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.

The perfect proof is one sentence. Maybe two. Sorry. Actually, really sorry

What? The proof is the proof and it's all there. I'm sorry if you can't understand it

Nobody runs it twice to change ev.

People run it twice to reduce variance vs ev.

(By definition, ev is “expected value” if you run it infinite times.)

That argument only works if you reshuffle every time. When you don't reshuffle you need the weighted average of the depleted decks to give you the same EV. But there is a simple logic proof that it always does. It's not just an intuitive thought. Its stone cold logic and it can be stated in a few sentences.

Yeah you didnt need math for this. The EV of cards #6 and 8 of the deck (second turn and river) are not changed by other cards being revealed, because this isnt a monte hall problem where you can change the action once you see some of the cards. You could reveal burn cards too, or even slowly reveal the rest of the entire deck for that matter if you want some especially exciting bit of showmanship at a home game, card #6 and 8 wont change, nor will the deck that it was randomly selected from.

I like to think of toy games, or at least to pick and isolate simpler spots of the true game. Let's say all in spots preflop HU in 6-max full tables where we had AA and villain had KK. If we always run it only once, after an infinite number of trials we will have won it 81.71% of the time.

What does this 81.71% number means? It means that for all possible combinations of flop+turn+river, you have 81.71% of them where you win.

In the scenario I formulated for the preflop action, after it goes thru, the dealer will always have 40 cards remaining to then form up to 8 totally distinct 5 card board textures. Logic dictates that, after an infinite number of different hands played, you will always go over all possible distinct board textures. No completely distinct board can appear more often than any other, even if card removal is a factor in any individual hand played.

The implication of the paragraph above is that, even though card removal is a thing on individual hands, it's not at infinity. Basically, card removal doesn't carry itself into the future, it is deleted as soon as the current poker hand ends.

The variance reduction effect is quite obvious. Given that variance doesn't exist over the infinite run, running it twice should make the infinite converge 2 times faster, running 3 times makes it converge 3 times faster etc.

The same spot for the KK, and then for any other possible spot, can be thought by just using analogy.

Please write down your stone cold logic mathematical proof. Because so far nobody in the ''this is just obvious'' camp has actually provided a rigorous proof, everyone just keeps repeating how ''trivial'' it is.

I did.

No, you didn't. The fact that cards 6 and 8 don't change doesn't mean that they are independent of each other. The 6th card is P(X given flop and hole cards) and 8th card is P(X given the whole first board - which includes card 6 - and hole cards). Also all this is under the assumption that are opponent's cards are random, otherwise it would be the 8th and 10th card, not 6th and 8th.

Wrong, Odds are calculated at the time all the money is placed into the pot, not at the time the card is about to be revealed, so the revealed cards are irrelevant to the calculation.

If i had a bag with 2 marbles in it, 1 red and 1 blue, and i said “i bet the 2nd marble i pull out is blue” what are the betting odds? Is it P(X) or P(X given the first marble)?

But you don't care just about the second marble, you also care about the first. You care about BOTH TOGETHER, about the COMBINATION OF BOTH. And the example with just 2 marbles is very bad because the color of one marble is automatically determined by the color of the other.

If you want a better analogy, suppose the bag has 100 ping pong balls in it and each ping pong ball is numbered from 1-100 (no two balls having the same number). You are interested in knowing the expected value of the ping pong balls selected (i.e., the average over all balls selected).

Then consider two different scenarios:

(1) Exactly one ball is selected;

(2) Exactly two balls are selected.

What is the expected value in each of the two scenarios *before* any balls are selected??

This is an oversimplification. This example that you have just provided is a somewhat good analogy for running it twice on river only, nothing more than that.

You are:

(1) stupid

(2) a jerk

Please stop posting in this thread.

You just described yourself. Your analogy wasn't even good, not to mention that analogies aren't mathematical proofs. If you weren't (1), you would most likely understand this, most people can.

Also the debate never was about whether running it twice changes EV, I'm pretty sure every single person in this thread thinks that it doesn't (some more blindly than others), the only debate was between people who believe hand waiving is a good substitute for rigor and people who understand that that is nonsense. The irony of you calling me stupid.

Imagine that five different people bet on five different runouts, cards 1-2, cards 3-4, cards 5-6 cards 7-8 and cards 9-10. Assume the odds each of them accepted was the fair odds for the first two cards. The five guys never even bother to look at anything but the cards they bet on. Would not each of those people have an expectation of zero before any cards were exposed? If the experiment was done a million times each of those gamblers would essentially break out even. Given that, it must be the case that the weighted averages of the cards exposed gives a variety of winning chances for gamblers 2, 3, 4, and five that average out t what the first guys chances were. If it didn't, then the four other gamblers would not have an EV of zero as they obviously do.

The example provided is no better than the ping pong example, the only difference is that you made some arbitrary groupings of 2. Neither your example or any of the previous ones take into account the fact that the boards in the real 'run it multiple times' problem can't be considered independent from each other. The gamblers are not betting on the runout of one specific board, but on whether they will win it zero times, 1 time, 2 times or n times. The example you provided, while mathematically true in isolation, has no relevance to the actual problem of running it multiple times, because, and I will repeat myself again, the gamblers are not betting on the runout of one specific board, but on whether they will win it zero times, 1 time, 2 times or n times, which makes the interactions between different runouts important.

This is exactly why math is done through rigorous proofs and not analogies.

The technique I used is perfectly rigorous. It is similar to a logic technique that will calculate the chances that a player with three dollars will eventually bust a player with five dollars if they flip a fair coin for a dollar a pop until one of them is busted. You can do that by summing some series but you don't have to. All you need to know is that if the coin is fair and they meet once a day to play this freezeout, their results will coalesce to dead even since a coinflip bestows no advantage. So, the three-coin guy wins three out of eight freezouts. The fancy equations will also come up with 3/8.

But the thing is that if we agree that the equations do indeed solve the problem, than I can get cute and say that I can solve the equation by invoking the logic even though I don't know how to solve the equation the normal way. Cute but completely rigorous.

Wrong. Determining that the EV of board #2 doesnt change by revealing cards beforehand (duh) means the EV of RIT doesnt change (duh) . The EV of n boards is the ev of the average of each individual board (duh), so if board 2’s ev doesnt change, then it has the same ev as board 1, and thus running it twice has the same ev. (Duh)

confidently telling a whole thread of people that their explanation isnt sufficient shows serious lack of introspection. Tell me honestly, did you ever stop for even one second and consider “is it possible the problem isnt the explanations, its me?”

Proofs can be done via natural language which is what my first post did. you just didnt understand it. I guess i couldve separated my postulates out, and included the even more obvious ones like that if board 2’s EV is the same as board 1 then rit’s ev doesnt change but even proofs reauire you to simply accept some base level facts like that 1+1= 2 (and other simple understandings of the world in natural language). Deduction still requires that the postulates be agreed upon, so someone dense enough could claim almost any proof to not be rigorous enough. I see no reason to bother as i genuinely think this isnt about you understanding, you just think saying rigor a lot makes you sound smart.

You realized you sound dumb and are now saying oh we all know rit doesnt change ev, whicn really just makes responding to you any further a fools errand.

How many times does someone have to tell you the difference between dependent and independent events? It's truly unbelievable how you stlll can't understand such a basic difference.

It's ironic, because you have just described yourself. Projection seems to be a big trend on this forum. I have thought about the possibility of not understanding something, but the same thing clearly can't be said about you, judging by your ''duh'' responses while not understanding a simple concept of probability.

Also, an argument from consensus is a logical fallacy and completely irrelevant to any serious debate. If the whole thread is wrong, that's not my personal problem.

1+1 is a super basic axiom - and even this can be derived from even more basic axioms. Trying to compare such basic axioms to a problem like running it twice is really quite something. Maybe we should just stop doing math and write ''trivial'' instead of a genuine proof.

Also, I never said that non-formal language can't be rigorous. And I'm using the word because it's relevant.

No, but keep assuming things without any kind of evidence, you are clearly very good at it.

Wrong again! Board #2 is “dependent” upon board #1 just as much as board #1 is dependent upon board #2, or the burn cards, or the final card in the deck, even in the circumstance a player runs it once. Board #2 is NOT dependent upon whether any card in the deck is revealed. Cards #6 and 8 do not change based on cards #2 and 4 being revealed, you simply have more information about the possibilities of their values. Expected value is calculated based on what possible cards could come out, however, expected value is calculated when the decision is made, which is before the first board is revealed, meaning the values of cards #2 and 4 are also irrelevant to this calculation, as the values were not known at the time of the decision.



Yes, its quite simple, and youre wrong about it. I already said, its the monte hall problem except you cant change doors. Do you know the monte hall problem?

3 doors, 1 has a car, 2 have goats. You select a door, monte hall reveals a goat, your odds of having a car behind your door are still 1 in 3 (meaning the remaining door is 2 in 3, but of course here you cant change doors, which means your EV of all 3 doors is 1 in 3, even if another door is revealed before yours,)

More erroneous reasoning from you. Argument from consensus is inductive reasoning, which has value. Youre right, everyone else COULD be wrong. But i suppose if literally every other person is right except you, that would be a you problem wouldnt it?



Yeah, im at the “deriving 1+1 from even more basic axioms” with you now that im forced to explain the timeframe EV is calculated. My god dude.