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The definitive proof that running it multiple times from Turn -> River doesn't change EV The definitive proof that running it multiple times from Turn -> River doesn't change EV
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05-20-2024 , 06:59 PM
#26
KlemenZ
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This looks like a valid equation for running it twice on river only. The first one in the whole thread. Because you took dependency into account. However I don't think the equation for RIT with more than one card will be this simple.
The definitive proof that running it multiple times from Turn -> River doesn't change EV Quote
05-20-2024 , 07:20 PM
#27
KlemenZ
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I think you made a mistake in the L1L2 part - it's not 43*44, but it doesn't matter because it's multiplied my zero anyways.
The definitive proof that running it multiple times from Turn -> River doesn't change EV Quote
05-21-2024 , 10:30 AM
#28
dude45
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Not sure how this is still an argument. Personally i never run it twice for one it waste time and second i dont care about someone's aversion to variance.
The definitive proof that running it multiple times from Turn -> River doesn't change EV Quote
05-21-2024 , 03:09 PM
#29
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Haizemberg93
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You should care about variance if you are winning player. Less variance is better for you.

As for the proof
Equity =(number of winning runouts)/(number of all runouts)
Let's write it like
EQ=W/T
If you run it twice
A)you win first run than your Equity for next one is
EquityW=(number of winning runouts-1)/(number of all runouts-1) or
EQW=(W-1)/(T-1)
B)If you lose first run formula is
EQL=(W)/(T-1)


Adding those two you get
EQ*[(W-1)/(T-1)]+(1-EQ)*[(W)/(T-1)]=
[EQ*W-EQ+W-EQ*W]/(T-1)=
(W-EQ)/(T-1)=
W/(T-1)-W/[T*(T-1)]

To get Equity over runing it twice
1/2[WT-W+WT-W]/[T(T-1)]=
[W*(T-1)]/[T(T-1)]=
W/T=EQ

So Equity of running twice is same as running twice.
The definitive proof that running it multiple times from Turn -> River doesn't change EV Quote
05-21-2024 , 04:19 PM
#30
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Quote:
Originally Posted by Haizemberg93
You should care about variance if you are winning player. Less variance is better for you.
Some have made the argument to willingly keep the variance high:

1) Running it once increases the chance of getting deep stacked, which increases the edge of the better player
2) Running it once increases the chance of a tilt-prone player going on tilt, which increases the edge of the calmer player
The definitive proof that running it multiple times from Turn -> River doesn't change EV Quote
05-21-2024 , 06:04 PM
#31
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Yes. I like increasing my edge after losing stack
The definitive proof that running it multiple times from Turn -> River doesn't change EV Quote
05-21-2024 , 06:15 PM
#32
dude45
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Quote:
Originally Posted by Haizemberg93
You should care about variance if you are winning player. Less variance is better for you.

As for the proof
Equity =(number of winning runouts)/(number of all runouts)
Let's write it like
EQ=W/T
If you run it twice
A)you win first run than your Equity for next one is
EquityW=(number of winning runouts-1)/(number of all runouts-1) or
EQW=(W-1)/(T-1)
B)If you lose first run formula is
EQL=(W)/(T-1)


Adding those two you get
EQ*[(W-1)/(T-1)]+(1-EQ)*[(W)/(T-1)]=
[EQ*W-EQ+W-EQ*W]/(T-1)=
(W-EQ)/(T-1)=
W/(T-1)-W/[T*(T-1)]

To get Equity over runing it twice
1/2[WT-W+WT-W]/[T(T-1)]=
[W*(T-1)]/[T(T-1)]=
W/T=EQ

So Equity of running twice is same as running twice.
I want to make life as difficult as possible for scared money shot takers plus a lot of people play differently if they know you'll only run it once. I play below my means so even a 40 Buy in down swing would not hurt me. Thankfully that's never happen though
The definitive proof that running it multiple times from Turn -> River doesn't change EV Quote
05-21-2024 , 06:15 PM
#33
dude45
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Quote:
Originally Posted by Zamadhi
Some have made the argument to willingly keep the variance high:

1) Running it once increases the chance of getting deep stacked, which increases the edge of the better player
2) Running it once increases the chance of a tilt-prone player going on tilt, which increases the edge of the calmer player
Yep
The definitive proof that running it multiple times from Turn -> River doesn't change EV Quote
Yesterday , 04:54 AM
#34
dchen
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EV from RIT/RIO are identical regardless of how many cards are to come. E(RIT) = E(run 1 + run 2) / 2 = (E(run 1) + E(run 2)) / 2, since E(run 1) = E(run 2) this is just E(run 1) = E(RIO).

The principle that E(X+Y) = E(X) + E(Y) holds true even if X and Y are dependent. Google "linearity of expectation" for more info
The definitive proof that running it multiple times from Turn -> River doesn't change EV Quote
Yesterday , 09:44 AM
#35
KlemenZ
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The linearity of expectations is a theorem that needed to be proven and no person in this thread before has quoted it and its proof. Like I've already mentioned, the point was never to argue that RIT changes EV, I was responding to people who were giving some analogies without proof.

You either have to quote the theorem and its proof or prove things in a more convoluted way taking dependency into account, neither of which the first posters in this thread have done.
The definitive proof that running it multiple times from Turn -> River doesn't change EV Quote
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