How Short Should You Go?

As I see it, there are two parts to this problem. One part is strictly math probabilities (which I can say a little about) and the other requires a lot of poker experience and judgement.

The basic math problem is, given that your probability of winning a tournament is P, what is the probability of winning 1 out of “n” tournaments? (You have four values of n, these are 1, 5, 20, 100)

Your chance of losing one tourney is (1-P). Therefore your chance of losing all n tourneys is (1-P)^n (that’s 1-P raised to the “n” power). So finally your chance of winning at least one out of n is 1-(1-P)^n. Multiply that by the prize and that’s your equity in the tourney, if it’s a “winner take all”.

If there are multiple prizes then you have to work out the same kind of formula for each prize, with different P values representing the probability of winning each prize. It’s a whole lot easier if you assume that it’s a “winner take all” tourney so you don’t have to do that.

That’s the easy part. Now you have to estimate P for the four cases of n. I assume that being short stacked is supposed to make your decisions easier, thereby having a positive effect on P. But then you have a shorter stack which has a negative effect. Whoever can work that out knows a heck of a lot more about poker than I do.

With a large bankroll 1×10000
With a smaller bankroll don’t play or try 5×2000 (though I think you are giving up something)

What about extending the range the other way? You are allowed to play 1 out of 5 of these with a starting stack of $50000. I think many players would take this.

I have no theoretical justification for this but with increasing blinds and relatively equal skill level, don’t you want to maximize the time you can play? With a short stack you will be shoving early and likely getting fewer calls by hands that you beat.

JJS,

The independent chip model assigns probabilities of winning as ratios of stack sizes and assumes that all of the players are more or less equal in skill. So, someones probability of winning is something like stack size/sum of all stack sizes. That seems like a reasonable estimate here. It then goes on to estimate how often you win, come in second, third, etc. There’s a problem from Harrington on Hold ‘em 3 that goes through the whole process.

Looking at this particular problem, you have about 16.67% equity in one tournament. With mre then one, your individual equity goes down, but not linearly. If you sum all of the equities, you pretty quickly approach and asymptote at 20% equity. I’m sure there’s a closed form solution that will yield that curve. The ability to produce that closed form solution has atrophied out of my math brain. I’ll have to leave it to someone else. So, in this case, you get the most equity from 100 entries, but it is only fractionally different from the solution with 5 or 20 entries. At the end of the day, don’t take the one. Once you get past 5 or so, it would have to be a lot of money for me to want to spend the time.

I think it strongly depends on the player’s skill of playing with a certain stack size. This is one side of the problem. The second is the math side… but I can’t imagine the details, yet I’m awaiting your (Ed’s ) input (shall I say ‘output’ ) impatiently, as it will be a good tournament lesson I suppose.
Pawel

I think it almost has to be 100 entries of $100. My argument is as follows:

Given that you will be all-in on one of the next 1 to 5 hands, you can choose your best hand, and have a big advantage over everyone else, since you won’t have to fold. That makes your chance to increase your stack size to $500 much better than 20%. Therefore option 4 must be better than option 3. With a similar argument, someone with $500 is more than 25% likely to get to $2000, therefore 3 is better than 2. Finally, someone with $2000 should get to $10000 more than 20% of the time. Thus 4 is best of all.

Todd,

Based upon the model stack_size/sum_stack_sizes that you gave, I get P values of 0.167, 0.038, 0.0099, and 0.002 respectively for the n values of 1, 5, 20, and 100. Working out 1-(1-P)^n for each of these, I get probabilities of 16.7%, 17.8%, 18.0%, and 18.1%.

So from a strictly mathematical viewpoint, it looks like you get a slight advantage from the smaller stack sizes but going to n larger than 5 does not give much advangage over n=5 (only 0.3%).

Well the 100 dollar starting stack is no good…with blinds starting at 50/100, you will literally win about 1/10 times……..because there isn’t much room for skill there.

I play short stacked in AC allot in live NL cash games (and I do good…..not inredible…..but good), and I play some tourney’s there. If any else here on Ed’s forum plays tourneys there, you’ll notice that at one point the blinds go up so high that It does become almost coin-flip-like scenarios.

Just looking at ed’s scenarios, I like number 2…..starting with 2000 dollar stacks with 50/100 blinds. It gives you a good starting short stack (20bb).

I like the idea of doing all the math also, and i think it would yield the same result.

Either way, I haven’t posted in awhile i just found this very interesting!

JJS,

Yeah, it’s very small after 5. It’s the exponential nature of the problem, I think. I stuck it into the ICM tool at http://www.poker-tools-online.com/icm.html and it gives me ~20% equity for a single payout with stacks of 5x10K and 1×100. There must be something to how they are determining the probabilities that I’m forgetting. At the end of the day, more tries are better. A lot more tries are only a little better.

I think I like option #4. I don’t know all of the math behind this scenario, but what I do know is it would take doubling up 5 times before the blinds go up to get out of being short-stacked relative to the blinds (100 X 2 ^ 5 = 3,200 or 32 BB). Given 100 chances to do this, I think I like my odds. I know you can’t be too selective going all-in initially with just 1 BB, but if you double-up 3 times, then you can be a little more selective after that, giving you a decent chance of doubling up 2 more times. Statistically, I guess you would win the first race 17% of the time (1 out of 6). Hopefully, this percentage would be greater if you go all-in with better than average hands before the big blind hits you the first round. I think 25 – 30 first-hand wins would be a realistic guess. This means you would have 25 – 30 shots with a stack of $200 – $600 depending on how many people were in the first pot. This is preferable to choice #3. After being able to be more selective, if you get a good hand, the larger stacks are going to be more likely to call all-ins, thinking it is not much of their stack. I would think a realistic guess of getting out of being short-stacked (relative to the blinds) from 25 – 30 chances with $200 – $600 would be half, or 12 – 15. Again, I am not basing all of this on exact math, but based on my limited experience, I like my odds with 100 buyins.

Here’s one way to look at it: how often will you reach the, say, 20,000 chip mark with these scenarios? I get one chance to play average-stacked against average opponents versus five chances to play short-stacked against those same opponents. Is it fair to say that in the first scenario you are relying on the chance that an opponent will make a big mistake against you (big relative to the stack sizes)? I don’t think that’s the case at all with the shorter stack. There, average players will make all kinds of smaller mistakes against you. Since average players will make more smaller to medium sized mistakes than larger mistakes, playing a shorter stack five times might be more profitable than one single average stack.

I don’t think this reasoning applies to the 500 stacks. There will definitely be mistakes made against you with that size stack as well, but they won’t be of the same magnitude as with the 2000 stacks. So, it’s 20 chances with a small-mistake-making stack vs. 5 chances with a small-AND-medium-mistake-making stack. I might be wrong about that, though I’m not very worried about it being an infinite regress.

Guys, your analysis is only correct if you accept the independent chip model as accurate. I would argue that it breaks down for very small values. Ed has been explaining that small stacks have an inherent advantage over multiple big stacks (but not against a single big stack) since the big stacks have to mix a large stack strategy with a small stack strategy and cannot maximize both. That gives the smallest possible stack an inherent advantage beyond what the independent chip model suggests.

VegasSocrates, there is in fact infinite regress going on here. Take your 500 chip example. You start 20 tournaments with 500 chips. How many times do you think you can build it into 2000 chips? Maybe 6 out of 20? Well, that is the equivalent of getting 6 entries that start with 2000 (assuming a blind structure slow enough to not be a factor). That must be better than 5 entries that start at 2000 by definition. Apply the same analysis to the 100 chip example.

Toonces,

I agree that the ICM only provides an estimate. I don’t think it’s a bad one for the situation described, but it surely is an estimate. I think you’re argument that you get some additional advantage being short is valid. At small values, though, I think it probably gets balanced a bit by the fact that you only get to see, at most, one orbits worth of hands before being forced to choose one. At the end of the day, for this problem, I think both arguments point to more chances are better. No?

Todd