math

http://www.liquidpoker.net/forum/viewmessage.php?topic_id=368084

I did this a while back. The general equation I put in is definitely accurate. Don't have time now to rework and supply the working. Hope it helps...

I also just realize I can keep reading beyond his examples durr
looks like I'll be up a bit longer to read through this and that post. Thanks!

I like this equation better, easier to understand

EV = 2x + (1-x){(1/3 * 6) + (2/3 * (-4))} => EV = 2x + (1-x)(6/3 - 8/3) => EV = 2x + (1-x) * (-2/3) I thought it would be something like: 1/3 of the time you draw out and win 6 , 2/3 you don't and lose 4, hence the formula I wrote down. So the results are the same, I just don't understand the way you calculate it...

armed with this, I shall try and continue my original example.

My example/math:

EV = 2x + (1-x){(9/44*116) + (35/44)*(-87)}
EV = 2x + (1-x){(261/11) + (-3045/44)}
EV = 2x + (1-x)(-1989/44)
EV = 2x + (-1989/44 + 1989/44x)
1989/44 = 2077/44x
1989=2077x
x = 0.957
So 95.7% time to be 0EV??

That cant be right.. can it?

I always do this by looking at how much you will lose when called first

200bb total pot, you win 20.5% of the time = 41bb won long run if he always calls. so long run you lose -87bb+41bb = -46bb if he calls

then you win 50bb if he folds, so if x is % of times he folds
(x)50bb - ( 1- x )46bb = 0 [eq1]
solve for x
..but that takes work so I like to just plug in random %'s for x and see how it comes out

eg if he folds 50% eq1 = 25bb - 23bb = 2bb profit
if he folds 40% calls 60% eq1 = 20bb - 27.6bb = -7.6bb loss

so you really want him to be folding at least 50% of the time.

looked up some old posts cuz i was curious myself and found this way of working it out:

EV (fold) = 50$
EV(call) = (0.205 * 200) - 87 = -46

p(fold) = EV(call) / [ EV(call) - EV(fold) ]
p(fold) = -46 / [-46 - 50] = -46/-96 or 47.92%

so u need him to fold like 47.92% of the time for it to be profitable

I haven't read through all the stuff, but in fail #2, x means the % he folds and thus you win 2, 1-x is the % you have to showdown your hand where you win 1/3 of the time 10 (this is the pot size, 2+4+4) minus your investment (4)

hope this helps if you haven't solved the problem yet

NM I misread the op apparently

Let x be the % of the time he folds.

When he folds (x % of the time) we insta win the pot (50bb). ==> x*(50bb)

When he calls (1-x % of the time), 20.5% of the time we win the pot (50bb) and his stack (66bb) and the other 79.5% we lose our stack (87bb) ==> (1-x)*((0.205*(116bb) - 0.795*(87bb))

Then:

EV(jam) = 0 = x*(50bb) + (1-x)*(0.205*(116bb) - 0.795*(87bb))
EV(jam) = 0 = x*(50bb) - (1-x)*(45.385bb)
EV(jam) = 0 = x*(95.385bb) - 45.385bb
==> x = 45.385/95.385 = 0.4758 = 47.58%