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

Before I post the proof, I just want to say I'm looking for any mathematicians or statisticians lurking around here. That is the main purpose of me writing this post.

I'm looking for any mathematicians familiar with poker, because I'm not sure how hard it would be prove that EV is unchanged if you run it multiples times from Flop->River or Preflop->River instead. Are there any proofs for this online anywhere?

I'm going back to university later this year to study physics, but I haven't studied statistics or probability formally beyond a high school level.

Time for the proof, the scenario is this:

Two players are heads up and all-in on the turn. Let's label them as "Player 1" and "Player 2".
We will prove that no matter how many times they decide to run it after being all in (whether once, twice, 3 times or 10 times), the EV for Player 1 (and hence Player 2 by extension) is unchanged.

They decide to run the river card "q times", where 1 ≤ q≤ N, and N represents the number of cards left in the deck for them to use. If someone wants to run it three times, set q=3. If someone only wants to run it once, set q=1.

We will create a function EV(q) which is the EV of player 1 if they run the turn to river card q times, and spoiler alert, the final simplified expression is independent of q. In other words, EV is unaffected by however many times you decide to run it.

The pot is size "P" after all players go all-in on the turn (assume there's no rake). The pot is split between the players in proportion to the number of runouts out of q they win.

If player 1 wins "x" out of q runouts , they win x/q of the total pot P, and will hence win x/q * P. So let's make "x" as a variable which denotes the number of runouts that player 1 wins out of q runouts.

Let "k" denote the number of cards in the deck out of N cards for which if they are dealt on the river, player 1 will win that board. When you run it multiple times, what you're doing is creating multiple boards. After running it q times, there will be q boards generated, and then you'll get a payout relative to how many x boards out of q you win.

If "k" denotes the number of cards in the N card deck remaining that player 1 will win if they arrive on the river, then "N-k" denotes the numbers of cards on the river for which player 2 will win that runout if that card appears.

For the of simplicity, I've assumed that there's no chop outs. I.e., there's a binary choice where each card in the N-card deck either makes player 1 win or makes player 2 win.

Constructing a probability function
Before calculating the EV of running the deck q times, the first step is to create a multivariable probability function " f(N,q,k,x)" which denotes the probability that player 1 will win x out of q run-outs. This function uses a lot of factorials and combinatorics/binomial functions.

The most compact form to write this function in on the end takes the form of what's called a "hypergeometric distribution". This proof uses standard theory related to that, where we end up taking the mean of this function later on.

After writing an equation for the EV, someone who was a mathematician told me that there was a "hypergeometric distribution" inside of it, and so I'll start off by writing this function.

If anyone has any questions about why this function models the situation, ask away

This function takes the form of something called a "hypergeometric distribution" . The deck is a finite resource which isn't replenished between multiple runouts, hence it keeps decreasing as you run out the deck more and more. And the outcome of each draw is a binary result of win or lose, and that's why the situation is modelled in this form.

An important observation to be used later is that:

With this, we can now write an expression for EV

After a bit of re-arranging, the EV is now equal to P/q multiplied by the mean of a hypergeometric distribution. Calculating the mean of this distribution is in standard theory:

In the first equality, I change the index of summation from (x=0 to x=q) to (x=1 to x=q) instead. Why? Because when x=0, the whole thing is multiplied by 0, and hence it can be removed from the sum. And this makes sense intuitively, because when you win 0 of the q runouts, you win 0% of the pot, and hence that tree can be removed when you're calculating EV. And EV is the sum of your winnings across all different possibilities. If you win 0, we can ignore that term

The other thing which happens in the first equality is re-writing combinatorics functions by using the algebraic property

Alright, so now we're left with a new summation to deal with, multiplied by (kq)/N.
However, it can actually be shown that after altering the combinatorics functions this way, with the above property and taking out some factors, we've generated a new hypergeometric distribution again, and all we're doing is summing all the probabilities, which makes that term = to 1

By substituting dummy variables, it has been shown that this is another hypergeometric distribution, and summing all the probabilities of those is therefore 1.

And now finally, to calculate the EV of running the deck q times.

In other words, your EV is k/N * P, which is your equity multiplied by pot size P.

In conclusion

This proof showed that EV unaffected by the decision to run it multiple times from turn to river. It uses a simple hypergeometric distribution.

Proving this for flop --> River seems a lot more difficult than turn --> River, and the above was for turn --> River

Considering running it multiples times from flop to river:

Let's say it's heads up again, and both players are all-in on the flop and they run it three times

Let's say N=45 for 45 cards left in the deck (doesnt matter if you make it 44 or 45)

You have 45C2=990 possibilities for the 1st board, then 43C2=903 possiblities for the 2nd board, 41C2=820 possiblilities for the 3rd board

It becomes a lot more complicated to calculate expected profit, because sometimes the turn card and river card work in combination work together to produce a winning hand for a player (so someone can get very lucky)

In a hypergeometric distribution, a singular card is always good for you or a singular card is always bad for you. This works for turn to river, where only 1 card peels off to finish the board.

But if you run it from flop to river multiple times, there's two cards in combination working together. Often times no singular card is good or bad for you, its the combination of two cards from a "fixed resource" (the deck) that counts as a success or failure. (think backdoor draws, or maybe you hit an out on the turn but you don't win because your opponent hits a river card which makes them win)

I'm assuming that the EV is unchanged from running the flop --> river multiples times, I just don't know why or where the proof is.

I can maybe understand intuitively that it makes no difference to EV however many times you run it from flop to river.
Because intuition wise, what often happens in probability scenarios is that when one event fails for a desired outcome, that outcome becomes more likely to happen in the next event. And then when you analyse the entire tree, it all balances out by symmetries and terms cancelling out in the numerator and denominator, to give you a final figure which matches the probability of the event occurring at the first available opportunity.