Theory of freeroll (equity)

[Perestroika]

Lets assume you are playing omaha and get all in with a complete freeroll such as this situation. http://twodimes.net/poker/?g=o&b=8s+...%0Ajc+qs+2h+3d

Usually its (%chance win)(pot size)-(%chance lose)(pot size) to calculate equity.

Is it correct therefore to lets say in an all-in 200 dollar pot to calculate your EV as (.615)(200)-(0)(200)=$130 or am I missing something?

[Vetgirig]

Well, in a freeroll you have a base line that is the same as a split pot.

So you already own half the pot - its the rest you are freerolling for - your original algorithm dont take into account the chance of a split pot.

Add +(%chance split)(pot size)

[tarheeljks]

i think the expression should be

(.615)(200)-(0)(0)

b/c you can never lose money in this scenario. (it's a bit nit picky, but you lose no money, 0% of the time)


also, the equation should give ~123

[ QUOTE ]
Usually its (%chance win)(pot size)-(%chance lose)(pot size) to calculate equity.

[/ QUOTE ]

a normal ev calc applies here, it's just looks a little different than a standard poker ev calc b/c you can never lose money in a free roll.

you could also say

(win freq)*potsize + (chop freq)*0.5*potsize =
(.229)*200 + (.771)*0.5*200 =

(.229 + .771*.5)*200

(.615)*200

so you get the same result