when we get stacks in vs that hand

I know I'm wasting my breath here but,

IF THE MONEY GOES IN AT ANY TIME IN THIS HAND ANY HAND THAT IS NOT A WHEEL BY MY CALCULATIONS IS GOING TO BE GETTING ZERO CLOSE TO 100% OF THE TIME. WHICH IS WHY IT DOESN'T MATTER ANYWAYS BECAUSE IF NOBODY HAS A WHEEL THE MONEY AINT GOING IN!

lol...that's why i posted the hand...if i raise i say i have the nut low whch in their eyes will be the nut hi so do u raise here or do u just call? why do u think i posted the thread?

we have the nut low here 98% of the time, and we scoop that low as many times as we are scooped by a4...when we get done with the times we scoop it's going to amount to getting scooped on average less then 1%...i want all 3 stacks in, how do i get them?

i mean shoot i ask what people would do if they actually ahd the nut low in the hand...ppl insist i don't as if that really mattered in the first place(which wan't even my question)...i prove that i do...and i'm the stubborn one who doesn't listen?

Do you mean the board after the flop? If so, then we look at all the distributions that would make two pairs or better with this flop. 52-12-4

hand	# ways	method
AA22	9	3*3
AA33	9	3*3
2233	9	3*3
AA23	27	3*3*3
A223	27	3*3*3
A233	27	3*3*3
AA2*	324	3*3*36
AA3*	324	3*3*36
A22*	324	3*3*36
223*	324	3*3*36
A33*	324	3*3*36
233*	324	3*3*36
AA**	1890	3*36*35/2
22**	1890	3*36*35/2
33**	1890	3*36*35/2
A2**	5670	3*3*36*35/2
A3**	5670	3*3*36*35/2
23**	5670	3*3*36*35/2
total	24732	total

Did I get them all? I think so. Now the only ways we can have a four and a five are with the last six in the chart.

hand	# ways	method
AA46	36	3*4*3
2246	36	3*4*3
3346	36	3*4*3
A246	108	3*3*4*3
A346	108	3*3*4*3
2346	108	3*3*4*3
total	432	total

OK. Now 432/24732=~0.0175 or almost 2%.

That's what is involved. But it's moot because your premise is wrong!!! What is to prevent an opponent from doing something similar to what you, yourself, are doing... playing 46** or 56**??? In addition, haven't you ever encountered an opponent who slow plays the flopped nuts for one round???

This is the end of my discussion of this. Hope I helped you. (And if not you, maybe someone else). Good luck to you.

Buzz

yeah it was cool,ty...was hoping u could help me come up with how often i scoop which is drectly correllated to the times they have two better hearts...because they can only have two pair or better my hearts are much stronger...

i don't think it really matters what they actually had tho..if i'm gambling that my assumptions are correct i want to know how much of the pot i own compared to mmy risk...it really doesn't matter what they have...they could play 45 the same way but i'm gambling that they don't...

YOU CAN'T AND IF YOU CAN YOU DON'T!


btw I give my advice in post 6 and again in 41.

ProPokerTools Omaha Hi/Lo Simulation

600,000 trials (Randomized)

board: a237

Hand	Pot equity	Scoops	Wins Hi	Ties Hi	Wins Lo	Ties Lo
5h6hqsks	48.38%	80,100	118,787	2,264	438,103	45,687
(a2,a3,a4,a5,a7,23,27,37)!45	25.79%	18,908	229,154	21,362	53,507	32,857
(a2,a3,a4,a5,a7,23,27,37)!45	25.84%	19,130	229,541	21,431	53,905	32,459

equity sim. FTW if thats not the range you are looking for plug the one you want it.

you scoop ~13% you get scooped ~6.3%

btw see post #58

thanks..if this thread were ridc there wouldn't be so many ppl following it and so many posts tho i most of em are mine...

because the ace of hearts is on board...and they have to have 2 pair or better, and can't have 45 exactly, this reduces the number of flush draws they actually have which makes their hands weaker and they will know it...so my hand is really really strong here and their hands are comparitively weak and they know it

and i don't think it's ridic...most ppl, u included, thought this hand was weak but it's very very close to the nuts....so why is that not interesting...and even more interesting is how to keep ur opponents in the pot and get their sstacks in, which no one is mentioning, and it seems to be because they don't want to admit they were fooled by the hand and wrong about it

and look even further this is why i thought the 50% was way off for other fds...because they have to have two pair or better, the other two cards have to be a heart and then another heart near randomly, because only the combinations including the 3 can be a heart...this shouldn't be too hard to figure out how often another flush draw is there...it's going to be somewhat negligible...remember i 3/4 min any 4 on the river...that will happen 9.7% of the time 4/41

Ok i'll just keep posting coz i'm thinking aobut things and i'm way way off on the number of times we are scooped...I forgot about 24 and 34 combos, they are the same as A4...so we are going to be scooped far more often than i thought...the math is gonna be a little wierd because the set combos will scoop us less frequently, i think about half the time, then the two pair combos, the two pairs will scoop us every time a five hits, and the set combos will scoop us only the times another a23 is part of their random xx cards

I beilive it will work out to about 1.5% or so, i did some rudimentary math to discover how often the two pair combos scoop us, which was pretty easy and came up with 1.05%...the set combos will only scoop us when it also holds a wheel card, so because one of them must necessarily be a four...the set combos will contribute to a scoop 1/2 as often as the two pair combos

therefore

we have the nut low 98%
and we will be scooped 1.5%

this is the top end because i didn't want to go into too much detail on the set combos, it doesn't really contribute that much but it's a good round number

i was pretty far off when i thought of this in game..in fact i didn't even think about the 24 and 34 combos until yesterday!!!

what a cool hand huh????