"Is it getting bubbly in here?" [Archive] - Two Plus Two Newer Archives

SumZero

09-09-2006, 05:48 AM

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Question: I am the big stack (8000 out of 20.000 total) in a sit go with 4 remaining players and the blinds are getting big (400/800).

Is it then a wrong strategy to raise the blinds (SB: 2000 BB: 2500) with any2 from the button, when I have them well covered?

I thought that the gain outwayed the % of times they would call when given a certain range (ex: A10+,22+).

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I think this is a question that the STT people are all over [pushbot galore], but I've hung out there in the past and can help answer it.

First of all you need to know the payout of the tournament. I'm assuming a 10 person tournament that pays 50/30/20 to the top 3 players (for ease of money let's say it is a $100 tournament so the prizes are $500, $300, and $200). Second I'm assuming that the amount for the blind is what they have pre-blinds and that it is folded to you on the button.

Note this isn't the ideal steal situation as ideally you'd want to have the short stack folded as the CO and the blinds be big enough that they can out wait the short stack but small enough that you cover them and aren't totally crippled if they call and win. If you have 8K maybe the CO with 2K SB with 5K and BB with 5K would be ideal stealing for 8/20K chips.

But still we have CO with 7500 chips, you with 8000, SB with 2000, and BB with 2500. Before the hand started ICM says you have $341.45 in equity.

If everyone folds their hand (the BB wins in a walk) then you'll have CO with 7500, you with 8000, SB with 1600, BB with 2900 which ICM says is worth $342.42 for you.

If you fold and SB pushes and BB folds then you have 7500/8000/2800/1700 which is worth $342.09.

If SB and BB end up all in and SB wins then you have 7500/8000/4000/500 which is worth $349.77 (and more even as you definitely push any2 next hand with the mini $500 stack).

If SB and BB end up all in and the BB wins then you have 7500/8000/4500 which is worth $355.61.

So once the CO folds here you are no worse off than you were at the start of the hand when you just fold (in general the more unequal the SB and BB end up the better for you - the worst out come when you fold is for the SB and BB to split the pot which leaves you with the same $342.42 expectation that you started the hand with).

The exact amount is hard to figure as it depends on what SB and BB are likely to do but if you figure BB walks 1/3, SB steals 1/3, SB pushes is called and wins 1/6, SB pushes is called and loses 1/6 as the very rough numbers you get:
1/3 * $342.42 + 1/3 * 342.09 + 1/6 * 349.77 + 1/6 * 355.61 = $345.73

But what of pushing?

For simplicity make the assumption that if you push at most 1 of the 2 short stacks will call. This isn't perfect as with the BB>SB the BB can be pretty liberal in calling all-ins once the SB has called all-in so hands as bad as TT or AQ seem fine to call with in the BB after the SB has called all-in. If you switch the SB and BB's stacks you'd have more protection but KK and AA are still likely to call in the BB with 2000 after the 2500 SB is all in.

If you push and no one calls then you have 7500/9200/1600/1700 which is worth $367.48 (and likely more as the next hand is another easy push all-in any2 from the CO).

If you push and the SB calls and loses you have 7500/10800/1700 which is worth $399.42 (and again an easy all-in any2 hand).

If you push and the SB calls and wins you have 7500/6000/4000/1700 which is worth $302.19 (maybe less since the CO can now abuse you by pushing any2).

If you push and the BB calls and loses you have 7500/10900/1600 which is worth $400.94 (maybe more since again an easy all-in any2 hand).

If you push and the BB calls and wins you have 7500/5500/1600/5400 which is worth $283.90 (maybe less since CO can abuse you).

So you need to know 4 numbers:
p(SB calls)
p(SB wins|SB calls)
p(BB calls)
p(BB wins|BB calls)

If I'm the SB I'm decently lose with my calls here, unless I have a read that the button and/or BB are idiots who have no clue on how to play bubbles, as I think button is raising aggressively here and that BB has me outchipped/in a race that depends on exactly how the blinds hit and when the level goes up. If the blinds never go up and both BB and SB fold from here on out you get "BB"/"SB" of 1700/1600 then 1300/1600 then 1300/800 then 500/400 then 100/400 then 100/0. So I might call with something like top ~50% of heads up hands (any pair, any A, any K, any Qxs, Q5o+, J6s+, J8o+, T7s+, T9o, 98s). This represents 49.3% of hands the SB calls with, all of which are >50% against random cards, as a whole the SB will beat a random hand 58.13% of the time when it calls with this range.

If the button isn't pushing any 2, but is still pushing aggressively I'd tone it down to something like (any pair, any 2 broadway, any A or K, Q8o+, Q6s+) which is 40.3% of hands and wins 59.57% against a random hand and 57.81% against the top ~83% of hands. If the button is only pushing about 1/2 the hands in this situation I'd call only with the top ~20% of hands (something like 44+, Axs, A4o+, K9s+, KJo+, QJs) which is 21% of the hands and which beats the top 50% of hands about 58.63% of the time.

But assuming the button is pushing any2 and that the SB calls with the 49.3% above then we have P(SB calls) = 49.3% and p(SB wins|SB calls) of 58.13%.

If the SB folds and I'm the BB and the Button is pushing aggressively I'm tighter than I'd be in the SB because we have the SB outchipped. But a double up is really quite valuable here for me as it puts me close to tied for 2nd with a really short stack so I'm still not uber tight. I'd say something like (any pair, any A, K6s+, K8o, QTs+, QJo) which is 28.5% of the hands and will beat any random cards 61.8% of the time.

So p(BB calls) = 28.5% and p(BB wins|BB calls) = 61.8%.

So the formula is:

[1-p(SB calls)-p(BB calls)]*EV(steal)
+ p(SB calls)*[p(SB wins|SB calls)*EV(SB calls&wins)+(1-p(SB wins|SB calls))*EV(SB calls&loses)]
+ p(BB calls)*[p(BB wins|BB calls)*EV(BB calls&wins)+(1-p(BB wins|BB calls))*EV(BB calls&loses)]
= [1-.493-.285]*$367.48
+ .493*[.5813*$302.19+(1-.5813)*$399.42]
+ .285*[.618*$283.90+(1-.618)*$400.94]
= $344.28.

And since $344.28 < $345.73 this suggests that if the SB and BB play as aggressively as I do that folding every hand is slightly better than pushing every hand.

If we instead leave the BB as I've described but make the SB have the tight top 21% of hand range then the p(SB calls)=21% and p(SB wins|SB calls)=63.65% and instead the formula gives:

[1-p(SB calls)-p(BB calls)]*EV(steal)
+ p(SB calls)*[p(SB wins|SB calls)*EV(SB calls&wins)+(1-p(SB wins|SB calls))*EV(SB calls&loses)]
+ p(BB calls)*[p(BB wins|BB calls)*EV(BB calls&wins)+(1-p(BB wins|BB calls))*EV(BB calls&loses)]
= [1-.21-.285]*$367.48
+ .21*[.6365*$302.19+(1-.6365)*$399.42]
+ .285*[.618*$283.90+(1-.618)*$400.94]
= $350.11.

And since $350.11 > $345.73 this suggests that if the SB and BB play like this pushing any 2 is better than folding any 2.

But clearly the best play is actually to push many cards, but not quite any 2. To show why pushing any 2 is wrong, even in the last example, consider 23o - the worst hand you can have to push with. If the SB and BB ranges stay the same then P(SB calls) and P(BB calls) stay the same but P(X wins|X calls) is much higher because you don't have a random hand you have 23o. p(SB wins|SB calls) = 71.69% and p(BB wins|BB calls) = 71.07% so the formula gives:

[1-p(SB calls)-p(BB calls)]*EV(steal)
+ p(SB calls)*[p(SB wins|SB calls)*EV(SB calls&wins)+(1-p(SB wins|SB calls))*EV(SB calls&loses)]
+ p(BB calls)*[p(BB wins|BB calls)*EV(BB calls&wins)+(1-p(BB wins|BB calls))*EV(BB calls&loses)]
= [1-.21-.285]*$367.48
+ .21*[.7169*$302.19+(1-.7169)*$399.42]
+ .285*[.7107*$283.90+(1-.7107)*$400.94]
= $345.38.

And since $345.38 < $345.73 then pushing 32o is wrong even when SB is on this tight 21% range [and remember we've been a little optimistic since we've assumed BB never calls when SB calls, an assumption that makes pushing seem slightly stronger].

Now in your post you had super tight ranges for the SB and BB of AT+ and any pair. If this is the range for both the SB and BB then pushing any 2 will be correct as pushing 32o will give us p(X calls)=10.7% and p(X wins|X calls)=76.62% which gives a formula of:

[1-p(SB calls)-p(BB calls)]*EV(steal)
+ p(SB calls)*[p(SB wins|SB calls)*EV(SB calls&wins)+(1-p(SB wins|SB calls))*EV(SB calls&loses)]
+ p(BB calls)*[p(BB wins|BB calls)*EV(BB calls&wins)+(1-p(BB wins|BB calls))*EV(BB calls&loses)]
= [1-.107-.107]*$367.48
+ .107*[.7662*$302.19+(1-.7662)*$399.42]
+ .107*[.7662*$283.90+(1-.7662)*$400.94]
= $356.91.

And $356.91 >> $345.73 so pushing any 2 cards is indeed correct against this tight of a range from the SB and the BB (in fact this is better than any result obtained by folding - even the BB knocking out the SB).

In fact, if you didn't even look at your cards and just pushed any 2 random is SB and BB were playing this tight you'd get p(X wins|X calls)=66.76% which gives a formula of:

[1-p(SB calls)-p(BB calls)]*EV(steal)
+ p(SB calls)*[p(SB wins|SB calls)*EV(SB calls&wins)+(1-p(SB wins|SB calls))*EV(SB calls&loses)]
+ p(BB calls)*[p(BB wins|BB calls)*EV(BB calls&wins)+(1-p(BB wins|BB calls))*EV(BB calls&loses)]
= [1-.107-.107]*$367.48
+ .107*[.6676*$302.19+(1-.6676)*$399.42]
+ .107*[.6676*$283.90+(1-.6676)*$400.94]
= $359.17.

Which again is better than any result obtained by folding - even the BB knocking out the SB.

So the long winded answer is definitely push any 2 if SB and BB are that tight. But they shouldn't play that tight, and if they don't play that tight, you should push most - but not all - hands in this situation (say 4 out of 5 hands).