If we're calling 1 to win 15 we're getting immediate odds to 
chase so working with those numbers is silly. 
Proof: 
Board: Kd Qs 2h
Dead:  
    
equity     win     tie           pots won     pots tied    
Hand 0:     14.148%      14.15%     00.00%               6723             0.00   { ATs, ATo }
Hand 1:     85.852%      85.85%     00.00%              40797             0.00   { 22 }
We have a 
gutshot against a 
set on the 
flop, we'll hit 14% of the time, so we need to be given 1:6.14 
pot odds.  Even if we only consider the 
turn and half that, we only need 1:12.28 
pot odds and we're given 1:15 so we can just blindly 
chase here.  
//
So...how to form an ev equation:
EV = 'what we can win' * 'how often we win' - 'what we lose' * ' how often we lose'
Simple right?  If we lose more often than we win then we need to win more money than those times we lose.  
Ok, lets apply it to this situation
A 
standard EV equation for a 
call would be made how i outlined above
EV = '8% to hit' * 'the current pot size + his bet + the 
effective stack' - '92% to miss' * 'his bet'
Using formula...
EV = 8% * (p+b+s) - 92% * b
Normally, we would isolate these terms so we end with 8/92 = b/(p+b+s) and find what those values need to be in order for us to make a +EV 
call.  With b=1, p=15, we find s can actually be negative here and still be +EV.  Which is exactly what we expect.
Ok, now add the condition that we lose 20% of the time we hit.  How does that change things?
Well, we are now only expected to win 80% of the time we hit right?  So 80%*8% = 6.4% of the time we'll win.  But also, that 20% of the time we hit and lose we're gonna lose our 
stack and the bet we called.  So lets put these in the EV equation
EV= 6.4%*(p+b+s) - 92% *b - 1.6%*(b+s) right?  (
Check it, an ev equation must take into account all outcomes.  So the probabilities of winning+losing must equal 100%.  6.4+92+1.6 = 100, ok good)
Ok, so what did we do with that last EV equation?  We tried to get those b,p, and s terms by themselves after setting the equation equal to 0 ya.  Lets try and do that.
6.4%(p+b+s) = 92%b + 1.6%(b+s).  since the equation now equals 0
Right now though, it seems like we have a few too many terms.  So lets simplify it since some cancel each other
6.4%p + 6.4%b + 6.4%s = 92%b +1.6%b + 1.6%s
Moving terms around we get
6.4%p +4.8%s = 87.2%b
We can bet b, p, and s bythemselves by being creative.  6.4%p +4.8%s is the same as 6.4%*(p+.75s).
So we have 
6.4%(p+.75s) = 87.2%b
and by moving things around again
6.4% / 87.2% = b / (p+.75s)
.073 = b/ (p+.75s)
Sweet.  So now lets say b=1, and p = 1.  what does s have to be (ie, what do we have to win in addition to the pot to make calling profitable?)
.073 = 1/ (1+.75s)...........so.......13.69 = 1+.75s,......12.69 = .75s........s =16.93 so with a 
gutshot draw we need 
villain to have 16.93* the pot if we wanted to 
call a potsized bet given his redraw 
outs.  
What was it before we knew he had a 
set and just assumed we'd win with our 8% outter?
Recall that we had 8/92 = b/(p+b+s), so if b=p=1 we'll find s=9.5.  
Thats kinda cool.  We needed about twice as much 
behind to 
call given that he could redraw on us.  
But does that hold for different bet sizes?  
Does that hold for different equities?  Suppose we instead were drawing to a 
flush in the same scenario, how does it change now?  
what if 60% of the time 
villain had a 
set, but the other 40% he had TP?  A 
draw that 
dominated ours?  
Answer these questions using what you learned from this 
post.  I took alot of time writing this mainly cuz im bored but dont feel like grinding...but if you dont actually put in effort to answering these questions then it basically says you have no interest in learning...so answer them.