The definitive proof that running it multiple times from Turn -> River doesn't change EV

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.

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.

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.

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.

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

Yes. I like increasing my edge after losing stack 😀

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

Yep

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 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.

I'm back fellas. I forgot to send any updates in this thread after any replies.

After David Skalanasky sent such a rude and unprofessional reply, I was off put from checking this thread again for a while. I don't know what mathematician talks the way that he does, it's clear he's far too prideful and didn't even engage with the topic or proof on its own merits.

The proof I posted for running it multiples from turn to river is completely correct. However, there's a general theorem about the "expectation value" of any random variable which is simpler, and also applies to the situations of running the deck multiple times from flop to river or preflop to river.

I proved it with brute force for turn to river, but there's a simpler theorem in probability and statistics which can be used, it's the "linearity in expectation". I found out about this later, because I don't know much about statistics or probability other than what I was taught in high school.

So for any poker players who want to see the proof which is true for running it multiple times from turn to river, flop to river, or preflop to river, look up a proof for "linearity of expectation"

You can look at the post I created as a demonstration of sorts of a specific example, showing the specifics of how it's true for turn to river.

For flop to river or preflop to river, it becomes almost impossible to demonstrate every branching tree in specifics, so for that you look at the abstract proof for linearity of expectation and then modify the variables appropriately for what's going in poker to get the desired statement at the end

Yep, this is it

There are arguments either way and it comes down to preference. Someone who is good may want to tilt other players, or they may want to run it multiple times to decrease [the standard deviation]/[variance] in how much their actual winrate deviates from their theoretical winrate.

I think that becoming the sort of the person who's happy running it once is a strength if you don't mind the "increased" (or just "normal amount of") risk and want to tilt people

You're both probably right but this was so long ago I've forgot all the details of this

The whole point of this thread was "proving it", not "disputing it" , which some of the posters here didn't seem to understand

Intuition is important and good, and the posters in this thread have intuition for why it's true. I have some intuition for why it's true but that's not an actual proof. Nobody is saying you can't have intuition, but proof relies on pure math.

The proof for the general case is in the linearity of expectation theorem