Testing ICM -- some questions for discussion

[the shadow]

I'm not sure I agree. Isn't that asking us to ignore the man behind the curtain?

If your argument were correct, why wouldn't it equally apply in a SNG? At the bubble, the big stack has 5,000 chips, the two medium stacks have 2,000 chips, and the small stack has 1,000 chips. We don't ignore the big stack's extra chips then. Why should we HU?

The Shadow

[marv]

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I'm not sure I agree. Isn't that asking us to ignore the man behind the curtain?

If your argument were correct, why wouldn't it equally apply in a SNG? At the bubble, the big stack has 5,000 chips, the two medium stacks have 2,000 chips, and the small stack has 1,000 chips. We don't ignore the big stack's extra chips then. Why should we HU?

The Shadow

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With 3+ players you won't always have the symmetry I used to guarantee each hand is 'fair'.

In your example, with stacks of 1, 2, 2 and 5, if we randomize seats and button position at the start of each hand (to avoid the complications of position - which is a much bigger change in 3+ players than with 2 anyway) then the next hand is effectively played with stacks of 1,2,2,2 .

This isn't symmetrical, so the small stack may be at a disadvantage (in terms of EV in chips) on this hand. And given the small stack may be at a disadvantage, the middle stacks must worry about becoming the small stack on later hands - when they suffer.

But with only 2 players, on every hand (with randomized button again), it's always symmetrical so each player can play a strategy which gets at least 0 EV in chips over that hand, and hence gets them at least the equity that the linear model suggests - so no S-shaped equity function.

There's another complication with 3+ players. The variance of our opponents effects our equity. In a 2 player game, the variance of each 'step' in the random walk our stack follows doesn't effect the linear equity function (provided every step has mean 0), but with 3 players, insanely loose agressive opponents increase our equity (for any sensible payout structure)

Marv

[eastbay]

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In my example of a freezeout tournament with antes but no blinds and where the players flip a coin for the button at the start of each hand, let's imagine we're the big stack at the start of a hand (before the coin flip). Here we will still be able to force oppo to 'play for the tournament', but our extra chips have no bearing on the up-coming hand as we'll never get to use them. Thus the hand we're about to play is a symmetrical game and we have no edge(*) in it.

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We do get to "use" them, in the sense that they prevent us from getting frozen out if we lose. This is the key point. Our N chips at risk in the hand have less value than small stack's N chips. At least, this seems plausible to me.

eastbay

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But we can't use the extra chips on that hand, so that hand (in isolation, forgetting about the subsequent hands in the tournament)


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Well, that's the whole point. I don't think you can forget about the rest of the tournament and arrive at a legitimate conclusion.

The hands in a tournament are just turns in the game, they are not independent games themselves.

eastbay

[marv]

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In my example of a freezeout tournament with antes but no blinds and where the players flip a coin for the button at the start of each hand, let's imagine we're the big stack at the start of a hand (before the coin flip). Here we will still be able to force oppo to 'play for the tournament', but our extra chips have no bearing on the up-coming hand as we'll never get to use them. Thus the hand we're about to play is a symmetrical game and we have no edge(*) in it.

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We do get to "use" them, in the sense that they prevent us from getting frozen out if we lose. This is the key point. Our N chips at risk in the hand have less value than small stack's N chips. At least, this seems plausible to me.

eastbay

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But we can't use the extra chips on that hand, so that hand (in isolation, forgetting about the subsequent hands in the tournament)


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Well, that's the whole point. I don't think you can forget about the rest of the tournament and arrive at a legitimate conclusion.

The hands in a tournament are just turns in the game, they are not independent games themselves.

eastbay

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OK, you're still not convinced so I'll try a slightly different tack:

First, note even if both players play an identical strategy (so they're certainly equally skilled), we may *not* get a linear equity function:

Example: suppose each player looks at his chips at the start of the hand, if it's <25% of the total, he just folds at his first action. If its 25%-75% he calls at every opportunity and if its >75% he makes a min raise at his first opportunity.

For these players the equity function looks like _/- it starts at 0, is horizontal up to 25%, then rises linearly to reach 1 at 75% and is then horizontal again until 100%.

So for linear equity we need some assumption about the players' common skill level.

Now suppose both players play the entire tournament optimally (in the game theory sense, so they each know each other's playing tendencies perfectly, in all situations and so play unexploitably).

Each player can guarantee an EV of at least 0 over each hand, as each hand (in isolation) is a symmetric game (I'm assuming the button is randomized). Thus each player can guarantee themselves least the equity given by the linear equity function, by just adopting the optimal strategy over each hand for all the remaining hands in the tournament.

But if player 1 can guarantee himself an equity of x% with x% of the chips, then the most player 2 can hope to get is (100-x)%. But he can also guarantee at least this by adopting an optimal strategy over each hand too, hence he should do this, and with (100-x)% of the chips, his equity is exactly (100-x)%, and we get the linear equity function.

Marv

[the shadow]

I've been thinking a bit more about your graph.

It may be trivial, but it seems to me that in the case of a HU freezeout, if the skills of the two players differ, the equity of player A cannot equal that of player B when the chips are equal. Do you agree and what can we infer from this?

The Shadow

[holdem2000]

I think bot data would be far better than human data for this - you're looking for situations in which player A and player B have identical strategies (which can of course vary with stack size, but the whole strategy for various chip sizes must be the same).

[eastbay]

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OK, you're still not convinced so I'll try a slightly different tack:

First, note even if both players play an identical strategy (so they're certainly equally skilled), we may *not* get a linear equity function:

Example: suppose each player looks at his chips at the start of the hand, if it's <25% of the total, he just folds at his first action. If its 25%-75% he calls at every opportunity and if its >75% he makes a min raise at his first opportunity.

For these players the equity function looks like _/- it starts at 0, is horizontal up to 25%, then rises linearly to reach 1 at 75% and is then horizontal again until 100%.


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Yes, I've pointed variations of this out as a counterexample to DS's "proof" many times.

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So for linear equity we need some assumption about the players' common skill level.

Now suppose both players play the entire tournament optimally (in the game theory sense, so they each know each other's playing tendencies perfectly, in all situations and so play unexploitably).


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That's not what game theory optimal is. No knowledge of the opponent's tendencies is required for game theory optimal play. In fact, that's kind of the whole point of game theory optimal - that you don't need to know anything about your opponent's tendencies to not lose to him. That's just an aside here.

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Each player can guarantee an EV of at least 0 over each hand, as each hand (in isolation) is a symmetric game (I'm assuming the button is randomized).

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Grr.

There is no reason to believe that optimal play for a single hand (to achieve 0 chip EV) is optimal play for the tournament. This is something you keep assuming and it's still unsubstantiated.

Giving up chip EV on one hand may allow the finding of a bigger edge later, producing better results for the tournament, even if it gives up something on some particular hand. The tournament is the game. The hands are not.

Let me repeat:

You have given no evidence that optimal play of each hand produces optimal play for the tournament.

Therefore, no conclusions about tournament equity functions for optimal play for the tournament can be derived from optimal play for each hand.

eastbay

[marv]

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Each player can guarantee an EV of at least 0 over each hand, as each hand (in isolation) is a symmetric game (I'm assuming the button is randomized).

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Grr.

There is no reason to believe that optimal play for a single hand (to achieve 0 chip EV) is optimal play for the tournament. This is something you keep assuming and it's still unsubstantiated.


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OK eastbay, here's a proof.

Suppose A and B are playing a HU tournament starting with 50 chips each. They play each hand with a randomized button, and at some point the tournament, at the start of a hand (before they flip a coin for the button) A has x chips, and B has 100-x.

Claim 1: A can play the remainder of the tournament in such a way that

P(A wins the tourney | A starts with x chips) >= x/100

no matter how B plays. A suitable strategy for A is to play each hand using an optimal strategy for the game G which represents the 'hand in isolation'. This is the game which starts with a chance node representing the selection of the button, ends at the end of the hand, and whose payout for each player is simply the number of chips they have at the end of the hand. Different hands will have a different game G, so A's strategy will change from hand to hand.

Claim 2: The equity function for a player who plays optimally for game which represents the remainder of tournament as a whole (i.e. the game which ends when one player has all the chips) lies on or above the linear equity function, and hence the equity function for a pair of players each using an optimal strategy for the remainder of the tournament as a whole is the linear equity function.


Proof of 1
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Let ca denote the number of chips A currently has (so ca=x) and cb the number of chips B has (so cb=100-x). Recall that G denotes the 2-player game which amounts to stopping the tourney at the end of the next hand and paying each player in proportion to their chips at that point.

If ca>cb, the extra chips A has play no part in the hand, except to increase the value of the game G by (ca-cb) since the extra chips never end up in B's stack. Similarly if ca<cb, B's extra chips just increase the value of the game in his favour by cb-ca since they'll always be his at the end of the hand.

Thus G is a symmetric game (other than one player getting a constant bonus payoff), so if A plays an optimal strategy for G, we have E(CA) >= ca, however B plays, where CA denotes the (random) number of chips A has at the end of the hand.

Now consider the number of chips A has at the start of each successive hand in the remainder of the tourney if he uses the method just given to dicate his play on each hand. Denote this sequence

CA_0, CA_1, CA_2, ...

Where CA_0 = x since A currently has x chips.

Since E(CA_{n+1} - C_n | all information up to the start of hand n) >= 0, we can write the process CA_n as

CA_n = CA_0 + M_n + Z_n

where M_n is an 'unbiased random walk' (a martingale) with M_0=0, and Z_n is a random, non-decreasing process with Z_0=0.

A standard result for martingales is that

M_0 = E(M_T)

for any suitable 'stopping time' T. An example of a suitable stopping time here is T={least n for which CA_n=0 or CA_n=100}, i.e. T is the moment after the last hand of the tournament.

Thus

E(CA_T) = CA_0 + 0 + E(Z_T)

and note that the LHS is

0*P(A loses the tourney|A starts with x chips) + 100*P(A wins|A starts with x chips)

since at time T we'll either have CA_T=0 if A lost or CA_T=100 if he won. Also note that the RHS is >= x since CA_0 = x, Z_n is non-decreasing and Z_0=0.

Thus we have

100*P(A wins | A starts with x chips) >= x

This compeletes the proof.


Proof of 2
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Any player with x chips can guarantee themselves an equity of at least x% no matter how oppo plays (from the claim), so a player playing optimally for the tournament as a whole must get at least this. In other words if A plays using a strategy which is 'tournament optimal' then

P(A wins|A has x chips) >= x/100.

This proves the first part of claim 2.

But if B uses the strategy indicated in the proof of the claim, then B's equity is at least

P(B wins|B has 100-x chips) >= (100-x)/100.

Hence as

P(A wins|A has x chips) = 1-P(B wins|B has 100-x chips)

we have

P(A wins|A has x chips) <= x/100,

and so

P(A wins|A has x chips) = x/100.

which proves the second part.

Marv

[eastbay]

There's a lot here for me to figure out and/or be convinced of. So I may not respond for awhile, but that doesn't mean I'm ignoring your demonstration. It looks very interesting and may settle some key points that are always in dispute when discussing this sort of thing.

Thanks,
eastbay

[eastbay]

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There's a lot here for me to figure out and/or be convinced of. So I may not respond for awhile, but that doesn't mean I'm ignoring your demonstration. It looks very interesting and may settle some key points that are always in dispute when discussing this sort of thing.

Thanks,
eastbay

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Marv,

I still haven't quite figured out your proof but I do believe it is correct. I'll post why shortly in a new thread.

eastbay