Button ($98) limps.
I complete in the SB
BB goes all in for $28.
Button calls
I shove all in and button calls.

BB has [Js] and is all in for $28
Button has [4s] and is all in for $98
I have [Qs] and cover them

How much equity do I have? I dont know how to figure it when theres 2 different pots.

The 3 way pot has $81 after rake
AQ...42%
KJ.....32%
84s....26%

The 2 way pot has another $140
AQ.....63%
84s....37%

Can I take my 63% of $81
and
63% of $140

for a total of $122?

And if so, how do I figure if I won or lost more than I should have if I won the $140 pot but lost the $81 pot?

If I get all in for $200 each with AA vs KK....and win the $400 pot, I won $80 more than I shouldve. $400 comes sliding my way but I had $320 in equity.

So can I say that since $140 came sliding my way and I had $122 total equity, that I won $18 more than I deserved? Am I even close?

Come on math people...help me out please.

Aisthesis? Where are you?

Its 42% of 81, but i guess thats a typo since your final numbers are correct.

After posting, I started having (wrong) second thoughts about this calculation because the two events are not independent--hence making it seem dubious just to add up equities.

So, here's a generalization:

Suppose p0 is the main pot and we have n side pots p1, ..., pn. The order is important!

The thing is, if you win p0, then you win all the rest. If you lose p0 but win p1, then you also win p2 thru pn.

It follows that e0 <= e1 <= ... <= en where ei is your equity in side pot i.

It should also be clear that your total EV here is:

e0*(p0+p1+...+pn) + (e1-e0)*(p1+ ...+pn) + (e2-e1)*(p2+...+pn) + (en-e[n-1])*pn.

It's easy to see that everything cancels out to leave (e0*p0) + ... + (en*pn).

The correct answer is 42%x$81 + 63%x$140.

While winning the main means that you win the side, you don't need to do all those extra 'possible worlds' calculations.

You have 42% equity in the $81 pot...so you take your EV from there.

You have 63% equity in the $140 side pot, which you add to your EV in the main.

Also, this probably belongs in the Odds forum.

You can do it that way. They're not interdependent. Winning Pot 2 does not require winning Pot 1 ... it just so happens that if you win Pot 1 you also win Pot 2.

In fact, there's logically no connection between the pots at all. It would be possible to do a variation of "running it twice" here: deal 5 cards to decide who wins pot 1 and then (shuffle up and) deal 5 different cards to decide who wins pot 2. That wouldn't change the EV calculation, but it gets rid of the illusion that the two pots are inherently interdependent.

You did it right the first time, and Aisthesis is too smart for his own good.