Bubble Aggression? (Long, math)

[sfustsh]

So after reading the article DarkDenim posted about playing ultra aggressive (pushing like every hand) on the bubble in order to win enough chips to place first instead of taking 2nd or 3rd, I've decided to do a little analysis of a hyper aggressive bubble strategy. This only includes bubble play and doesn't take into account what happens after that.

Basically, this is based on the assumption that you will push a certain range Rp (in terms of decimal percent, that is, if you push the top 75% or hands your Rp is .75) and your opponents will call with another range, Ro. Your opponents would, I'm sure, raise their range after a while but that's not included yet.

With these assumptions, you push Rp percent of hands. Of those, you win 1-3Ro hands uncontested. That is, there is a 3Ro percent chance that one of your three opponents will have a playable hand. Ro of course is different for each player, I'll explain that in a second. Then, for each hand you win uncontested you get the blinds.

So:
Money in = (Blinds)*(Rp)*(1-Ro)^3*n

Occassionally, you'll get called and win. If you push at a range of Rp your average hand is Pa = 1-Rp+Rp/2. For example, if you push the top 75% of hands, your average hand is ranked 62.5 out of 100. Your opponents average hand is figured similiarly: Oa = 1-Ro+Ro/2. I'm not entirely sure what your chances of winning according to the relative ranks, so I made something up: Your odds of losing is about equal to the quotient of your opponents average hand and your average hand, so that Oa/Pa = odds of winning. The function explodes near 0 but you won't push anything close to that range so I think it's (reasonably) accurate. For example, if your opponent pushes the top 8 percent of hands his Oa = .96 and if you push the top 75% of hands your Pa = .625. Therefore, Oa/Pa = 1.536 and your chance of winning is about 40%. This function might need refining.

So, when you get called, you win 1/(Oa/Pa+1) percent of hands and you win the opponents average stack size, let's call it Os. You will get called according to the chance that your opponent has a playable hand, but only when you push which happens 1-Rp percent of the time. Add this to the money in.

Money in: (Blinds)*(Rp)*(1-Ro)^3*n + (Rp)*1/(Oa/Pa+1)*Os*(1-(1-Ro)^3)*n

Simple enough. But then you will lose money as this happens of course. Every 4 hands you pay blinds.

Money out: (Blinds)*(n/4)

Then, you will get called and lose exactly as often as you "don't win" so you lose 1-1/(Oa/Pa+1) percent of the time. Again, you lose your opponents' average stack and this can happen only on occassions that you push which is 1-Rp percent of the time. So subtract this money.

Money out: (Blinds)*(n/4) + (Rp)*(1-1/(Oa/Pa+1))*Os*(1-(1-Ro)^3)*n

Okay so I figured big stacks and small stacks have a higher range than average stacks so I made up a function to describe the range of your opponents:
Ro=1/sin(Os/(sigma(a=1,p,Oa)*Pi)+5/100
Basically, your opponents push a higher percent of the time with a short stack and a large stack. The function approaches infinity as the player's stack goes to either zero or to the total number of chips. The sine function is symmetric about Pi/2 for 0<x<Pi so it is okay to assume that you will get called the same number of times by a big stack as you will a short stack, so on average you will get called by an "average" stack and therefore lose or win an average stack's worth of chips on every call.

Just to let you know, Ro = about 6% for a stack that represents 25% of the total, which seems reasonable to me, though it's not really accurate for very large or very small stacks. Of course, these numbers can be changed.

I decided not to make your range dependent on your stack size (I could if you wanted to but it defeats the purpose!).

Okay so we need to decide if this is +EV, right? That's easy, take the difference of money in and money out and divide by the number of hands (n, if you haven't figured it out yet).

Money in: (Blinds)*(Rp)*(1-Ro)^3*n + (Rp)*1/(Oa/Pa+1)*Os*(1-(1-Ro)^3)*n
Money out: (Blinds)*(n/4) + (Rp)*(1-1/(Oa/Pa+1))*Os*(1-(1-Ro)^3)*n

Money in - Money out = [n(Rp)][(Blinds)(1-Ro)^3+(1-(1-Ro)^3)Ro/(Oa/Pa+1)]-[(Blinds)*(n/4)+(Rp)(1-1/(Oa/Pa+1))*Os*(1-(1-Ro)^3)*n]

Divide by n.

Money in - Money out per hand = [(Rp)][(Blinds)(1-Ro)^3+(1-(1-Ro)^3)Os/(Oa/Pa+1)]-[(Blinds/4)+(Rp)(1-1/(Oa/Pa+1))*Os*(1-(1-Ro)^3)]

Yeah that looks pretty useless but it tells you exactly how much money you expect to make per hand. If you would put numbers in there you could solve for Rp (your push range). That, obviously, is the most useful part. Use percent of total chips for the "blinds" number.

Edited:

Let's take an example, if the blinds are 200 and 400 (600 total, or 3% of total 20000), your opponents are tight (will call top 8% of hands) and they all have 4500 in chips then your EV is maximum if you don't push any hands. Increase blinds to 1500 and your EV positive push range is maximum at 20% of all hands. Now let's say your opponents will call with the top 15% instead of top 8%. Again you can't push any hands.

Does anyone notice any huge errors with this model? I think I need to refine the odds of winning / losing a little more, since premium hands seem to me to be certainly qualified to be pushed.

[ugignadl]

I haven't read the whole thing, but the odds of 3 opponents not calling is not 1-3Ro. It is:

(1-Ro)^3 percent.

Odds that at least one opponent will call is then:

1-(1-Ro)^3 percent.

[sfustsh]

Thanks I'll change that later, I could never remember that one damnit ...