Negreanu's tournament theory regarding big pots.

[aces961]

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Thus, you have not doubled your winning chances by doubling your stack; you have increased them from 0.6Y to Y.


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fine. the part that i do not understand is why we stop the process here.

so now we are at Y...and we want to get to Z.
and we are going to go from 0.6Z to Z since we are still 60% badass.
but Z = 2Y. so 0.6Z != Y which is where we thought we were after the first double up.

something has to give, right?

or to try to put it in words. why are we restricting a player from increasing their chances of winning the torunament since they will have the same double-up opportunity at the next level? or is this the flaw in my thinking since we are not assuming they will have this same opportunity

-tpir

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Why does Z=2V. V was defined as your chance of winning with the stack doubled from the initial starting stack.
Now with this doubled stack you will have a probability of doubling up of say .6 again(this probability could be different from .6 depending on how well you play with a stack size of different proportion from your opponents) Now you get that .6Z=Y, or Z=5/3 times Y.

To generalize this suppose you are playing in a heads up tourney that started with 32 people. All 31 other people in this tourney are of equal skill and you are better than them all in the sense you will win 60 percent of your matches against any one of them.

Thus your chances of winning the individual match will be .6, to win the tourney you must win 5 in a row or .6^5. Once you have won the first match your chance of winning the tourney hasn't doubled but has been multipled by 1/.6. so if you have won the first match now your chance of winning the tourney is .6^4 and so on until you get to a .6 chance of winning if you have won your first 4 matches.

[KidRoute66]

Re: S-Curve Hypothesis

Two questions that may/may not increase my understanding of this particular issue, and/or increase my ability to understand and apply what I've learned in the future.
1. Is the S-Curve Hypothesis based on the assumption that all players are dealt the same pocket (which is an impossibility in such situations there are more than four players at any table), or is it based on the assumption that a larger stack has a greater opportunity to survive a longer run of "unplayable" cards, or neither?
2. How might position influence the sublinear function addressed on this thread?
Thank you.

[tpir90036]

my point is that if we have a better than average chance of making it to the next level... when we get to that level... we *still* have a better than average chance, so why are we limiting the growth of our chances.

X = chances of avg. player winning the tourney with 10K
Y = chances of avg. player winning with 20K
Z = chances of avg. player winning with 40K
2X = Y
2Y = Z

so when an average player is at X, we are at 0.6Y. so when we get to Y...we aren't doubling our chances -- i understand that.

*but* if our edge still stands (which might be the flaw in my thinking) then we aren't increasing to Y... we are actually increasing to 0.6Z, 10% past Y. i understand the crux of this argument.... but the way this was laid out seems to not make sense if we follow it through to the next level.

-tpir

[aces961]

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my point is that if we have a better than average chance of making it to the next level... when we get to that level... we *still* have a better than average chance, so why are we limiting the growth of our chances.

X = chances of avg. player winning the tourney with 10K
Y = chances of avg. player winning with 20K
Z = chances of avg. player winning with 40K
2X = Y
2Y = Z

so when an average player is at X, we are at 0.6Y. so when we get to Y...we aren't doubling our chances -- i understand that.

*but* if our edge still stands (which might be the flaw in my thinking) then we aren't increasing to Y... we are actually increasing to 0.6Z, 10% past Y. i understand the crux of this argument.... but the way this was laid out seems to not make sense if we follow it through to the next level.

-tpir

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X, V, and Z should be the chances of the above average player winning the tourney with 10,20,40k. If at any point these were equal to the chances for a average player we would have a contradiction to the premise that the player is above average.

[gergery]

That post did it, thanks. Your last one and Sklanksy's just confused me.

However, that assumes you are more likely to double up than go broke, which means you are a better player than average, and my understanding of the premise of the thread is that you are supposed to be dead equal to the field.

[Eldog605]

I've been thinking a lot about it, and I think I came up with a pretty good analogy. Lets take major league baseball. Ichiro Suzuki's batting average is .378 and his on base percentage is .417. This means that whether the count is 3-0 or 0-0, Ichiro has about the same chances of getting to first base. As the pitcher misses the plate, and the count runs to 3-0, Ichiro's chances of getting to first base are only enhanced partially. I think we can compare chip counts in poker and ball count in baseball for this example.

On the other hand, there is a player like Lance Berkman, whose BA is only .313 (small in comparison) and whose on base percentage is .455. For the analogy, we'll equate getting to first base to winning a tourney in poker.

But contrary to Ichiro, Berkman's chances of making it to first base rise dramatically with each pitch that misses the plate (or for the poker player, each chip he gains). We know this, because his on base percentage is .455...meaning he walks an incredible amount of times. So obviously, the ball (chip) count matters much more to the lesser of the two hitters, Berkman. When the ball count is 3-0, Berkman's chances of getting to first have risen so high that his numbers can compete with Ichiro's.

Lance Berkman needs the pitcher to miss the plate so that he can compete with Ichiro Suzuki. Just like lesser poker players NEED the chips in order to compete with world class pros. As for Ichiro, he is a danger any time he's at the plate.

Of course, then there is Barry Bonds, whose BA is .373, and whose on base percentage is a staggering .611. I guess we could compare him with a world class pro who also happens to be devastatingly lethal, and unrelenting, with a big stack of chips (3-0 or 3-1 counts, for the baseball analogy). Any thoughts on which famous pros might fit into these three categories?

[gergery]

I love the S-curve post.

But in practice, the shortstacks don't play more conservatively, and the bigstacks play more aggressively.

How do you reconcile that with the hypothesis?

-g

[aces961]

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I've been thinking a lot about it, and I think I came up with a pretty good analogy. Lets take major league baseball. Ichiro Suzuki's batting average is .378 and his on base percentage is .417. This means that whether the count is 3-0 or 0-0, Ichiro has about the same chances of getting to first base. As the pitcher misses the plate, and the count runs to 3-0, Ichiro's chances of getting to first base are only enhanced partially. I think we can compare chip counts in poker and ball count in baseball for this example.

On the other hand, there is a player like Lance Berkman, whose BA is only .313 (small in comparison) and whose on base percentage is .455. For the analogy, we'll equate getting to first base to winning a tourney in poker.

But contrary to Ichiro, Berkman's chances of making it to first base rise dramatically with each pitch that misses the plate (or for the poker player, each chip he gains). We know this, because his on base percentage is .455...meaning he walks an incredible amount of times. So obviously, the ball (chip) count matters much more to the lesser of the two hitters, Berkman. When the ball count is 3-0, Berkman's chances of getting to first have risen so high that his numbers can compete with Ichiro's.

Lance Berkman needs the pitcher to miss the plate so that he can compete with Ichiro Suzuki. Just like lesser poker players NEED the chips in order to compete with world class pros. As for Ichiro, he is a danger any time he's at the plate.

Of course, then there is Barry Bonds, whose BA is .373, and whose on base percentage is a staggering .611. I guess we could compare him with a world class pro who also happens to be devastatingly lethal, and unrelenting, with a big stack of chips (3-0 or 3-1 counts, for the baseball analogy). Any thoughts on which famous pros might fit into these three categories?

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Here is some information you might want to know after the count has reached 3-0 this season ichiro has a OBP of 848. His average in these sitations is only 375 though( he is 3/8) This is a lot higher than his OBP of 417.

Berrkman on the other hand has a obp of 808 after the count has reached 3-0 at some point in the plate appearacne.

In reality Berkman and Ichiros chances of making it to first increase dramaticly as the count goes from having 1 to 2 to 3 balls. The reason's Ichiros obp is lower than berkmans is mainly for two reasons. one Ichiro may swing at balls that are out of the zone that berkman doesn't not swing at. Thus berkman is more likely to recieve a ball for that reason. A second reason is that pitchers may be more afraid to throw a strike to berkman since he has a better chance than ichiro of hitting a home run. Hence Berkman will be thrown more balls allowing him to get on base without a hit.

It is not that berkman needs the pitcher to miss the plate, it is that the pitchers miss the plate out of fear of what berkman will do if they throw the ball over the plate.

Same thing is true of bonds in the sense that pitchers fear what he will do if they throw him a srike even more than berkman. If bonds was thrown strikes every time he went to the plate his average may be say 400 but his obp would only be this 400 as well. Though he might hit 90 homers. the pitchers allow him to get to first the extra 200 percent of the time in order to keep him from those extra 50 homers.

The same thing is true for Berkman to a lesser extent.

[SossMan]

just curious...what's bonds' OBP when 3-0? It must be darn near 1.000.

[aces961]

bonds has had the count reach 3-0 against him 143 times this season. of those 143 times he has walked 137 times got a hit 5 times and been retired once. So his obp in plate appearances where the count reached 3-0 is 142/143=.993