Since you brought it up, would you care to show us your longterm tourney results? In pokertracker, I assume.
I subscribe to the punkymisha form of result tracking. IE, I don't.
When you have 2.5 BB, are you really focused on winning the event at that point?
Why not? You are only one double up away from having a stack that will make people think before they call you. Once you have that, you can push those chips around.
the technically best play is not the best in all the circumstances.
The 'technically best play' by which I think you mean the biggest +EV play is always the correct play. Anything less is scared poker.
Sure I set up the numbers! My whole point is show me a plausible set of numbers that has a different conclusion. You aren't willing to do that because it might take a whole 5 minutes and there's a chance you might have to admit you are wrong.
I subscribe to the punkymisha form of result tracking. IE, I don't.
Why did I expect this? Very convenient.
The 'technically best play' by which I think you mean the biggest +EV play is always the correct play. Anything less is scared poker.
Depends on which EV you are talking. chips or $. They are not the same thing. And of course it depends on your goal winning or winning $. Again, not the same thing.
Someone said
I just know there are a ton of lurkers on this message board, so when I see bad logic being passed off as good, I have a need to answer.
Your logic is in question, you have done no grunt work to support your position nor do you have any results to demonstrate credibility. Thanks for coming out.
SirWatts has credibility with MTTs. His method looks interesting. I'm curious how applicable it is or if it is just something you can use as a quick and dirty calculation?
My method is not perfect but it is probably the best approximation available and i suspect it is fairly accurate. To get a very good approximation would take a lot of mathematical modeling and simulation that I think very few people have ever bothered to take on, and the ones that have (I would not be surprised if Sklansky or someone of similar stature may have) probably see no reason to let their hard work go to waste by telling everyone the results. Certainly I've never seen anything like it print. The fact that the result from my approximation isn't even close should be enough to convince you that pushing is correct though.
My method is not perfect but it is probably the best approximation available and i suspect it is fairly accurate.
Good point. In this case you are assuming AK has 66% equity if you push on the bubble. That's fine but you're ignoring the potential for any double up for the scenario where you fold your way into the money and then go all in shortly thereafter with a hand that has at least 50-60% equity. With little difference in your chip stacks at the point you go all-in, the EV shouldn't be drastically different except for the fact that you are now guaranteed $, so it should be lower variance. The only difference I can see between pushing pre and post bubble is you may be more likely to get called post bubble by someone on the short or medium stack.
My concern with simple models is they can break down at the extremes. I don't know much about the factors that go into this one but I would expect on the bubble where the ratio between $ before and after the bubble is infinity would represent a challenge. I think once you are into the money it is probably a very good tool. (I'd also expect this falls apart with the Sklansky problem with Aces and an extremely short stack at the final table where you've already got the medium stacks all-in. With a simple calculation you can show calling is +EV but in reality it is an extremely bad move.)
To eliminate tany concerns about the bubble we can just assume you stall out your time bank as necessary to ensure you make the $$ assuming you survive the AK hand. The cases where you double up after folding into the $$ are considered in the EV calculation as a whole. Given how short your stack is and how few hands you have before the blinds it's unlikely you will get the $$ in as a favourite in this case though. There IS a point where folding into the money is correct here though, but with the M and Q as in this example we have way too many chips to give up.
Good 'ol results oriented thinking. I expect nothing less from you. How do my results factor into a question about the EV potential of two decisions?
Your logic is in question
SW justified my logic. I believe that chip equity is a concept straight out of TPFAP.
That's fine but you're ignoring the potential for any double up for the scenario where you fold your way into the money and then go all in shortly thereafter with a hand that has at least 50-60% equity
Read Harrington. Red Zone play. You let your stack get to 0.75BB then pick up ACES! You double up to.... 1.5BB... and blind off next orbit anyway.
There IS a point where folding into the money is correct here though, but with the M and Q as in this example we have way too many chips to give up.
Another concept straight out of TPFAP. But it only comes into play when you are desperately far behind and I don't think we're there yet.
A rebuttal of an earlier post (this thread has sure been lively):
So with 4 all-ins, he is around 6% to make it to an *average* stack. Once he gets there, he has a 1-2.5% chance (again generous) to actually get to the serious money.
I have problems with these numbers since you seem to disregard any fold equity he gets once he doubles the first time getting his M into a more respectable level. As well, I'm not sure where the 1-2.5% chance comes from since many people may have busted by the time hero gets to a decent stack. I don't take that much stock in getting to an average stack anyways, since M is a bigger deal than Q. He only really needs a couple of doubles and the odd steal or 2 to get out of the red zone...
I doubt that the chip count method for dividing the prize pool applies here.
This is a method that works for heads-up equal-skilled players, and is suggested (by Sklansky) as a "good approximation" for 3-handed situations. Though I doubt anyone will be able to prove or disprove it, my gut feeling is that the chip count rule is not a good approximation to use in a 300+ handed situation.
There is still lots of play left in the tournament (which neutralizes the "equal-skilled" hypothesis), and my feeling is that the vagaries of being a short stack right now are also going to make the chip count approximation inaccurate.
FWIW the average stack here is only slightly above 25K so we only need to double up once more after our AK holds up to get to average. I'm assuming hero is average to above average skill so that that hypothese isn't a big deal for us, if he's not then I'm not sure why he's in the tournament in the first place. Being shortstacked probably hurts a bit but everyone is under pressure so it can' t be that bad. I'll throw it out there that my approximation is good to within $100, that's my intuition as a math guy. From personal experience 12.5K with 1K/2K blinds is really not that bad and several tournaments I've won I was in much worse shape than that at one point.
One quick point in terms of the equal-skill hypothesis.
I wasn't really thinking about the skill of the hero necessarily, although this is an important component. The fact that the entire tournament is full of players of varying skill levels is going to affect the results too.
For example, if the 30 (or so) most skilled players happen to be at different tables right now, your chances of winning the tournament go way down. They go way up if the 30 best players are at the same 3 or 4 tables. The important question is, obviously, what happens on average? I'm thinking that, on average, you're more likely to find the 30 best players fairly spread out, say 1 or 2 per table, rather than mostly lumped together at a few tables.
The heads-up equal-skilled chip count model is based on the fact that the opponents are simply playing a fair game (they could essentially even be tossing coins) with different sized bankrolls. Analyzing a 300+ player, short-stacked, MTT poker scenario with such a tool is, IMHO, stepping too far outside the methodology of the model you're trying to use.
Point taken on the skill hypothesis problem, I also just realized that at this point it is likely that more skilled than poor players remain so our relative skill might not be as high. I still don't think the methodology is that far off though. Our chance of winning are based mostly on how many chips we have and how skilled we are compared to our opposiiton at any point in the tournament. There are many other factors but these 2 must outweigh them all. so any estimate on our EV will be based mainly on our skill and chip stack with some error term due to all the other things. It is hard to put an exact number on how much the skill factor influences things but the average stack is short enough here that I don't think it can be too drastic. Luck is going to be the prevalent factor.
Our chance of winning are based mostly on how many chips we have and how skilled we are compared to our opposiiton at any point in the tournament. There are many other factors but these 2 must outweigh them all. so any estimate on our EV will be based mainly on our skill and chip stack with some error term due to all the other things.
I totally agree with this.
However, it seems unlikely to me that a player's expected win is well approximated by a linear function of his chip count.
This kind of approximation might even be okay for the ultra-late stages of tournaments (i.e. the deal making period) where blinds are big and skill differentials are decreased. However, expected payouts from a scenario where there are lots of players, and a diverse collection of skill levels, player types, and stack sizes, aren't going to be easy describe as a linear function of chip counts.
Again, these kinds of statements are pretty difficult, if not impossible to prove. Certainly an interesting discussion though.
Ultimately, we may have to make a compromise in strategy on this one. It seems that the correct play is going to be an in-between strategy. It must be the case that making it $4,000 to go is the right play here.
No more math today... Pokerforum is supposed to be my break from this stuff. Doubling the blind is so brilliant it makes my head spin. Make sure you do it with AA as well so they can't afford to risk calling the times you only have AK.
The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side!
I don't think so. You want to prove this?
I heard it on the Simpsons once, so it must be true.
QED
Homer: (wearing glasses) The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side!
Man: (in a cubicle) That's a right triangle, you idiot!
Homer: D'oh!
The best part is that it's still wrong after the guy corrects him.
Please forgive me for my lack of humor. I have two degrees majoring in math. I teach high school math. I have been in two days of mind crippling computer software seminars with one day still to go. I have been on the forum during the whole thing just to keep from wanting to kill myself. Help me, people, before it is too late!
After putting on Dr. Kissinger's glasses, Homer says "The sum of the
square roots of any two sides of an isosceles triangle is equal to
the square root of the remaining side," to which the man in the
bathroom stall goes, "That's a right triangle, you idiot!"
Marc Dreyfuss comments, "Not only is what Homer said not true for
isosceles triangles, but also, it is in fact the SQUARES of two
sides and the SQUARE of the remaining side. And not just ANY side,
but the sides that subtend the 90 degree angle. I'm afraid the
Simpson's might have really shown us the depth of math-illiteracy in
this country -- even the person in the stall who screamed `idiot!'
couldn't get the problem right."
Ron Carter says this is a reference to "The Wizard of Oz", in the scene
where the Scarecrow is given a brain and says the same thing. He
says, "And the guy in the booth is right; it -is-a right angle -and-
they -did-get it wrong in the movie; I never noticed."
Scarecrow: The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side. Oh joy! Rapture! I got a brain! How can I ever thank you enough?
Wizard of Oz: You can't.
Comments
I subscribe to the punkymisha form of result tracking. IE, I don't.
Why not? You are only one double up away from having a stack that will make people think before they call you. Once you have that, you can push those chips around.
The 'technically best play' by which I think you mean the biggest +EV play is always the correct play. Anything less is scared poker.
Why did I expect this? Very convenient.
Depends on which EV you are talking. chips or $. They are not the same thing. And of course it depends on your goal winning or winning $. Again, not the same thing.
Someone said Your logic is in question, you have done no grunt work to support your position nor do you have any results to demonstrate credibility. Thanks for coming out.
SirWatts has credibility with MTTs. His method looks interesting. I'm curious how applicable it is or if it is just something you can use as a quick and dirty calculation?
My concern with simple models is they can break down at the extremes. I don't know much about the factors that go into this one but I would expect on the bubble where the ratio between $ before and after the bubble is infinity would represent a challenge. I think once you are into the money it is probably a very good tool. (I'd also expect this falls apart with the Sklansky problem with Aces and an extremely short stack at the final table where you've already got the medium stacks all-in. With a simple calculation you can show calling is +EV but in reality it is an extremely bad move.)
Sounds like the subject for another thread.
Good 'ol results oriented thinking. I expect nothing less from you. How do my results factor into a question about the EV potential of two decisions?
SW justified my logic. I believe that chip equity is a concept straight out of TPFAP.
Read Harrington. Red Zone play. You let your stack get to 0.75BB then pick up ACES! You double up to.... 1.5BB... and blind off next orbit anyway.
Another concept straight out of TPFAP. But it only comes into play when you are desperately far behind and I don't think we're there yet.
I have problems with these numbers since you seem to disregard any fold equity he gets once he doubles the first time getting his M into a more respectable level. As well, I'm not sure where the 1-2.5% chance comes from since many people may have busted by the time hero gets to a decent stack. I don't take that much stock in getting to an average stack anyways, since M is a bigger deal than Q. He only really needs a couple of doubles and the odd steal or 2 to get out of the red zone...
This is a method that works for heads-up equal-skilled players, and is suggested (by Sklansky) as a "good approximation" for 3-handed situations. Though I doubt anyone will be able to prove or disprove it, my gut feeling is that the chip count rule is not a good approximation to use in a 300+ handed situation.
There is still lots of play left in the tournament (which neutralizes the "equal-skilled" hypothesis), and my feeling is that the vagaries of being a short stack right now are also going to make the chip count approximation inaccurate.
ScottyZ
Pffff...
What do math guys know.
Just kidding of course.
One quick point in terms of the equal-skill hypothesis.
I wasn't really thinking about the skill of the hero necessarily, although this is an important component. The fact that the entire tournament is full of players of varying skill levels is going to affect the results too.
For example, if the 30 (or so) most skilled players happen to be at different tables right now, your chances of winning the tournament go way down. They go way up if the 30 best players are at the same 3 or 4 tables. The important question is, obviously, what happens on average? I'm thinking that, on average, you're more likely to find the 30 best players fairly spread out, say 1 or 2 per table, rather than mostly lumped together at a few tables.
The heads-up equal-skilled chip count model is based on the fact that the opponents are simply playing a fair game (they could essentially even be tossing coins) with different sized bankrolls. Analyzing a 300+ player, short-stacked, MTT poker scenario with such a tool is, IMHO, stepping too far outside the methodology of the model you're trying to use.
Okay, it wasn't that quick.
ScottyZ
I totally agree with this.
However, it seems unlikely to me that a player's expected win is well approximated by a linear function of his chip count.
This kind of approximation might even be okay for the ultra-late stages of tournaments (i.e. the deal making period) where blinds are big and skill differentials are decreased. However, expected payouts from a scenario where there are lots of players, and a diverse collection of skill levels, player types, and stack sizes, aren't going to be easy describe as a linear function of chip counts.
Again, these kinds of statements are pretty difficult, if not impossible to prove. Certainly an interesting discussion though.
Ultimately, we may have to make a compromise in strategy on this one. It seems that the correct play is going to be an in-between strategy. It must be the case that making it $4,000 to go is the right play here.
ScottyZ
I figured you'd like this play.
The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side!
Mwahahaha
ScottyZ
I don't think so. You want to prove this?
I heard it on the Simpsons once, so it must be true.
QED
Homer: (wearing glasses) The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side!
Man: (in a cubicle) That's a right triangle, you idiot!
Homer: D'oh!
The best part is that it's still wrong after the guy corrects him.
ScottyZ
I do love the Simpsons.
Are you sure you have the quote right though? (I remembered the Simpson's line sort of) and it makes sense if the quote is:
The sum of the squares of any two sides of an isosceles triangle is equal to the square of the remaining side!
Actually that's still wrong because it depends on which sides you're talking about... Nevermind...
http://www.snpp.com/episodes/1F08.html
Pythagorean error
After putting on Dr. Kissinger's glasses, Homer says "The sum of the
square roots of any two sides of an isosceles triangle is equal to
the square root of the remaining side," to which the man in the
bathroom stall goes, "That's a right triangle, you idiot!"
Marc Dreyfuss comments, "Not only is what Homer said not true for
isosceles triangles, but also, it is in fact the SQUARES of two
sides and the SQUARE of the remaining side. And not just ANY side,
but the sides that subtend the 90 degree angle. I'm afraid the
Simpson's might have really shown us the depth of math-illiteracy in
this country -- even the person in the stall who screamed `idiot!'
couldn't get the problem right."
Ron Carter says this is a reference to "The Wizard of Oz", in the scene
where the Scarecrow is given a brain and says the same thing. He
says, "And the guy in the booth is right; it -is-a right angle -and-
they -did-get it wrong in the movie; I never noticed."
See also:
http://www.imdb.com/title/tt0032138/quotes
Scarecrow: The sum of the square roots of any two sides of an isosceles triangle is equal to the square root of the remaining side. Oh joy! Rapture! I got a brain! How can I ever thank you enough?
Wizard of Oz: You can't.