Negreanu's tournament theory regarding big pots.

[Pensive Gerbil]

In an article in the current Card Player magazine, I was surprised that Negreanu advocates sometimes CALLING with a draw in a big pot with slightly inadequate effective odds. If I recall, he said this is a significant difference between tournament and ring game play; top tournament players know it is correct to chase big pots with inadequate odds when it only costs a small portion of their stack.

It seems to me that top tournament players (such as Negreanu) should be less willing to chase pots with marginal or inadequate odds than less skilled players. I am also skeptical about Negreanu's suggestion that you should be more willing to chase pots with inadequate odds when it will just cost a small portion of your stack. Might not it be better (or less bad) for a desperately short-stacked player (who is not close to making the money) to chase a big pot despite slightly inadequate odds?

-PG

[Ryner]

Wasn't that article on limit hold-em, though? Seems to me chasing and going for draws can occasionally be much more do-able in limit than no limit, even when the odds are against you.

[Pensive Gerbil]

Although there are more opportunities to chase draws with marginal or only slightly inadequate odds in limit hold'em, the opportunity arises in no limit as well. Opponents often underbet pots, and implied odds tend to be greater in no limit. Nevertheless, I do not think this is relevant to the theoretical issues in my post.

-PG

[AtlBrvs4Life]

Do you have a link to the article?

[uw_madtown]

Given that the more chips you have, the less each chip is actually worth (theoretically), I think Negreanu's idea makes sense. I'll usually employ a similar strategy in 2-table SNGs at Party, actually. If I have an adequate stack and the right situation arises where I'm getting not-so-hot odds to chase a flush, I'll usually be willing to take a shot at chasing it with the idea being that winning this big pot when the flush comes it will solidify me as the chip-leader -- and if I don't make the flush, I'll still have enough chips to have a chance to out-maneuver my opponents.

Might just be bad strategy on my part.

- UW

[Pensive Gerbil]

http://www.cardplayer.com/poker_magazine/archives/showarticle.php?a_id=14222

[Pensive Gerbil]

[ QUOTE ]
Given that the more chips you have, the less each chip is actually worth (theoretically), I think Negreanu's idea makes sense...
- UW

[/ QUOTE ]

It is also true that the theoretical value of each chip you would be risking is greater than the value of each chip you might win.

-PG

[Vannek]

The move is -CEV but is +$EV at certain
points of the tournament

[gergery]

This only makes sense to do if
1) you are getting close to not quite the correct odds to call, and
2) the extra value you get from being able to use a bigstack effectively in tournament play more than offsets the loss in EV from calling when you dont have the odds

and maybe toss in alittle psychological benefit from putting o's on tilt if they see what you played

[Pensive Gerbil]

[ QUOTE ]
The move is -CEV but is +$EV at certain
points of the tournament

[/ QUOTE ]

At which points in a tournament do you think this is true? Why?

[Pensive Gerbil]

Your explanation makes sense. I wonder if that's what Daniel N. was thinking.

Should players who are more skillful with a short-stack than a large-stack avoid chasing draws with marginally adequate odds?

-PG

[TStoneMBD]

i am not a game theory specialist, but i think the answer to this debate is clear to me. let me try to explain:

saying that by chasing with improper odds with gold chips to get bronze chips is being used in the wrong context. the reason for this is that you do not want your chips to be gold, you want them bronze, which is reverse logic but true.

a big stack with 40k in chips is going to win a tournament MORE than twice as often as a player with 20k in chips. the shorter the stack you are, the worse the odds of you winning the tournament become. now i am not good with math, but i am sure someone could assemble an algebraic equation (if they really wanted to) that i could barely understand to further portray the reasoning behind this theory. if you are chasing a 5:1 draw with 4:1 odds, you would have to assume that winning the pot makes up for the improper odds you receive by the advantage of possessing more chips.

with blinds of 1k/2k and a 4k stack... folding on the turn with these incorrect odds leaves you with 4k in chips (with virtually no chance of winning), while if you call and lose you are eliminated, is there really any difference? However, if you win the pot you would suddenly have 20k and are suddenly in fantastic shape. clearly, calling with improper odds becomes correct in this scenario.

im not going to go into further math here as ill just end up confusing myself, but i think ive clearly described the logic behind negreanus article.

[cferejohn]

[ QUOTE ]

a big stack with 40k in chips is going to win a tournament MORE than twice as often as a player with 20k in chips.

[/ QUOTE ]

Actually, everything I've ever seen argues that a 40K stack will win EXACTLY twice as much as a 20K stack (assuming equally skilled players). That is why in a winner-pays-all tournament you should play for straight up chip EV. Major tournament adjustments as for as adjusting your odds come from the stratified pay structure which can make decisions +chipEV and -cashEV or vice versa when you are close to or in the money.

Now, certain players may play better as big stacks or small stacks. If someone is truly a much much better player as a big stack than a small-to-medium stack, this may make sense. I don't think the vast majority of players are *that* much better with a big stack (or *that* much worse with a small-medium one)...

[skydancing8]

I don't know how many of you have read Phil Hellmuth's book, "Play Poker like the Pros"? In the book he talks about Negreanu doing this. He does it often and busts a lot of people with it, however it it also causes a big time fluctuation in your bankroll.

[Bernas]

I believe this has a lot to do with his style of play. He is a very good aggressive player and as such it is much easier to play aggressively with a larger stack. Therefore the chips in the pot mean that much more to him.

I doubt a player like Dan Harrington would make this play but then again, he probably wouldn't be involved in these pots to begin with.

[David Sklansky]

a big stack with 40k in chips is going to win a tournament MORE than twice as often as a player with 20k in chips.


LESS, not more, if you are a good player. To prove this one need only to ask the probability question "what are the chances that a good player with 20K will double to 40K before going broke". I'll let others elaborate.

[La Brujita]

[ QUOTE ]
a big stack with 40k in chips is going to win a tournament MORE than twice as often as a player with 20k in chips.


LESS, not more, if you are a good player. To prove this one need only to ask the probability question "what are the chances that a good player with 20K will double to 40K before going broke". I'll let others elaborate.

[/ QUOTE ]

I guess I don't understand the wording of the post. Would you mind clarifying what you mean by "LESS, not more, if you are a good player?"

[gergery]

[ QUOTE ]
a big stack with 40k in chips is going to win a tournament MORE than twice as often as a player with 20k in chips.

LESS, not more, if you are a good player. To prove this one need only to ask the probability question "what are the chances that a good player with 20K will double to 40K before going broke". I'll let others elaborate.

[/ QUOTE ]

This doesn't make sense to me. Mainly because the risk of ruin is much higher for the 20k player. Imagine a scenario where you flip a coin -- heads you win .75 cents, tails you lose .50 cents. Great game, except that if you only start with 50 cents, half the time you're out. Whereas if you start with $3.00 you are better off. In this scenario, the player with $3 has EV MORE than 6x higher than the player wtih .50. It's this same principle that applies to the 40k and 20k player example above, and why the ChipEV=$EV early in tourneys is a good guideline, but somewhat flawed, IMHO.

To say nothign of the fact that having a larger stack influences the play of your opponents in ways favorable to you (they fold more, you steal more, etc.)

But, as of last count, I've written no books on poker, so maybe someone can enlighten me /images/graemlins/smile.gif
--Greg

[pzhon]

[ QUOTE ]
Imagine a scenario where you flip a coin -- heads you win .75 cents, tails you lose .50 cents. Great game, except that if you only start with 50 cents, half the time you're out. Whereas if you start with $3.00 you are better off.

[/ QUOTE ]
True.

[ QUOTE ]
In this scenario, the player with $3 has EV MORE than 6x higher than the player wtih .50.

[/ QUOTE ]
False. It's less than 4 times as much.

This may be counterintuitive, but try the thought experiment Sklansky suggested. How frequently will the player with $0.50 make it to $3.00? More than 1 time in 6, or less than 1 time in 6? Even if you count falling to $0.25 as going broke, and do not distinguish reaching $3.00, $3.25, or $3.50, you reach $3 more than 1/6 of the time without going broke if you start with $0.50.

Under these assumptions, the probability of making it to at least $3 without dropping below $0.50 are as follows:

$0.50: .291 = 41/141
$0.75: .348 = 49/141
$1.00: .512 = 73/141
$1.25: .582 = 82/141
$1.50: .695 = 98/141
$1.75: .745 = 105/141
$2.00: .816 = 115/141
$2.25: .872 = 123/141
$2.50: .908 = 128/141
$2.75: .936 = 132/141

There are some odd results from assuming that dropping to $0.25 is just as bad as going broke. In general, a player with an advantage over the field has more than a 1/2 chance to double up.

[cferejohn]

[ QUOTE ]
I don't know how many of you have read Phil Hellmuth's book, "Play Poker like the Pros"? In the book he talks about Negreanu doing this. He does it often and busts a lot of people with it, however it it also causes a big time fluctuation in your bankroll.

[/ QUOTE ]

I thought usually when Negraneau calls with marginal hands in situations like this it is with the following assumptions:

a) If my draw hits I will be able to extract more extra chips than the average player.

and/or

b) If my draw misses, I will be able to pick up the pot with a bluff more often than the average player.

If one or both of these are true (i.e. you really are that much better than your opponents and/or you can read them that well) than calling may in fact be +EV.

We've seen players like Hansen and Forest make calls on the flop with pretty much nothing because they think that they will be able to buy thepot often enough to make the move +EV. Are they right? I don't know, but its certainly a more +EV (or less -EV move) for them than for 99.99% of players out there.

[Che]

[ QUOTE ]
To prove this one need only to ask the probability question "what are the chances that a good player with 20K will double to 40K before going broke".

[/ QUOTE ]

Maybe I'm just showing my ignorance here, but I think the size of the blinds also needs to be considered.

If the blinds are negligible, then a good player will certainly double more often than he busts out. But when the blinds are significant (e.g. an orbit costs 1/4 of the stack or more), even a skilled player is probably more likely to bust out than double since opportunities to double up as at least a 51% favorite do not come around every orbit.

I hope someone will elaborate on the flaws in my thinking.

Che

[SossMan]

[ QUOTE ]
I hope someone will elaborate on the flaws in my thinking.


[/ QUOTE ]

There aren't any. David is ignoring this important consideration for some reason.

[cferejohn]

[ QUOTE ]
[ QUOTE ]
To prove this one need only to ask the probability question "what are the chances that a good player with 20K will double to 40K before going broke".

[/ QUOTE ]

Maybe I'm just showing my ignorance here, but I think the size of the blinds also needs to be considered.

If the blinds are negligible, then a good player will certainly double more often than he busts out. But when the blinds are significant (e.g. an orbit costs 1/4 of the stack or more), even a skilled player is probably more likely to bust out than double since opportunities to double up as at least a 51% favorite do not come around every orbit.

I hope someone will elaborate on the flaws in my thinking.

Che

[/ QUOTE ]

Well, on average shouldn't you have the best hand *exactly* once every orbit? If there are (say) 7 players, you are going to have the best hand, in the long run, exactly 1 time in 7.

The variation is gigantic, of course, and sometimes you'll be in a situation where it didn't matter which hand you put your money in with, you were destined to lose. Otoh, sometimes you will get good cards and double up twice or more.

The problem is that there is a pretty good chance you will have to put your money in as an underdog sometimes, but remember that you will rarely be *that* big of an underdog, even when you are one.

I can't see where David is going with the idea that the small stack is actually at an *advantage* but if the small stack plays correctly (and in this high blind context playing correctly often involves getting in with marginal hands), they should have a chance of winning equal to their stack size.

Maybe this will halp. Imagine you were heads up and the blinds were so high you were forced all-in every hand. In that case, it is pretty easy to see that your chances of doubling up are identical to your chances of busting.

[aces961]

[ QUOTE ]
a big stack with 40k in chips is going to win a tournament MORE than twice as often as a player with 20k in chips.


LESS, not more, if you are a good player. To prove this one need only to ask the probability question "what are the chances that a good player with 20K will double to 40K before going broke". I'll let others elaborate.

[/ QUOTE ]


Lets see if this helps some of you understand why what David says is true. Let 2X be your chance of winning the tourney with a 40k chip stack. Now we ask which happens first to a player with a 20k chip stack, does he double to 40k or go broke, one of these two will eventually happen. Say for a good player his chance of doubling first is 60% and his probability of going broke first is 40%.

Now if he gets to 40k first he will have a 2X chance of winning in this case which happens 60 percent of the time.
If he goes broke first he can't win. .4*0+.6*2X is 1.2X. Thus with a 20k stack this player has a 1.2X chance of winning.

We can see that for the player who will double up before going broke 60 percent of the time he will win less than twice as much with twice as big of a stack since 2 is less than 2.4.

As a side note the case when the 40k in chips gives you twice as high a chance of winning as having 20k chips occurs only if you are equally likely to go broke or double up first when you start with 20k chips.

Also if you will go broke before doubling up from 20k chips over half the time your chances will be higher than twice as much of winning the tourney if you have 40k chips than 20k

[TStoneMBD]

maybe i am incorrect in my assumptions as i am clearly debating with sklansky here, but i think it is far too presumptious to assume that a good player will double up to 40k 60% of the time without concerning many table factors including blind structure. if a player has 20k in chips, is UTG, and the blinds are 5/10k at this point he is not doubling up any more often than a poor player. id like to see a professional double up here more 60% of the time. this is clearly silly to presume this as he will be forced to push with any hand that is greater than the average blind hand. i find it difficult to believe that under these circumstances, a better player with a short stack will win more than half the time as a player of equal skill with a large stack. clearly in this UTG position, if he were to push with say, K4o, he is going to be called by a hand that is at least a 60/40 favorite to his, assuming at least 1 ace was dealt to the table. you are not doubling up 60% of the time here, but 40%, which is why having more chips allows you to survive blinds.

[Bigwig]

I read the article after I read the thread, and don't feel like reading the posts again, so if this has been said, I apoligize.

Daniel is playing limit poker, and therefore, when he calculates his pot odds, he knows that nobody can push in a bet that will cost him the tournament in that particular hand on the river. I don't see how this factor can be ignored when calculating the odds. In no-limit, drawing to the non-nuts, you may still be beaten, and it will cost you more than one extra bet.

[Chaostracize]

Negreanu also makes some weak calls on the turn sometimes... he himself stated "I know I wasn't getting odds, but there's power in that pot!".

A few things: If you're opponent doesn't know you, and sees you showdown a hand the was beaten in the flop and turn, more action in the future. And obviously, if you're drawing to the nuts, and the pot is big which means you've been getting action already, you get paid off when you hit.

[aces961]

First of all the number 60 percent was chosen randomly, as long as it is any number over 50 the result that the good player wins less than twice as much when he has 40k than 20k holds up. Secondly part of being a good player is not letting your stack get to two big blinds. In no limit especially I don't see many good players left with only 2 big blinds unless they have just lost an all in to bring them to that. Even just stealing according to the advanced system chart it would be pretty unlikely to get blinded down to 2 big blinds.
Keep in mind the tourneys that pros play in are usually nothing like the low buy in multitables online that start you off with 100 big blinds if you are lucky.

[Eldog605]

This is a reaction to this entire thread. I would think that if one got double the chips, he'd have much more than double the chances to win the tourney. Of course the downside is that you need to either play a draw with inadequate odds, or call a coin flip to double up and get that edge. Both are risky ventures.

Lets say Player X has a tough choice...he has 20,000 and if he calls a big bet and loses, hes out. Lets say he folds this hand and then gets dealt aces vs an opponent's kings, and NOW he doubles up to 40,000.
But another possibility is that Player X calls the original bet for the 20,000, and he wins and doubles to 40,000. If he then gets dealt aces vs kings on the next hand, he will double up to 80,000. If a player's chips can grow exponentially, and they can, I would think it is pretty clear that doubling up more than double's one's chances at winning a tournament. Just my two cents.

[aces961]

[ QUOTE ]
This is a reaction to this entire thread. I would think that if one got double the chips, he'd have much more than double the chances to win the tourney. Of course the downside is that you need to either play a draw with inadequate odds, or call a coin flip to double up and get that edge. Both are risky ventures.

Lets say Player X has a tough choice...he has 20,000 and if he calls a big bet and loses, hes out. Lets say he folds this hand and then gets dealt aces vs an opponent's kings, and NOW he doubles up to 40,000.
But another possibility is that Player X calls the original bet for the 20,000, and he wins and doubles to 40,000. If he then gets dealt aces vs kings on the next hand, he will double up to 80,000. If a player's chips can grow exponentially, and they can, I would think it is pretty clear that doubling up more than double's one's chances at winning a tournament. Just my two cents.

[/ QUOTE ]

I really need to shoot down this arguement immediatly. This arguement never said anything about player skill level so lets assume they are all equally skilled. Say we have 1000 people in a tourney. They each have a 1/1000 chance of winning the tourney at the start, also say they each have 1000 chips. Now say a few hands later 500 people have been knocked out and all 500 people left have exactly 2000 chips. So now we see that the player has a 1/500 chance of winning. Now 1/500 divided by 1/1000 ia 2. So with all players of equal skill the double up in this situation has resulted in a doubling of the chances of the player winning the tourney. So the arguement that doubling up more than doubles one's chances at winning a tourney is false in the situation of equal skill level.

If you look at my previous posts in this thread you will see the only time that doubling up more than doubles ones chances of winning a tourney is when you are a below average player compared to the average player in the tourney.

[David Sklansky]

"This is a reaction to this entire thread. I would think that if one got double the chips, he'd have much more than double the chances to win the tourney."

Listen carefully. If, for whatever reason, your chances of doubling up is greater than 50%, than your chances of winning the tournament, if you do double up, is less than twice what is was than before you did. That statement is ALWAYS true and you don't have permission to leave your computer, or even read another post, until you are sure you understand why.

[Piers]

[ QUOTE ]

If the blinds are negligible, then a good player will certainly double more often than he busts out. But when the blinds are significant (e.g. an orbit costs 1/4 of the stack or more), even a skilled player is probably more likely to bust out than double since opportunities to double up as at least a 51% favorite do not come around every orbit.


[/ QUOTE ]

If you are very short stacked in relation to the blinds, then you only need to steal the blinds once or maybe twice to double up; much easier then doubling though someone with a big stack.

Admittedly skill is much more important to doubling up when you have a large stack, but I do not see why that is relevant. I am afraid DS is right again /images/graemlins/frown.gif .

[fnurt]

[ QUOTE ]
"This is a reaction to this entire thread. I would think that if one got double the chips, he'd have much more than double the chances to win the tourney."

Listen carefully. If, for whatever reason, your chances of doubling up is greater than 50%, than your chances of winning the tournament, if you do double up, is less than twice what is was than before you did. That statement is ALWAYS true and you don't have permission to leave your computer, or even read another post, until you are sure you understand why.

[/ QUOTE ]

For anyone who doesn't get it, the point is that if you have a better than 50% chance of doubling your money, that fact is already calculated in to whatever your chances are of winning the tournament, before you actually get the double-up.

Assume the following. You are in a 100 player tournament and have a 1% chance of winning. But if you double, you are so good at playing a big stack that even though you have 1/50 of the chips in play, you have a 1/40 (2.5%) chance of winning.

Since you are a good player, let's assume you have a 60% chance to double up before you bust. But wait! If you have a 40% chance of busting, and a 60% chance of doubling into a position where you will have a 2.5% chance to win the tournament, your actual starting odds were not 1% at all, they were 60% * 2.5% which is 1.5%.

[eMarkM]

Thanks fnurt, now I can finally get up from my computer and go pee.

[Rushmore]

I must admit to not caring at all.

If I am a favorite to double up, and there are no extenuating circumstances (payout structure, bubble, tiny stacks/blinds/blind size, etc), and I DO, indeed, double up, I then feel good about having doubled up. I do not make a decision about exactly how pleased I am to have done so.

I will then attempt to double up again and again, as often as possible, until I have as many of the chips as possible, thusly earning as high a prize as possible.

I don't mean to give lip to a noted poker theorist/scientist/brainiac, but I seem to be doing it, so...

Why should I care if I am 3-1, 7-2, or 5-2 to win the tournament?

I have enough voices in my head as it is, without the booming voice of Oz telling me I am not allowed to leave my computer if I don't fully grasp each and every iota of poker esoteria.

Damn the voices in my head.

[37offsuit]

What if you're heads up, you can double up through your opponent leaving him only 1K in chips to your 200K if you win this next hand. You hold AA and he holds A7o. You push, he calls.

Result
http://twodimes.net/h/?z=75840
pokenum -h ad ah - ac 7h
Holdem Hi: 1712304 enumerated boards
cards win %win lose %lose tie %tie EV
Ad Ah 1591406 92.94 96820 5.65 24078 1.41 0.936
Ac 7h 96820 5.65 1591406 92.94 24078 1.41 0.064

Is that statement still ALWAYS true?

[SossMan]

lol...my thoughts exactly, Rush. good post

[BobboFitos]

Assume the following. You are in a 100 player tournament and have a 1% chance of winning. But if you double, you are so good at playing a big stack that even though you have 1/50 of the chips in play, you have a 1/40 (2.5%) chance of winning.

Since you are a good player, let's assume you have a 60% chance to double up before you bust. But wait! If you have a 40% chance of busting, and a 60% chance of doubling into a position where you will have a 2.5% chance to win the tournament, your actual starting odds were not 1% at all, they were 60% * 2.5% which is 1.5%.

fnurt, thank you so much... I sort of understood what this meant, (and felt a little silly when DS said I shouldn't leave my computer until I understood it fully) but now I get it 100%. Thank you.

[Rushmore]

[ QUOTE ]
b) If my draw misses, I will be able to pick up the pot with a bluff more often than the average player.

If one or both of these are true (i.e. you really are that much better than your opponents and/or you can read them that well) than calling may in fact be +EV.

We've seen players like Hansen and Forest make calls on the flop with pretty much nothing because they think that they will be able to buy thepot often enough to make the move +EV. Are they right? I don't know, but its certainly a more +EV (or less -EV move) for them than for 99.99% of players out there.

[/ QUOTE ]

This is the good stuff.

Although it is true that the EV of any imminent bluff is card-dependent, just as a draw is, it is indeed a factor which one may only grow into. There's no way in the world for the inexperienced or inferior player to consider this dynamic, at least not properly.

[Eldog605]

Aces and David,
Let me reiterate Aces logic. 1000 people in a tourney, 1000 chips each, everyone has a 1/1000 chance of winning. Agreed. Fastforward...500 players left, 2000 chips each, 1/500 chance to win. Agreed. You have now doubled up and you have TWICE the chances to win, by your own admission.

Now take it another step. 250 players left, 4000 chips each. 1/250 chance to win. Agreed? So you went from a 1/1000 shot to a 1/250 shot. Call me stupid, but you have effectively quadrupled your chances of winning the tourney from 1/1000 to 1/250. By doing what? By doubling up twice. Let me repeat, you have doubled up twice, and in doing so, you have quadrupled your chances of winning. Chip growth is exponential, so is one's chances of winning with them.

[fnurt]

Doubling twice is the same as quadrupling. Now here's a point that everyone can understand.

[aces961]

[ QUOTE ]
Aces and David,
Let me reiterate Aces logic. 1000 people in a tourney, 1000 chips each, everyone has a 1/1000 chance of winning. Agreed. Fastforward...500 players left, 2000 chips each, 1/500 chance to win. Agreed. You have now doubled up and you have TWICE the chances to win, by your own admission.

Now take it another step. 250 players left, 4000 chips each. 1/250 chance to win. Agreed? So you went from a 1/1000 shot to a 1/250 shot. Call me stupid, but you have effectively quadrupled your chances of winning the tourney from 1/1000 to 1/250. By doing what? By doubling up twice. Let me repeat, you have doubled up twice, and in doing so, you have quadrupled your chances of winning. Chip growth is exponential, so is one's chances of winning with them.

[/ QUOTE ]

This example I gave was for players of equal skill. Only when all the players are of equal skill will all of them having the same number of chips mean that they each have an equal chance of winning.

The idea of being a good player is that it is easier for you to gain chips than it is for the average player in the tourney, and thus the percent of time you double up before going broke will be above 50% regardless of your chip stack. From the post I made earlier if this percentage is above 50 percent then you will be less than twice as likely to win the tourney after doubling up.

If you do not believe that you will be less than twice as likey to win the tourney after doubling up perhaps it is true. This would simply mean you have less than a 50 percent chance of doubling up before going broke and therefore you would be a worse than average player in the tourney.

[JNash]

In a ring game, you certainly think about pot-odds versus the probability of making a hand, right? In a ring game, the chip EV = $EV, so that's the right arithmetic. In a tournament, the two can differ significantly.

Being able to properly estimate the relationship between the two strikes me as a very important point to being able to play good tournament poker.

[JNash]

Aces, the example you provide, while true, is not compelling to me. Let me explain.

Assume equal skill. 1000 people in a tourney, 1000 chips each, everyone has a 1/1000 chance of winning. Check. 500 players left, 2000 chips each, 1/500 chance to win. Check again. You have doubled up and now have twice the chances to win. Still 100% correct, but...

The more interesting question is this: at any one point in the tournament, how much should an outside player pay to take over your stack--and how much should he pay for a stack twice as big? The answer to that question requires a different logic.

[Equal]

"If, for whatever reason, your chances of doubling up is greater than 50%, than your chances of winning the tournament, if you do double up, is less than twice what is was than before you did."

Try as hard as I can, I have no idea what you are trying to say here David.

[aces961]

First of all the example I used with the 1000 person tourney all of equal skill and half the people doubling up and the other half losing was only used as a specific counterexample to Eldog605's statement.

Now as to your question of "The more interesting question is this: at any one point in the tournament, how much should an outside player pay to take over your stack--and how much should he pay for a stack twice as big? The answer to that question requires a different logic. "

Let X equal the lower stack size and 2X the higher one. Let Y be the percentage of the time the outside player would double up before going bust if he had a stack of size X. Now let Z be the expected prize payout value of a stack size of 2X in this tourney for the outside player. The outside player should then pay Z dollars for the 2X stack size. Now consider what he should pay for the X stack size. (Y is refered to as a percentage but for any multiplication by Y it is a probability from 0 to 1). Y percent of the time the outside player will be able to turn his stack into size 2X so the value from this portion of the time is YZ. Now the rest of the time he goes bust and there is 0 equity in that. So the player should be willing to pay Y*Z for the stack of size X.

The ratio between what he should pay for a stack of size 2X to a stack of size Z is going to be Z/(YZ) or 1/Y. So if Y is 50 percent you get 1/.5 or a ratio of 2. If Y=60 percent, 1/Y=1/.6=1.66666... so you see the better the player is the less the ratio is in the value of the 2X stack to X stack.

In the above example it is assumed that the tourney is winner take all. In a non winner take all tourney this is not the case since you can gain ev simply by not losing chips if other people are eliminated. However if you early in the tourney the above calculations will still be very close to being accurate, only when you get very near the bubble will their be major ev difference from considerations along this line.

[aces961]

[ QUOTE ]
"If, for whatever reason, your chances of doubling up is greater than 50%, than your chances of winning the tournament, if you do double up, is less than twice what is was than before you did."

Try as hard as I can, I have no idea what you are trying to say here David.

[/ QUOTE ]

Simply put he is saying if you can double up with ease then having twice as much chips isn't as valuable to you as someone who finds it very hard to double up.

He is also saying that if you can double up before you go broke exactly 50 percent of the time then your chances of winning the tourney will be exactly twice as high with the doubled up stack as they were before the double up.

I hope that helps.

[Rushmore]

Obviously, I agree that it's important to know the ramifications of any decision one makes at the table in a tournament.

I simply do not believe that this exercise is pertinent to the overall thinking necessary to make solid decisions.

Again, if we're down to three and I'm in second and the guy in third is drunk or absent or having a seizure, I need to consider the odds of my immediate play, tempered by the other information I have available.

But generally-speaking, a player making a solid, relatively straightforward immediate decision will not have too terribly much tempering to do based upon just how much better overall position he will be in the tournament.

All things being equal, that is.

[eastbay]

[ QUOTE ]
Aces and David,
Let me reiterate Aces logic. 1000 people in a tourney, 1000 chips each, everyone has a 1/1000 chance of winning. Agreed.

[/ QUOTE ]

Unless this was intended to be a non-sequitur, go back and read the "Listen carefully" part again.

eastbay

[eastbay]

[ QUOTE ]
[ QUOTE ]
"This is a reaction to this entire thread. I would think that if one got double the chips, he'd have much more than double the chances to win the tourney."

Listen carefully. If, for whatever reason, your chances of doubling up is greater than 50%, than your chances of winning the tournament, if you do double up, is less than twice what is was than before you did. That statement is ALWAYS true and you don't have permission to leave your computer, or even read another post, until you are sure you understand why.

[/ QUOTE ]

For anyone who doesn't get it, the point is that if you have a better than 50% chance of doubling your money, that fact is already calculated in to whatever your chances are of winning the tournament, before you actually get the double-up.

Assume the following. You are in a 100 player tournament and have a 1% chance of winning. But if you double, you are so good at playing a big stack that even though you have 1/50 of the chips in play, you have a 1/40 (2.5%) chance of winning.

Since you are a good player, let's assume you have a 60% chance to double up before you bust. But wait! If you have a 40% chance of busting, and a 60% chance of doubling into a position where you will have a 2.5% chance to win the tournament, your actual starting odds were not 1% at all, they were 60% * 2.5% which is 1.5%.

[/ QUOTE ]

No, that is not it at all. You are one of those who doesn't get it.

The point is that AFTER the double-up you are NOT more than twice as likely to win it. You are LESS than twice as likely. And that's AFTER the double-up, not before.

eastbay

[fnurt]

Before the double up, you were 1.5% to win the tournament. After the double up, you were 2.5% to win. That sure sounds like less than twice to me.

[JNash]

Hello David

I understand your point, but would like to pose a slightly more complex and perhaps more realistic question.

The question I am after is the relationship between stack size and fair value of the stack AT A PARTICULAR POINT IN TIME, i.e assuming same # of players left, same distribution of chips, etc. If an outside player of a given skill level could buy a 20000 chip stack or a 40000 chip stack, should he pay more, less or exactly 2x for the latter?

You point out that if he has above average skill, the answer is that 40K is worth less than 2x the 20K stack. Fine so far.

But suppose that your probability of winning the next hand for any given 2 cards you are dealt CHANGES depending on your stack size relative to field???

If you are short stacked and enter a pot, you are more likely to get picked on--you have an adverse selection problem. Your chances of winning a bluff are less than if your stack were bigger, your chances of winning blinds are less, etc. For a big stack, the reverse is true--you are more likely to win bluffs, steal blinds, etc.

This leads me to what I'll call the "S-curve hypothesis."

Assume everyone has equal inherent skill. By the middle of a tournament, some players have bigger stacks than others simply as a result of luck/random walk.

Now plot a curve that has the "fair value" (FV) of a stack on the y-axis and the stack size on the x-axis. What is the "fair value" FV that an outsider should pay for different stack sizes? My hypothesis is that initially that relationship would be convex, and later on concave. It would be an S-curve, with the inflection point somewhere near the average stack size.

My heuristic proof is that if your chip count is quite low relative to the average stack size, not only is your chance of winning the tournament low (i.e. the FV is low), but your chances of winning the next pot with any given hand are lower than average (it's as if your skill were suddenly lower than average). From these low levels, doubling your chip count will MORE than double the FV of your stack (i.e. the function is convex here).

Conversely, if you have a big stack, your chances of winning the next pot are BETTER than those of the average stack. Even though you have the same inherent skill as all the others, your "effective" skill is now greater than average. Doubling your chips now less than doubles your FV (per your point).

If the S-curve hypothesis is true, a number of things follow logically:
1) If you are a short stack, you would be mathematcially correct to take some "gambles" where your CEV is negative--because the chips you win have a higher $EV than those you lose.
2) If you are a big stack, you would be mathematically correct to avoid some positive chip-EV situations because the chips you lose are worth more $EV than those you lose.

Please note that this does not necessarily imply that a large stack should play fewer hands than a small stack. Remember, the probabilities a big stack would use to calculate the chip EV are better than those of a short stack--even if you have been dealt identical cards. While the shape of the FV-chip count relationship would lead a big stack to be more conservative, the fact that his chip-EV is much better than average may overcome this and lead him to correctly play more hands than a short stack.

Carrying this thinking further, one can speculate how the shape of this S-curve changes as the tournament progresses, and as the size of the blinds changes relative to the average stack size. Obviously, as fewer players are left, the y-axis intercept moves up (e.g. if you're in the money even a small chip count is guaranteed something). As the blinds get big relative to the average stack size, I would think that the curve would get "more bowed"--i.e. it becomes more like a step-function--low chip levels are virtually guaranteed not to advance much further, while big stacks are virtually guaranteed to do well. At the low end of the curve, it has become more convex, and for above-average chip counts it becomes more concave.

Sorry to ramble on for so long, but this has been bouncing around my head for quite some time...

[David Sklansky]

My point is not about tournaments per se. It is pure irrefutable logic and it is unacceptable if anyone on this forum doesn't understand it. I'm hoping I don't have to explain it.

[fnurt]

You are in the wrong forum to be so smart /images/graemlins/wink.gif but what you say sounds correct to me.

[JNash]

Hi Aces.

Thanks, your explanation is clear and compelling.

Now, take a peak at my othe post regarding the S-Curve Hypothesis in this thread. What if the probability of winning a hand (your "effective skill level") changes depending on your stack size? I believe the answer then becomes that for low stacks, doubling up is worth more than 2x, while for large stacks it is less than 2x.

[Rushmore]

I would have posited that your point was that the more chips one has, the less value each chip has, thusly diminishing the overall value of one's stack relative to a smaller stack, and therefore diminishing the odds of winning the tournament relative to the percentage increase in chips achieved.

But then you made that disclaimer about tournaments per se, and now I'm in the weeds again.

You are rolling this out like it's the easiest thing in the world.

I seem to recall a similar tone in TPFAP ("...if you do not know how I got this number, you are not ready for this book...").

I read the book anyway.

[eastbay]

[ QUOTE ]
Before the double up, you were 1.5% to win the tournament. After the double up, you were 2.5% to win. That sure sounds like less than twice to me.

[/ QUOTE ]

Huh?

You said: "You are in a 100 player tournament and have a 1% chance of winning."

Ok, 1%.

"But if you double, you are so good at playing a big stack that even though you have 1/50 of the chips in play, you have a 1/40 (2.5%) chance of winning."

Last time I checked 2.5% is more than 2x 1%.

eastbay

[fnurt]

You shouldn't have stopped reading. I went on to demonstrate why the initial 1% chance of winning, while it might be what most people would assume, is inaccurate if you have a 60% chance of doubling up.

[eastbay]

[ QUOTE ]
You shouldn't have stopped reading. I went on to demonstrate why the initial 1% chance of winning, while it might be what most people would assume, is inaccurate if you have a 60% chance of doubling up.

[/ QUOTE ]

So let me get this straight. You started from a false premise, and then from that, logically derived a conclusion which was, in fact, the correct premise? QED?

Uh, ok.

eastbay

[Eldog605]

yeah, I would have to say that was one of the smartest, yet best explained "theoroms" I've read in some time. Its nice to hear a mix of math and the english language.

[fnurt]

My post was clear to others, I apologize if it wasn't clear to you.

Let me try to explain it in mathematical terms.

At the beginning of the tournament, your chance of winning is X. If you successfully double up, your chance of winning will be Y.

We have arbitrarily assumed that your chance of doubling up before busting is 60%. What David Sklansky was pointing out was that this assumption, in and of itself, defines the mathematical relationship between X and Y. Since you have a 60% chance of increasing your winning chances from X to Y, X must be 60% of Y. Thus, you have not doubled your winning chances by doubling your stack; you have increased them from 0.6Y to Y.

By the same token, if you have less than a 50% chance of doubling up before you bust, then doubling up will always more than double your chances of winning the tournament. As Sklansky said, this is simply a mathematical fact.

[eastbay]

[ QUOTE ]
My post was clear to others, I apologize if it wasn't clear to you.

Let me try to explain it in mathematical terms.

At the beginning of the tournament, your chance of winning is X. If you successfully double up, your chance of winning will be Y.

We have arbitrarily assumed that your chance of doubling up before busting is 60%. What David Sklansky was pointing out was that this assumption, in and of itself, defines the mathematical relationship between X and Y. Since you have a 60% chance of increasing your winning chances from X to Y, X must be 60% of Y. Thus, you have not doubled your winning chances by doubling your stack; you have increased them from 0.6Y to Y.

By the same token, if you have less than a 50% chance of doubling up before you bust, then doubling up will always more than double your chances of winning the tournament. As Sklansky said, this is simply a mathematical fact.

[/ QUOTE ]

Ok, I do see what you meant. I do, however, think it was written in a misleading way.

I apologize for jumping on you.

eastbay

[Slo Pok]

J Nash commentary:

Interesting combination of left brain and right brain application. Beyond the linear and sequencial thought patterns of left brain logic in my opinion.

Thanks /images/graemlins/smile.gif

[eastbay]

Nice post.

There does appear to be a self-balancing effect in what you describe, doesn't there?

The advantages you describe of the big stack I think can be called aggressive tactics - more bluffing opportunities, more blind stealing opportunities.

But the consequence is that the big stack should play more conservatively - pass up small +chipEV situations, as you say.

It seems there is a balancing effect - if the bigger stack has an advantage in being aggressive, the conclusion is that he should play more conservatively?

Can you give any more detail about how you reconcile this?

eastbay

[TStoneMBD]

I fully understand DS's last post, but I agree with JNash's standpoint here and hope that DS has the time to argue if JNash's logic is flawed. JNash's standpoint was what I was trying to define earlier, but am not as sophisticated with the terminology as he is.

[aces961]

[ QUOTE ]


My heuristic proof is that if your chip count is quite low relative to the average stack size, not only is your chance of winning the tournament low (i.e. the FV is low), but your chances of winning the next pot with any given hand are lower than average (it's as if your skill were suddenly lower than average). From these low levels, doubling your chip count will MORE than double the FV of your stack (i.e. the function is convex here).

Conversely, if you have a big stack, your chances of winning the next pot are BETTER than those of the average stack. Even though you have the same inherent skill as all the others, your "effective" skill is now greater than average. Doubling your chips now less than doubles your FV (per your point).


[/ QUOTE ]

The initial comment David made was for winner take all tourneys, and we were commenting on the probabilities of winning the tourney not about ev in a tourney with a very complicated pay schedule. When this pay schedule becomes very complicated as say a party 1000 person tourney with 100 different payouts then things can change. I would ask you to note the flatness of most payout schedules after they have jumped to zero to right above the buy in. Most of them stay pretty flat until you get to the final table, online at least. The winner take all tourney calculations are a pretty good approximation of how you should play if you aren't nearing the final table or nearing the initial start of payouts in these type of tournaments.

In non bubble situations I believe that the tourney should be played similarly to a cash game, thus it should still be possible for a smaller stack to be able to play well enough to be able to double up over half the time. Now if a player is truly worldclass in any reasonable situation his chances of doubling up before going broke should be above .5 unless he is playing with other worldclass players. While this world class player may only have a chance of doubling up with the smaller stack of .501, while with the larger stack above the average stack it is .6.

Of course it is going to be possible for a player to have a chance of doubling up with the small stack of under .5 and with the large stack of over .5, and in this case obviously what you say will be correct. I'm just saying that this player might think there is nothing he can do about this fact but as long as the blinds are still reasonable in comparison to his stack and we arn't in a bubble situation with work his game should be able to improve to get this probability of doubling up with a smaller stack to above .5.

[pzhon]

[ QUOTE ]

My heuristic proof is that if your chip count is quite low relative to the average stack size, ... your chances of winning the next pot with any given hand are lower than average (it's as if your skill were suddenly lower than average). From these low levels, doubling your chip count will MORE than double the FV of your stack (i.e. the function is convex here).

Conversely, if you have a big stack, your chances of winning the next pot are BETTER than those of the average stack. Even though you have the same inherent skill as all the others, your "effective" skill is now greater than average. Doubling your chips now less than doubles your FV (per your point).

[/ QUOTE ]
I disagree. It's a good thought experiment, but I think your "heuristic proof" is specious.

If your chip count is extremely low, you don't have much bluffing power. However, it is very easy to value bet, and if you expect to be all-in preflop you don't have to worry about getting pushed off a weak draw later. Your chance of winning the next pot may be lower than average (that is not clear), but when you win you may more than double up. It is not uncommon for a small stack (say, with 1-3 BB) to 5-tuple up by pushing after limpers.

If you have a small stack, you can play optimally to take everyone else's chips. The large stacks have to worry about each other, so they can't act as accurately to take your stack. You can play hands that do well with short stacks, e.g., large unsuited cards gain in value and small pocket pairs lose value. The large stacks can't make the same adjustments or else they can lose to the other large stacks.

Imagine that you have a small stack, and observe the other players while ignoring any chips in excess of your own. The other players play terribly! They often fold while all-in, even with clear chances to win. They call based on implied odds that clearly are not present. Sometimes these "mistakes" only help the other big stacks, but often you benefit.

If you don't ignore the extra chips, you get an information advantage: Large stacks have the option to bet or raise more than the size of your stack. That they make large bets or fail to do so gives you extra hints about where you stand.

If you have a small stack, you should expect to gain chips on average. This suggests that the probability of winning is a sublinear function of your stack size. In addition, with a small stack you should get a disproportionate share of second place and lower prizes. This also argues that the chip value is sublinear.

If you have a large stack, you have the ability to bluff other large stacks off small pots. However, that comes with the weakness that you can get bluffed off a small pot. You can't value bet as easily, and you suffer from implied odds. You may win more pots, but lose a lot when you lose.

Though the probability of winning the tournament may be a superlinear function of your stack size when your stack is large, you get a disproportionately small share of second place and lower prizes. Since these effects are in opposite directions, it is not clear whether chip value should be superlinear or sublinear when you have a large stack.

[GimmeDaWatch]

Wow, this is a pretty long thread for a simple logical exercise. David's post was just a reply/contradiction to someone else's. Anyway, his point is about a "good player", i.e. someone with over a 50% chance of doubling through to 40K. How much more likely the player is to win the tourney once he gets to 40K is simply 1/% chance of his making it there. If he were poker god #1 and was 100% to double through, then he would be equally likely to win the tourney at 40K as he was at 20K (and he would be the inevitable winner of course). If he was 51% to double up, then he would be slightly less than twice the favorite once he got to 40K, and so on.

[SossMan]

[ QUOTE ]
If he were poker god #1 and was 100% to double through, then he would be equally likely to win the tourney at 40K as he was at 20K (and he would be the inevitable winner of course). If he was 51% to double up, then he would be slightly less than twice the favorite once he got to 40K, and so on.


[/ QUOTE ]

That's pretty much how I thought about what DS said. Take it to the extremes, and it makes it easier.

[tpir90036]

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


[/ QUOTE ]
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

[aces961]

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


[/ QUOTE ]
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

[/ QUOTE ]

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]

[ QUOTE ]
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

[/ QUOTE ]

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]

[ QUOTE ]
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?

[/ QUOTE ]

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

[SossMan]

[ QUOTE ]
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

[/ QUOTE ]

i'd say that's darn near 1.000.

[Eldog605]

Today I had the chance to triple up. I had 10's and I knew one opponent could have anything, he was a maniac. And I figured the other had two over cards. We all called all in (i had short stack). Crazy man flipped over A-7 offsuit, other guy flipped A-Q of diamonds. So, I'd like to first of all know what my chances were going into the flop of winning this hand.

Second, if we assume I am a good player (this is wholly debatable), and I have a 60 % shot at doubling before going broke, what would be the correct percentage to seek in order to try and triple up? A 55%, 75%? Basically, I'm asking if my move made sense, considering the fact that I had 60 % to double anyway. Would a player who doubles up 80% of the time before going broke make the same move?
Thanks in advance

[aces961]

[ QUOTE ]
Today I had the chance to triple up. I had 10's and I knew one opponent could have anything, he was a maniac. And I figured the other had two over cards. We all called all in (i had short stack). Crazy man flipped over A-7 offsuit, other guy flipped A-Q of diamonds. So, I'd like to first of all know what my chances were going into the flop of winning this hand.

Second, if we assume I am a good player (this is wholly debatable), and I have a 60 % shot at doubling before going broke, what would be the correct percentage to seek in order to try and triple up? A 55%, 75%? Basically, I'm asking if my move made sense, considering the fact that I had 60 % to double anyway. Would a player who doubles up 80% of the time before going broke make the same move?
Thanks in advance

[/ QUOTE ]


You win the hand 53 percent of the time. (could vary by a slight amount due to exact suits that your tens and the a7 were)

When we say that a good player has a 60 percent to double up before going broke this is a generalization included in his play. Therefore the all these situations you can bring up are part of what gives him that 60 percent chance. You shouldn't change how you play for the reason of trying to get to that 60 percent, but you should change plays you make because they aren't the correct plays.

Now going all in here is with the short stack is a very good play and somewhat obvious against the maniac.

Now if a good player has a 60 percent chance of doubling up. He is going to have a .6^1.5 chance of tripling up this comes out to about 46.5 percent of the time.

[sdplayerb]

sorry, but you are way overestimating the intelligence of people.

[Charon]

This is simply Bayes Theorem.

Let x be that a player doubles up and
y that the player wins the tournament. Then:

p(y|x)=p(x|y)*p(y)/p(x)

Since p(x)>1/2 follows:

p(y|x)<2*p(x|y)*p(y) and since p(x|y) is 1 (because if he wins the tournament he had to double up there (otherwise he would have busted out) follows:

p(y|x)<2*p(y)

So..guess I saved Sklansky some work.

[Pensive Gerbil]

Thank you for contributing...sorry I missed the fun! I have not yet read any of the posts that follow yours. I thought it was common knowledge among well-read tournament players that if everyone is equally skilled, then the chance of taking first place is generally proportional to the percentage of the total chips that are in your stack.

As you point out, this simple principle does not hold for players whose skills are above average or below average. To elaborate by addressing your question: The probability that an above-average player with 20K will double to 40K before going broke is greater than 50% (except perhaps if he is so short-stacked that he is in danger of being blinded off). Therefore, his chances of winning with 40K must be less than twice what it would be with 20K. To further illustrate, suppose our above-average hero has a 55% chance of doubling-up. If the total chips in play were 80K, then 40K would give our hero a 55% chance of victory. If our hero controlled 20K in chips, his chance of doubling to 40K would be 55% and his chance of doubling again to 80K would be 55%. Therefore, 20K would give our hero a .55 x. 55 or 30.25% chance of victory. As you can see, 55% is less than twice as much as 30.25%. [Due to rising blinds, our hero's chance of doubling up may decline over time. As long as it does not fall below 50%, however, this principle remains valid.]

It is easy to see that in the case of a below-average player, 40K will provide more than twice the chance to win the tournament as 20K in chips.

I'm afraid I won't have time to read or respond to more posts until possibly Sunday.

Regards,

PG

[Pensive Gerbil]

Now that I (and presumably others) have elaborated, I would much appreciate any further comments you may have on the original topic of this thread. I expected you to agree that above-average players should be generally less inclined to chase big pots with marginal odds than below-average players. Do you also agree that this practice would be better (or less bad) for good players who are desperately short-stacked (since they will likely have less opportunity to benefit from their skill edge)?

What do you think of the argument that some above-average players are justified in chasing big pots with slightly inadequate odds because their skill edge with a large stack is greater than their skill edge with a lesser stack? I suspect that in reality, this potential skill edge differential rarely justifies chasing pots with inadequate odds, due to the offsetting value that superior players realize by minimizing the variance of their tournament play.

Regards,

PG

[David Sklansky]

Agreed. Daniel is wrong (for himself). He would never realize it however merely from experience since it is no big deal either way. Notice however that the concept is correct as regards that rare bird who plays well with a big stack but is less than even money to double up with a small stack.

[JNash]

IF my hypothesis is correct..(and I will need to post it in the theory forum to get more comments), I believe that for a big stack there are two opposing forces at work:

1) Part 1 of the hypothesis is that the relationship between chip count and fair value (FV) of those chips (I'll call this the "payoff function") is concave for big stacks, and convex for small stacks, and the payoff function has its inflection point (switching from convex to concave)at the average chip count.

Incidentally, saying that the payoff function is concave is the same as saying that each chip you win is worth less than each chip you lose. (A commonly accepted statement). Saying that the payoff function is convex is the same as saying that each chip you win is worth MORE than each chip you lose. I claim that this is the case for short stacks. (This latter statement is, I believe, more controversial.)

If this is true, then a 50/50 situation (with a chip-EV of zero for both sides) will have a negative EV as measured in fair value terms for the big stack, and a positive one for the small stack. The implication is that short stacks should seek out chip-EV coinflips, while big stacks should avoid them. Or, to put it differently, short stacks can be correct from a FV-EV perspective to play hands that have negative chip-EVs. (Just how negative it can be and still be justified depends on how convex the payoff function is). [This is also the part of this post that relates to the original subject of the Negreanu thread--i.e. why and when it might make sense to go for big pots when you're seemingly getting the worst of it.]

2) The second part of the hypothesis is that big stacks have an a priori (i.e. before the cards are dealt) higher probability of winning the next hand than a small stack. That's because they get more respect, and so their bluffs and blind-steals have a higher probability of being successful.

If this is true, then big stacks can and should play more aggressively (open-raise pre-flop, bluff-raise or semi-bluff raise, etc.) and are actually correct (in the sense of having positive chip-EV) even though a short stack in the same situation might have a negative chip-EV.

These two forces point in opposite directions: avoiding 50/50 bets sounds like a "conservative" thing to do, while playing more hands (and raising them aggressively) sounds like a more "aggressive/loose" way to play.

While these two forces seem at odds, they can actually co-exist. If the big stack has a higher probability of being successful with bluffs and aggressive plays, then the big stack can enter more pots (i.e. situations that might have been negative chip-EV for a short stack become positive EV for the big stack.) As a result, the big stack is correct to play looser and more aggressively.

Now in situations where the big stack's clout has no value (e.g. when calling an allin bet from a short stack), the only thing that matters are the objective odds based on the cards. In this case, the concavity of the payoff function for the big stack (and the convexity of the function for the short stack) says that a 50/50 situation (with zero chip-EV expectation for both sides) are bad for the big stack and good for the short stack.

So while the two forces go in opposite directions, they do not contradict each other. Hope that helps...

[JNash]

[ QUOTE ]


Of course it is going to be possible for a player to have a chance of doubling up with the small stack of under .5 and with the large stack of over .5, and in this case obviously what you say will be correct. I'm just saying that this player might think there is nothing he can do about this fact but as long as the blinds are still reasonable in comparison to his stack and we arn't in a bubble situation with work his game should be able to improve to get this probability of doubling up with a smaller stack to above .5.

[/ QUOTE ]

That is the essence of the S-curve hypothesis. Even if all players have equal skill, their chip count influences their probability of winning. You state it in terms of the probability of doubling up, but I believe the point is even more general: a priori, before the cards are dealt, the short stack has a negative chip-EV and the big stack a positive chip-EV. Incidentally "by symmetry" (since this is after all a zero-sum game) if there were only two players, the amount of positive chip-EV for the big-stack is exactly equal to the amount of negative chip-EV of the small stack.

The practical question is when in a tourney this S-curve effect actually becomes pronounced, versus when the payoff function is simply linear. I believe that when the blinds are low compared to the average chip count, the curve is pretty much linear, and you ring-game EV calculations are correct. But as the blinds increase relative to the average stack size, I believe the curve becomes more S-shaped--certainly around bubble-time, but even beyond that.

Further, I believe that my hypothesis holds true no matter how steep the payout structure--i.e. it would include winner-take all situations. Mind you, the exact shape of the payoff function will certainly depend on the particulars of the payout structure, but the general convex-concave feture should not be affected by this.

I even believe (gasp--dare I say it?) that this applies to heads-up situations in no-limit winner-take-all settings, as long as the blinds continue to increase regularly. [I say this knowing that this sounds heretical and contrary to all accepted gospel, but I'll just need to go to the theory forum and take my licking...]

[JNash]

You make some very good points, such as the fact that a short stack is far more likely to be able to value-bet his good hands and get called. My counter to this is that the short stack gets into positions where he wants to value-bet so infrequently that this is not worth as much as the big stack's better chances at winning with bluffs.

I have purposely called this a "hypothesis" because I can't prove it, yet believe it to be true. The crux of the discussion is whether a short stack, beyond the obvious point that his stack is worth less than a bigger stack, is at a tactical advantage or disadvantage compared to a bigger stack.

I claim that because "big stacks get more respect" the big stack is in a better tactical position--i.e. a priori (before the hand is dealt) he has a positive chip-EV, while the short stack has a negative chip-EV.

You have listed some good reasons why a short stack might have an advantage, so the question is "net-net" which of these factors is more important.

I'll give you two things to ponder--one empirical, the other theoretical.

The empirical point is simpy based on what I've observed happen at various stages in a tournament. For example, at bubble time, there can be little question that short stacks play very conservatively (whether this is optimal or not is beside the point, it's simply a fact). At this point in a tournament, big stacks can (and often do) mop up lots of chips. This is consistent with saying that their a priori chip-EV is positive at this stage of the tourney.

Empirically, I would also say that even post-bubble, short stacks can be "forced" to play hands that they don't want to simply because the rising blinds are putting pressure on them. The "don't want to" means that they play even if they face negative chip-EVs.

My second argument is purely theoretical. It has been stated in many books (including TPFAP and Gambling Theory) that the payoff function is concave -- i.e. the chips you win are worth more than the chips you lose. Since the tournament is a zero-sum game, when you compute the fair value of the chips across all players, they have to add up to the total prize pool. So if player A has a concave function, there must be some player B who has a convex function. My contention is that the big stacks all have concave functions, and the short stacks convex ones. (This assumes equal skill -- with unequal skills this would be much messier, but I believe still directionally correct.)

[JNash]

It's definitely not based on the idea of two players being dealt the same cards in the same hand. It is simply a mental experiment that given any given set of two cards, a big stack can make more of it (has a higher chip-EV).

I believe that the ability to survive a run of "unplayable" cards is indeed part of the argument. A big stack can actually "play" some cards that a short stack wouldn't want to play. If you add the big stack's chances of winning the pot "fair and square" (i.e. play it to showdown) to his ability to force the opponent away from the pot, his probability of doing the latter allows him to play far more pots and thus do better even when the cards are unplayable.

At this point I can't see that position has much of an influence on the shape of the payoff function--except in extreme cases like when your stack is down to one big blind. In that case, the fair value of your chips is higher if you still have 8 hands to get lucky or whether you're forced to play the next hand.

[JNash]

It somewhat depends on exactly what you mean by "conservative" and "aggressive." You may want to read my answer to eastbay's post above.

My short answer is that the fact that the payoff function is convex for the short stack means that he is CORRECT (positive fair value-EV) to take 50/50 chances, or even some negative chip-EV situations. In other words, he "should" gamble it up.

However, the short stack can't afford to play as aggressively as the big stack (in the sense of trying to bully people out of a pot with sub-par cards), because he'll get called, so he needs to be far more selective as to when he picks his spots. His only saving grace (see pzhon's post) is that when he does enter a pot with premium cards, he's quite likely to get full value.

[Pensive Gerbil]

When I read Negreanu's article, I was surprised by what seemed to me to be his misunderstanding of basic tournament theory. I suppose it shows that excellent math/logic skills are not that important.

Sorry...I couldn't resist! /images/graemlins/wink.gif I believe Daniel N. is a very talented and courageous player who may have a couple of leaks in the area of logical/theoretical thinking. Perhaps he's been lucky that his particular logical leaks have not led to more costly errors.

-PG

[Ezcheeze]

Whether or not having a larger stack helps you win the next hand is irrelevant. If your chances of doubling up are greater than .5 then, as many poeple have shown previously in the thread, your chances of winning the tourny do not double after you have doubled your chips.

It seems like what you might be saying is that when a skilled player has a small stack then his chances of doubling up decrease. I doubt that this is true, and it would have to decrease to less than .5 in order for him to want to take -EV situations in the hopes of doubling up. If this was true then the stack size it is true for is surely so small that the whole concept is useless anyway.

-Ezcheeze

[REL18]

I have a quick question for you probably not gonna answer but only takes a yes or no answer. Do you know Barry Carpenter