An applied pobability theory problem

[AtticusFinch]

Consider the following problem:

You hold tt in an unraised pot. The board is jjx. You are considering a check-raise. (Leave aside general notions about the wisdom of this for now.)

Case 1: there are 5 limpers. You're the CO. Everyone checks in front, you check, the button bets, and it's folded to you.

Case 2: You're the only limper -- you complete in the SB. You check, and the BB bets.

Assume that in each case you and your opponent both have enough chips to make the larger pot size in case 1 immaterial. All other things being equal, in which of these two cases is the check-raise more likely to succeed? Or are they essentially identical?

Discuss.

[Che]

Is having everyone fold success?

Or, is having worse hands call (or even raise) success?

[AtticusFinch]

[ QUOTE ]
Is having everyone fold success?

Or, is having worse hands call (or even raise) success?

[/ QUOTE ]

OK, let me make the focus sharper: in which case do you think it's more likely you're beaten?

[Che]

[ QUOTE ]
OK, let me make the focus sharper: in which case do you think it's more likely you're beaten?

[/ QUOTE ]

Against 4 limpers, there is a 31.5% chance of at least one opponent having a J. A pair of x's and slowplayed AA/KK are also possibilities.

SB v BB, there is an 8.4% chance that the BB has a J and the chance of a pair of x's is obviously much less. AA/KK is almost unthinkable.

But even if I didn't know all of the above, just the fact that you're against 4 as opposed to 1 makes it common sense that you're more likely beaten in the first scenario, doesn't it?

Later,
Che

[locutus2002]

For my edification, how did you come up with 31%?

[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

[AtticusFinch]

[ QUOTE ]


But even if I didn't know all of the above, just the fact that you're against 4 as opposed to 1 makes it common sense that you're more likely beaten in the first scenario, doesn't it?


[/ 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>

[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

[MLG]

Che, given that you can never get a better hand to fold in these two situations then getting called or raised by a worse hand has to be the answer to that question because if everybody folds its clearly a bad thing.

[Che]

[ QUOTE ]
Che, given that you can never get a better hand to fold in these two situations then getting called or raised by a worse hand has to be the answer to that question because if everybody folds its clearly a bad thing.

[/ QUOTE ]

I understand what you're saying, MLG, but that doesn't mean that that is what the OP meant. I'm just trying to clarify since some will think taking the pot now is "success."

Later,
Che