Re: [0,1]- all-in or fold two player tournament
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.
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[...] 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 [...]
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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?
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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.
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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...
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