An applied pobability theory problem

[IgorSmiles]

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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?

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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

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Thanks Che. Actually, that is a perfect explanation. Which leads to the next question, let's say we get an inferior hand to call, how do you perfectly play the turn and river? Afterall, if your check raise is called, how do you not put the guy on a Jack? I guess you can make some kind of value bet and pray but if he senses weakness a good player may try and take you off your hand.

[A_PLUS]

I think the concept Finch was trying to get across here was the following (or maybe its the concept I want to see people discuss, I dont remember).
Also, I know this belongs in the probability thread, but its good to hear less technical views as well. I have had a lot of higher level math and some of those guys are WAY over my head.

You are the SB, and it is folded to you. Is the quality of hand you face from the BB changed by the fact that there are 8 players that folded vs. a standard heads up match?

In this case, I would argue yes, b/c we are looking at a specific card (J). But in general, I would say probably not. Here is why (just my gut, no math done):

The fact that you recieve two independent cards, changes things alot.

Lets say that instead of being dealt two cards, you were dealt one card, that said "AJo, 84s, 66, etc" and you were dealt these cards with the same probability that you would be dealt the combo from a regular deck. If that was the case, the fact that 8 other players folded, would have a big impact on the BB hand. For example, if you are dealt ATo, you are only worried about AA-TT, Ak-AJ. You know that everyone will play these hands when dealt them, and that there is an x% chance that one or more of the 9 players will be dealt such a hand. This is where you must use some Bayesian stats and calulate the probability that the BB has one of those hands give 8 players didnt.


BUT....
I think things change when you play with two cards. Assuming that players will not play A6, or K8, etc. Having 8 players fold will not have the same effect in this case. B/c many of the cards that make up a playable hand (99 is a playable hand, but someone will fold 97) will be folded.

If I remember correctly, someone did some analysis on this and found that in hold em, b/c of that fact no 'clustering' occure. In games like Razz, clustering is big. It is even prevelant in 08, b/c of the value of suited Aces, etc.

thoughts?

[Che]

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Which leads to the next question, let's say we get an inferior hand to call, how do you perfectly play the turn and river?

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I can't, most of the time, in this situation.

Perhaps someone could play this perfectly in a live game, but consistently playing this particular hand perfectly online is pretty much impossible IMHO.

So, having inferior hands fold is not what you want theoretically, but it's about the only way to insure you don't make any errors on the hand.

Later,
Che

[IgorSmiles]

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. They obviously do not have a J. so there are 2 out of 41 cards that the button is holding that may be a J. In the second scenario, there are 2 out of 47 cards that the BB is holding that may be a J.

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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.

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Your logic seems inherantly flawed. Just because you wouldnt limp in with J/4 doesnt mean an opponent wouldnt (or hasnt). The more players to a flop, the more likely one of them holds a Jack. This is so elementary. I think you're over analyzing.

[pfkaok]

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Your logic seems inherantly flawed. Just because you wouldnt limp in with J/4 doesnt mean an opponent wouldnt (or hasnt).

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Well, thats why I said that of course, in the real world, it depends on the limping standards of the player on the button.

When you say:

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The more players to a flop, the more likely one of them holds a Jack. This is so elementary. I think you're over analyzing.

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You are considering that all of them still COULD hold a J, but the fact that all except the button have folded means that the only one who could still hold a J is the button. If they hadn't folded, so any of them could have the J, then yeah, it would be WAY more likely that at lesat one of them holds a J (or AA, KK, QQ for that matter). But it just seems like since the others have folded, its close enough that a very small change in assumptions could sway the answer to the other way.

For example, If i only took out J2o, J3o, and J4o, it would sway it so that the button is now less slightly likely to have a J than the BB.

perhaps I am overanalyzing, but in reality there will be so many conflicting variables that its usually impossible to find a simple, exact answer in these type of spots.

[A_PLUS]

Ok, I did some calculations. The important factor we have been ignoring is the frequency that the last to act player will bet when he does not have at least one Jack.

Lets assume that your check raise will cause any hand that is does not have a Jack to fold. Simplifying things greatly.

So heads up:
opponents possible hands: 1081 [C(47,2)]
Hands with at least one J: 91 [C(47,2)-C(45,2)]
Hands bet without a J: X=(1081-91)*raise w/o J frequency
***I will assume 50% raise w/o J for now
Hands bet without a J: 495=(1081-91)*.5
Total hands bet= 586=495+91
Odds player has a Jack: 15.5% = 91/586

Now if you take into account that 6 non-jacks have been folded.
opponents possible hands: 820(41,2]
Hands with at least one J: 79 [C(41,2)-C(39,2)]
Hands bet without a J: 370.5=(820-79)*.5
Total hands bet= 449.5=370.5+79
Odds player has a Jack: 17.6% = 79/449.5


The fact that 3 players have folded doesn't change the odds to the point where the value of a raise vs. fold would change much. BUT...I think the probability that a player will bet without a Jack is significantly higher heads up than it is 5 handed. Most good players will fear a check raise (with good reason) more with 4 opponents than a solo opponent.

Personally, I would put my %'s as
80% heads up
10% 5 handed

That would change the % of time the player does have a jack to:
Heads up: apx 10%
5-way: apx 51%

The single most important factor in the hand, is the number of players, but not for the reasons suggested. It matters b/c players will bluff into a solo opponent with a much greater frequency than they will into 4 players.

thoughts?

[AtticusFinch]

Good post AP. Still more reasons for why you're not really "heads up" here, even though there are only two players left.

[pfkaok]

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The single most important factor in the hand, is the number of players, but not for the reasons suggested. It matters b/c players will bluff into a solo opponent with a much greater frequency than they will into 4 players.

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good point.

Most people in BB will probably bet close to 100% of time regardless of hand, but few will bet with a field of 5 as a bluff. Your % estimate is probably pretty reasonable, at least for most players.