[0,1]- all-in or fold two player tournament

[well]

I'll keep it short but will post more numbers if anyone seems to be interested...

This is about a well-known poker game simplification where each player is dealt a

sample of the uniform [0,1] distribution. In a show-down, the highest number wins.

There have been loads of posts about this, with different betting structions, but to

my knowledge never in a tournament set-up. I have done some calculations for this

situation and the main result is this:

The probability of winning the tournament is not equal to the player's chip share.

The correct strategy assuming the opposite is exploitable.

These are the betting rules:

- the SB posts 1
- the BB posts 2
- both players recieve a "hand"
- the SB decides to fold or go all-in
- if the SB folds, he loses his 1 to the BB
- if the SB raises, the BB either calls or folds
- if the BB folds, he loses his 2 to the SB
- if the BB calles, the highest "hand" wins the pot

For the tournament version:

- all-in means raising to the amount of chips the shortstacked has
- the SB and BB roles are switched after each hand
- the player who loses all his chips loses the tournament

At the beginning of the tournament, each player recieves t tournament chips; so

there are t big blinds in play. A coin is tossed to decide which player starts

in the BB.

I calculated the cash equilibria ($), where EV's are in terms of chip amounts,

tournament equilibria (T), where EV's are in terms of winning probabilities and

the best strategies against the cash optimal strategy (^); for the values

t=4, 5, ..., 20.

Furthermore, I ran the tournament strategies against each other (TT), the

tournament strategie against the cash strategy (T$) and the cash

counter-strategty against the cash strategy (^$).

For t=20,

- the probability of player I winning is:

TT: .50
T$: .52
^$: .57

- with a 1:3 chip disadvantage, in the SB:

TT: .24
T$: .30
^$: .38

- in the BB:

TT: .25
T$: .30
^$: .39

- 3:1 chip advantage, SB:

TT: .75
T$: .76
^$: .79

- BB:

TT: .76
T$: .77
^$: .80

Are these numbers too close to .50, .25 and .75 to matter? Not in my opinion. I don't

know (yet) how these results compare to a HE version but hey - it's someting!

All comments and questions are appreciated (by me),

Regards,

Well.

[pzhon]

Interesting work.

The cash game strategy ignores the value of position in future hands. For example, if you are getting 2:1 odds on an all-in call, the normal (positionless) assumption is that you need to win 1/3 of the time to call. However, the actual outcomes might be to have 0% of the chips, 10% of the chips in a particular position, or 30% of the chips in a particular position. Since the advantage/disadvantage of position depends on your stack size (primarily your effective stack size), the equities of these outcomes may not be spaced proportionately, and it might be right to call with a different threshold than 1/3.

I don't think this is important in practice because of the following:

[img]/images/graemlins/diamond.gif[/img] When the blinds are small with respect to the stacks, the value of position drops in proportion. You can improve your position by folding at most one big blind. Your simulation found large difference because you used very large blinds.

[img]/images/graemlins/diamond.gif[/img] Your [0,1]-game overestimates the effect in Hold'em, since even the worst hands win a substantial fraction of the time in Hold'em.

[GIER]

Hi Well,

Thank you for your post. It's a very interesting piece of work. One question though: can you explain in words why the probability of winning the tournament is not equal to the player's chip share. In other words: what are the reasons or assumptions that make it differ from eachother?

Regards,

GIER

[Siegmund]

My first guess - without getting out a pencil and paper - is that you're seeing something of an edge effect because of the short stack being right at 50% or 25% of the chips. Did you do the analysis when the stack sizes were at various other points in between? (Say every size from 10:10 to 19:1 in the 20-chip version?)

Intuitively, if I have more than 1/4 but less than 1/2 of the chips, I have to win twice to win, and lose once to lose. My winning chances should be very nearly constant at very nearly 1/3, for a range of stack sizes from about 30% to 45%. If the stacks were deep enough for the blinds to be 'small' compared to the stacks, I would expect a similar plateau of winning chances ~ 1/6 for stacks between 1/8 and 1/4.

When you are right on the knife-edge of having exactly half of the chips, you are in a situation where transferring a single chip causes a huge swing in your equity (moving you into either the 33% or 67% region) and the effect of who has to post which blind next (=can you win one or two chips if your opponent folds?) is going to show up as a significant equity difference.

I would guess that if you were in a situation where a swing of five chips didn't cross a big barrier (say there are 40 chips in play split 15-25) the difference from 33%/67% equity, and the difference between SB and BB, would be tiny.

Makes me wish I had a week to sit around and play with this problem myself. I did some work on [0,1] poker once upon a time, but never did consider a tournament. Great idea.

[well]

Thanks for your respose Siegmund!

I did the analysis for every integer stack count in the range [0,2t], here are the numbers for t=20. This means that each player starts with 20 chips, the SB is 1 and the BB is 2. In the first column are your chip fractions (e.g. .100 stands for a 1:9 disadvantage), in the second colomn the winning probability if you are in the SB, the third when in the BB.

.025 0.023646039 0.026782933
.050 0.053565866 0.047292078
.075 0.076811512 0.072038246
.100 0.099934306 0.099955267
.125 0.124034093 0.125878666
.150 0.14782364 0.151476556
.175 0.171644385 0.177177071
.200 0.195575556 0.202863407
.225 0.219981592 0.22840145
.250 0.244374407 0.25351313
.275 0.268434212 0.279322041
.300 0.293308127 0.304763669
.325 0.317884499 0.32979939
.350 0.342203619 0.355592867
.375 0.367408072 0.380827502
.400 0.391983475 0.405838729
.425 0.416456541 0.431985499
.450 0.442252849 0.456925133
.475 0.466643044 0.48191271
.500 0.491156036 0.508843964
.525 0.51808729 0.533356956
.550 0.543074867 0.557747151
.575 0.568014501 0.583543459
.600 0.594161271 0.608016525
.625 0.619172498 0.632591928
.650 0.644407133 0.657796381
.675 0.67020061 0.682115501
.700 0.695236331 0.706691873
.725 0.720677959 0.731565788
.750 0.74648687 0.755625593
.775 0.77159855 0.780018408
.800 0.797136593 0.804424444
.825 0.822822929 0.828355615
.850 0.848523444 0.85217636
.875 0.874121334 0.875965907
.900 0.900044733 0.900065694
.925 0.927961754 0.923188488
.950 0.952707922 0.946434134
.975 0.973217067 0.976353961

Note that all the numbers are quite close to what is generally assumed.

Because each player starts with t chips, other fractions than the above will never be reached. I do not see the plateaus you wrote about, and as for the tiny differences: if you compare P(win from SB with k chips) and P(win from BB with k chips), the first one is greater only for k=2, 3, 37 or 39.

An interseting fact about the equilibrium strategies for the tournament in comparison to the ones for the cash version:

For t=20, the SB strategy is tighter for k=2, 12, 13, ..., 20 and looser for the other values, whereas the BB strategy is tighter for k=4, 5, ..., 20 and again looser for other k.

Hope it helps, if you want any more information, please let me know...

Regards, Well.

PS paste the table into notepad and you can read it...

[well]

Hey GIER,

if you want to decide what's the best play, you'll always concider the possible outcomes of a hand, then do some EV calculations. In a cash-game, the EV's are in amounts of chips, that's what you want to maximize. In a tournament, you want to maximize your probability of winning the tournament. If the winning probabilities of the tournament are equal to the chip shares, the strategies would be equal, but there's little reason to assume so: it will lead to a contradiction! The roles of the blinds play an important role, one has an advantage over the other (except in a cash game where the raise is to 2 big blinds); at the beginning of one of these tournaments the player starting in the BB has an advantage, hence not a winning probability of 0.5. Since all other treshold values depend on the winning probabilities of the situation you'll land in after that hand, these deviations work through!

Hopefully it's clear to you,

Regards,

Well.

[well]

Thanks for the response.

I indeed used quite large blinds, but otherwise the all-in or fold strategy would not be good (I know these were just the rules of the game, but they are intended to represent the (assumed to be) correct strategies with large blinds). By the way, they are calculations, not simulations. As for your other point: you might be right, but I am not convinced. I have some trouble calculating the strategies for HE, but when I have overcome them you're sure to read about the results!

Regards,

Well.

[Ikke]

Nice post.

I have a few questions. First, at what stacksize is the positional advantage of the BB negated and will the SB have the advantage because he puts in less dead money (it could be that this will never be the case, since maybe at very low stacksizes, the optimum strategy will be all-in/call with any hand).

Also, the advantage of BB over SB seems to be in the 1% for t=20. Is this about constant or does it significantly deviate for different t's?

Secondly, I wonder how the maximizing strategy against the cash game matches up against the optimal tournament strategy. I'm not sure you've done these calculations, but they could be interesting.

And, if you have some spare time to waste:

Do you think it is possible to have some kind of a critical "test" to find out at which stacksize t solely preflop play will not be optimal anymore, and you have to take postflop play into account without actually calculating every scenario? Could the existence of such a test be possible, or can you say for sure that we are just left with calculating the optimal strategy for the most complex game (with postflop play, allowing reraises etc).

Thanks

[well]

Ooops...

While answering your questions and scrolling to the numbers, I found a serious error in my calculations... Hence I shut down my PC, and only now found the courage to redo the calculations; of which the results are a lot less interesting than before. DAMN. I calculated the cash game strategies to quickly/thoughtless: for instance, if the SB has 25 I concidered a raise to 25 instead of to 15.....

Here are the new, less shocking, results:

For t=20,

- the probability of player I winning is:

TT: .500
T$: .500
^$: .502

- with a 1:3 chip disadvantage, in the SB:

TT: .244
T$: .244
^$: .246

- in the BB:

TT: .254
T$: .254
^$: .255

- 3:1 chip advantage, SB:

TT: .746
T$: .747
^$: .749

- BB:

TT: .756
T$: .756
^$: .757

At least we now can say (if my calculations have any credability left) that the proportion assumption isn't that disastrous... For t=5 the winning probability from the BB can be inreased by .07% points, from the SB the most disastrous result is .03% point.

Anyway, I'll answer your questions now.


[ QUOTE ]
[...] at what stacksize is the positional advantage of the BB negated and will the SB have the advantage because he puts in less dead money [...]

[/ QUOTE ]

The EV in the equilibrium is equal to -(k^2-5k+4)/(k+2)^2, with a SB of 1, a BB of 2 and a raise to k. This value is positive for 2<k<4 and negative for k>4. As k increases, the forced bet becomes relatively less. When the raise is to an amount larger than 2 BBs, the advantage of having more information (position) becomes more essential than the disadvantage of having to put a bigger forced bet in.


[ QUOTE ]
[...] the advantage of BB over SB seems to be in the 1% for t=20. Is this about constant or does it significantly deviate for different t's?

[/ QUOTE ]


The winning probabilities when starting from the BB, when both players use the tournament strategies, are between .5088 and .5124 for 4<=t<=20.


[ QUOTE ]
[...] I wonder how the maximizing strategy against the cash game matches up against the optimal tournament strategy.

[/ QUOTE ]

Let's look at t=20 again.

The cash game player in the SB plays too loose if he has 2, too tight from 3 to 11, too loose from 12 to 34 and too tight from thereon.
In the BB he plays too tigth if he has 3, too lose from 4 to 20, too tigth with 21 and 22, too loose with 23 and 24, too tight with 25, too loose with 26 and 27 and too tigth from thereon.

When you are in the BB against the cash game player, you should tighten up when he's too tight and loosen up where he's too loose; if you're in the SB then you should play looser with 2, tighter with 38 and do the opposite of your opponent except for k=30. Note that if your correct counter strategy here (except for k=2) is loosening up, it means be a maniac, i.e. raise with everything.

Note that all these results are for the [0,1] games, which means that they are not transferable without thought to HE. As for your last question, I think I ought to put my spare time into more calculations, making money and watching football...