Simple Common Tournament Situation

[sdplayerb]

My assumptions:
For PP, only AA-99 will call (yes, blinds could call a little lower since nobody behind to act for the BB, but not a big factor)
For other, only AK-AJ. (i do think utg+1 wouldn't play with AJ and blinds may call with AT..i'll say it cancels out).

38% chance of getting called by at least one, if called by one a 29% chance of still winning.
It is seldom the hand ends up three handed, so my model doesn't include that.

but it comes out at an EV of +44.

Unless the have seen me pushin a ton of times, i don't hestitate but pushin, and have done so with QJ in an almost identical situation...everybody folded.

[Lurshy]

I am thinking about the math. Now my brain hurts. Thanks.

[West]

Keep in mind that there are no guarantees that it will be folded to you in steal position on this orbit. And even if it is, once the blinds go through you, the BB will be getting 11 to 6 to call you, knowing that you will likely move in with a wide range of hands. With the average stack at 4000, chances are the BB is not going to sweat losing the 600 it will cost him to call in the event that you do get the opportunity to steal. SB too. It may actually be very difficult for you to "steal" the blinds once your stack is at 800.

[CrisBrown]

Hi David,

[ QUOTE ]
At this point it is all math. Yet almost none of you tried to do this math. HOW DARE YOU? What makes you think that this question should be attempted by the seat of your pants? Were you just lazy? Or is it because you do not know how to do this relatively simple problem? If you don't, you better learn now because otherwise you are almost certainly destined to go broke.

[/ QUOTE ]

In all fairness, I don't think it's fair to assume that the respondents aren't considering the math because they didn't write out the calculations. As you said in another thread, these aren't the kind of calculations you perform in the heat of the game. Instead, you extrapolate from situations you've worked through before. And I think at least some of the posters probably did that.

But yes, an answer is better (and more instructive) when the poster explains how that answer was derived.

Cris

[jwvdcw]

I easily fold here. I've had many similar discussion to this in the MTT board. I think a major issue that everyone is forgetting here is this: How good of a shortstack player do you think you are? Are you comfortable dwindling down to only 2-3x BB?

For me, this isn't even an issue...fold.

[CrisBrown]

Hi David,

[ QUOTE ]
Why? Because almost none of the answers attempted to do any math.

[/ QUOTE ]

Pc = Current Pot ($)
By = Bet(s) you will add to pot ($)
Bo = Bet(s) your opponent(s) will add to pot ($)

pF = Probability that opponent(s) will fold if you bet
pW = Probability that your hand wins at showdown

Ef = Fold Equity = (pF x Pc)
Es = Showdown Equity = (pW x (Pc + Bo) – ((1-pW) x By)
En = Net Equity = Ef + ((1-pF) x Es)

If En > 0, you will win money over the long run. If En < 0, you will lose money over the long run. The break-even point is En = 0.

For a given set [Pc, By, Bo, pW], what is the break-even point for pF?

Compute Es = (pW x (Pc + Bo) – ((1-pW) x By)

(pF x Pc) + ((1-pF) x Es) = 0
(pF x Pc) + Es - (pF x Es) = 0
(pF x Pc) – (pF x Es) = -Es
pF (Pc – Es) = -Es
pF = -Es / (Pc – Es)

Example: It is midway through a tournament, and your $2000 stack is down to $1100. The blinds are $100/$200. You are UTG with QTs. You are the best player at your table, but there are no live ones. Should you move all-in with QTs?

Estimates: At most one opponent will call, and you will be a 3:1 underdog if called.

Pc = $300
By = $1100
Bo = $1100
pW = 0.25

Es = (pW x (Pc + Bo) – ((1-pW) x By)
= (0.25 x $1400) – (0.75 x $1100)
= -$475

pF = $475 / ($300 + $475)
pF = 0.61

Answer: Given these estimates, you should bet if you estimate that you will be called less than 39% of the time.

A more detailed answer can be found by estimating the range of hands with which your opponents would call your all-in bet. You can then determine pF as the probability that no hand in that range has been dealt, and your weighted pW vs. that range of hands (using Bayes Theorem).

For example, you assume that your four EP/MP opponents will not call unless they hold a pair 99 or higher, or AK (46 possible hands) while the CO, Button, SB, and BB would call with any of the above plus AQ, AJ, and KQ (totalling 82 possible hands). There are 1326 possible starting hands.

The probability that any given one of the first four opponents will be dealt a calling hand = 46/1326 = 0.035, thus the probability that none of them will be dealt a calling hand = (1 - 0.035)^4 = 0.867. The probability that any of the last four opponents will be dealt a calling hand = 86/1326 = 0.065, thus the probability that none of them will be dealt a calling hand = (1 - 0.065)^4 = 0.765. The probability that none of your opponents will be dealt a calling hand is 0.867 x 0.765 = 0.663. Thus, you will win $300 (the current pot) 66.3% of the time, so Ef = $198.90.

After that, it gets absurdly complicated, because the standards for calling will change with the number of prior callers. That is, while the Button might only call you with AA-99, AK, AQ, AK, or KQ if no one else has entered the pot, his range of calling hands will change if one or more of the earlier players has called.

You could work out a precise estimate of calling hands in each position for any given number of callers (and including reraises from AA, KK, etc., etc., etc.), and the Bayesian pW vs. the mean number of callers and the range of hands with which you stand to be called, but it would take a whole lot more time than I'm willing to put into it.

Cris

[TheGrifter]

Math schmath. This one ain't close.

[Dominic]

David,

Personally, my responses were to a "real world" tournament situation, and not a math problem...I sized up the situtation, figuerd what I'd have left after the blinds if I didn't play the QJ, and made my decision accordingly.

Just like the emminent pros you mentioned.

Sorry, but it was a fairly easy call to move all in once I thought about for a scond.

I'll take making the right move - just like Allen Cunningham, etc., based on game experience, proper tournament strategy and basic common sense than having to calculate my EV everytime I have a decision to make.

The math is important. But it ain't everything.

[gergery]

[ QUOTE ]
[ QUOTE ]
Basically you move in if the EV of your resultant stack is more than the $1100 you have now and not otherwise.

[/ QUOTE ]

This is a dangerously incorrect statement.

Do you see why?

[/ QUOTE ]

Delicious. I love it!

[Paul Phillips]

[ QUOTE ]
Pc = Current Pot ($)
By = Bet(s) you will add to pot ($)
Bo = Bet(s) your opponent(s) will add to pot ($)
pF = Probability that opponent(s) will fold if you bet
pW = Probability that your hand wins at showdown

[/ QUOTE ]

Don't forget these other important variables:

Pcb = probability that you have any idea about how to calculate a meaningful result
Pfc = probability I will have asian food tonight
Phb = probability that you remind me of a dilbert character

It's difficult to express the entire equation for calculating Pcb without a proper mathematical formatting language, but thanks to a few well-located zeroes we can skip it and express the unique trivial result.

Don't you inflict enough disinformation when writing in conceptual terms? Mathematics is wonderful and pure and innocent and pained by your touch. In other posts you've adequately demonstrated you don't have even the barest grasp of simple gambling concepts like expectation and inference. Assigning pretty names to a bunch of variables isn't going to help you here. You should stick to a mode of expression that offers you more weasel room.

You spent all that time laboriously manipulating symbols and then handwaved at every part that might help you, oh, I don't know, actually DECIDE WHAT TO DO.

[ QUOTE ]
Estimates: At most one opponent will call, and you will be a 3:1 underdog if called. [...] Answer: Given these estimates, you should bet if you estimate that you will be called less than 39% of the time.

[/ QUOTE ]

So just to clarify this a little for me: if we wildly guess that we'll be called by exactly one person and also wildly guess that we'll be a 3-1 dog, then if we wildly guess that we'll be called less than 39% of the time, then we should move all-in? Sounds good!

One of my favorite aspects of the original star trek was that whenever spock was called upon to give an answer, he would have about sixteen decimal points ready. Even when the source data involved wild guesstimations and incomplete tricorder readings, he'd happily (ok not happily, he's vulcan) predict the chances of success right down to the billionth part. This was supposed to make him sound smarter for the folks at home but it just made him sound silly to anyone who had made it through high school science. It's not a perfect analogy but do you see where it's going?

Incidentally, as to the original question: "move in." I offer no mathematical justification but I bet this answer is still way more useful than the one I just finished needling.