An applied pobability theory problem

[pfkaok]

[ QUOTE ]
With 4 checks in #1, the button would be less likely to bet with so many to act afterwards and for fear of being CR'd (highly likely with a J out there). In 2, the BB will be more likely to be on a steal/semi bluff type bet hoping u fear the board.

[/ QUOTE ]

I think this point probably means that in an actual game, the button might be more likely to have the J, since he usually won't be a bot, and won't bet all that often without the J... however, from a pure numbers standpoint, i still believe that the BB has it more often in the blind war.

[A_PLUS]

[ QUOTE ]
[ QUOTE ]
Sounds a little 'monte hall'ish to me. But the 4 limpers definitely make you worse off.

[/ QUOTE ]

That's the exact analogy I used when explaining this to a friend. Bravo.

[/ QUOTE ]

I try. Im pretty sure people have done some work about "clustering" in hold em. By "clustering" I mean that if there are 10 people in the hand, and it is folded to you in the SB, does the range of possible hands for the BB change. From what I have seen, this is an issue in Omaha, and O8, but not so in hold em. Havent realy looked into it, but would like your view.

[Che]

I'm not going to recalculate everything, but this is how to do the calculations as I explained it to SossMan a while back. Note: In the original case, there were 2 opponents, not 4.

[img]/images/graemlins/grin.gif[/img] [img]/images/graemlins/grin.gif[/img] [img]/images/graemlins/grin.gif[/img]

Here’s the formula for determining if your opponents have a specific card:

(aCx)(d-aCo-x)/(dCo)

a= number of cards available
C is the symbol for “combination”
x= number of the specific cards your opponents have
d= cards still in the deck
o= total number of cards the opponent(s) hold(s)

This is confusing so here’s how you would work through the example from the post that started all this:

2 opponents so opponents hold 4 cards total (o=4)
We can see our 2 cards plus the 3 board cards so there are 47 unseen cards (d=47)
We can see two 10’s so there are two available to our opponents (a=2)
We will start by assuming that the opponents hold exactly one 10 among their four cards so x=1

2 C 1 = 2 = number of ways one 10 can show up
47-2 C 4-1 = 45 C 3 = 14190 = number of ways the other 45 cards can show up 3 at a time
47 C 4 = 178365 = total combinations of the 47 available cards taken 4 at a time

So, odds of opponents having exactly one 10 = 2*14190/178365 = 28280/178365

But, opponents could also be holding both 10’s. x=2 now but other variables are unchanged. Probability that opponents hold exactly two 20’s equals:

(2 C 2)*(45 C 2)/47 C 4 = 1*990/178356=990/178365

So, the total probability that our opponents have at least one 10 equals:

(28280/178365)+(990/178365)=29370/178365=16.5%

If you want to try one with check figures, the probability of one opponent having at least one 10 on this flop is 91/1081 or 8.4%.

Hope this helps.

BTW- This reply is confusing reading for me and I understand it so feel free to let me know if you have questions and I'll try to explain the fuzzy parts a little better.

Later,
Che

[Che]

[ QUOTE ]
<devil's advocate> But in the first scenario, all the other hands have folded, so it's just you and the button left. Doesn't that mean the two are essentially the same? </devil's advocate>

[/ QUOTE ]

I haven't read the other replies, but the two scenarios are very similar.

a. Only one opponent that might have a J.
b. The opponent is more likely to be betting his position than betting a J.

There are slight differences, though:

In the multilimper scenario, there are more non-J dead cards so the button bettor is a little more likely than the BB bettor to have a J.

Also, the BB is a little more likely to bluff the SB headsup than the button is to bluff into multiple players.

So, they're very similar, but not exactly the same. Maybe that's what *essentially the same* means - I don't know.

Later,
Che

[Che]

[ QUOTE ]
Can you guys elaborate a little more here? It seems to me that if you're called, you are likely beat. And if you do get called or raised, particularly in a multiway pot, you'd be in a very tough spot. So taking the pot down now is a success. No?

[/ QUOTE ]

If you check-raise and everyone folds, you had them beat anyway so you had the best hand, but were unable to extract any further value from it. On this board, that's just the way it goes.

Theoretically, though, you would prefer to have inferior hands call your check-raise without proper odds to do so.

So, taking the pot down now is not bad, but being called by an inferior hand with improper odds would be better (assuming you play perfectly on the turn and river).

Ehh. Not a very good explanation, actually, but I can't think of anything better right now. Sorry! [img]/images/graemlins/smirk.gif[/img]

Later,
Che

[AtticusFinch]

[ QUOTE ]
from a pure numbers standpoint, i still believe that the BB has it more often in the blind war.

[/ QUOTE ]

Not to be rude, but this is simply false. Check out edfilan's post in this thread to see why.

Whether the BB is more likely to have a jack when he actually bets, for psychological reasons, etc., is a nother question. But from a strictly probabilistic point of view, the person in scenario 1 is much more likely to have it.

[AtticusFinch]

[ QUOTE ]
From what I have seen, this is an issue in Omaha, and O8, but not so in hold em. Havent realy looked into it, but would like your view.

[/ QUOTE ]

Clustering is obviously much less of an issue in Hold'em just because so few cards are dealt out. But it stands to reason that it makes some difference. 16 cards is still a hefty chunk of the deck, and while they're not all liable to be low, more than the fair share probably are. Thus I think it just isn't +ev to play as wildly in a blind battle as you must in a true heads-up match, even in a side game. Blackjack pros make their entire living off of similarly small clustering edges.

That being said, I don't think it should change your game much. Actually, I'm arguing for the opposite: I'm saying don't make too many extravagant changes in the way you play a hand just because you're "heads up." Stick to your gameplan. Go ahead and bluff on a board full of blanks, but don't get too fancy. It's not the same thing as a real heads-up match.

As for stealing, it's just one more small factor along with the others for why I don't like stealing as much from the BB and the CO: the main ones being that they look too obvious and the blinds start taking it personally. I pay for it in variance BC there's more risk with more people behind, but I think I make it up in better table image.

[pfkaok]

[ QUOTE ]
Not to be rude, but this is simply false. Check out edfilan's post in this thread to see why.


[/ QUOTE ]

well, did you actually read my previous post, in which I plugged in numbers???


[ QUOTE ]



OK... so out of 41 cards, that means that there's 820 possible hand combos for button to have. Lets say he'd limp with AJ, KJ, QJ, JJ, JT, J9s, and J8s (is that reasonable). SO then that gives him 33 possible holdings that contain a J. 33/820 = 4.0%



In the 2nd one, out of 47 cards, he's got 1081 combinations, but he could have ANY J. J2 - JA = 91, which would be 8.4%... i guess you could say that he would probably PF raise from BB with his bigger hands in this group (maybe JTs+), but even still he'd be more likely to have a J than button in other example.

I stand by my answer.

[/ QUOTE ]

I guess if the button is limping with EVERY hand, then you're right, but I think my estimation is reasonable... i'm assuming that he's foldin Jxo here.

[AtticusFinch]

[ QUOTE ]

well, did you actually read my previous post, in which I plugged in numbers???


[/ QUOTE ]

I had read it but didn't realize you were the same person. [img]/images/graemlins/smile.gif[/img] Anyway the points you make are well taken. I'm not so much trying to spell out a specific strategy as to illustrate a simple abstract probability concept. Your thinking is good -- if anything it's just more advanced than the point I'm trying to make.

That being said, with that many limpers, button's hand is probably closer to any two than you seem to think. Especially if they're sooooted. [img]/images/graemlins/wink.gif[/img]

Cheers,
AF

[pfkaok]

[ QUOTE ]
That being said, with that many limpers, button's hand is probably closer to any two than you seem to think. Especially if they're sooooted.


[/ QUOTE ]

that's true, i'm primarily a limit player, so i certainly know how good those soooooted cards look to most players [img]/images/graemlins/grin.gif[/img]

to actually KNOW the answer, you'd need to know the range of hands that the button would limp in that spot. YOu're probably right about how lots of soooooted hands will limp there, but that might be countered by the fact that he'd probably also raise PF with a lot of the stronger J hands to chase out the limpers.

To be honest, when i first read the ?, i thougth that it was sort of a trick ?, and since the limper solution seemed obivous, i just wanted to formulate a logical reason for why it should be the other way around.