Table VPIP to Seen Flop %, math question.

[OrianasDaad]

Not considering the blinds.

For example, the table VPIP outside the blinds is 25%, and I'm on the button. There's 7 players to act before me, so there's about an 86% chance that at least one person will come into the pot behind me [1-(1-.25)^7].

How do I figure out the probability of one player coming in behind me, or all seven coming in?

[OrianasDaad]

I just thought of another way to put the question.

Assume there are four events, and each has a specific probability of happening.

#1 - 25%
#2 - 35%
#3 - 50%
#4 - 70%

Each event is counted separately. On average, how many events will happen? I could write a simulation program to tell me, but I'd rather know the math.

[BruceZ]

[ QUOTE ]
I just thought of another way to put the question.

Assume there are four events, and each has a specific probability of happening.

#1 - 25%
#2 - 35%
#3 - 50%
#4 - 70%

Each event is counted separately. On average, how many events will happen? I could write a simulation program to tell me, but I'd rather know the math.

[/ QUOTE ]

0.25 + 0.35 + 0.5 + 0.7 = 1.8. This is always true, regardless of whether the events are independent.

[BruceZ]

[ QUOTE ]
Not considering the blinds.

For example, the table VPIP outside the blinds is 25%, and I'm on the button. There's 7 players to act before me, so there's about an 86% chance that at least one person will come into the pot behind me [1-(1-.25)^7].

How do I figure out the probability of one player coming in behind me, or all seven coming in?

[/ QUOTE ]

Assuming that they are independent (as you did with [1-(1-.25)^7] ), the probability that exactly 1 comes in is C(7,1)*(0.25)*(0.75)^6 =~ 31.1%, and the probability that all 7 come in is (0.25)^7 =~ 0.0061%.

[OrianasDaad]

I figured it out while sleeping.

A better stat than table VPIP might be: 1-(table fold flop%), which should provide a table seen flop %.

Thanks!