First of all, I know that running it twice (or more times) never affects EV. And having a degree in math, I’m sure I can work out a proof. For example, with only one card to come, suppose we win the first runout - then we lose a card on the second runout that would’ve been a win. But it’s fairly easy to show that the chances we lose on the second if we won the first are exactly offset by the chances we gain on the second if we lost the first.
It’s not so simple, however, for runouts of two cards (both turn and river) or of whole boards, and especially when we’re running out three or more times - where we have to consider all two card or five card combinations and how those combos are reduced with each runout. Still, I have no doubt I could prove this mathematically in a cumbersome way for any given scenario (doing it generically regardless of number of cards to runout and number of runouts would be more difficult I think).

What I’m looking for is an intuitive, easy to see, reason showing why running it twice doesn’t affect EV. Does such an explanation exist?

I’ve looked at this, and there is no intuitive proof.

The reason seems to me to be that a set of two card runouts using the whole deck may deviate significantly from EV.

This is contrary to the fact that a set of one card runouts using the whole deck will always deliver very close to your EV (most live poker) or exactly your EV (most online poker).

For example if you are playing live HU Holdem and looking for any club to win (having two in your hands and two on the board) the 7-9 clubs remaining in the deck can be split across 9 runouts (out of 21) or 4 runouts (out of 21).

There is.

It should be enough to show that the first run has the same EV as each additional run, since it trivially follows from that that running it n times doesn't affect EV.

It's easy to see that the first run has the same EV as the nth run because there is no information to differentiate cards by their order. Each unknown card, as well as each group of unknown cards of the same size, therefore has the exact same probability distribution, so the EV of each run is identical.

I'm sure someone could construct a simple and short formal proof from this if they wanted.

I wrote this math proof way back then. I thought it was right then and assume it's still right .

Two players are in a showdown situation, with b board cards yet to be seen (b=1, 2, or 5). If a player has a winning probability of W, then, prior to any board cards being dealt, W is the probability that each set of b cards in the remaining deck will give the player a win. For example, the second set of b cards is equivalent to the dealer burning off b+1 cards before dealing instead of the usual one burn card. If the showdown is run twice, then the player’s expectation can be found in the following way:

Let
P = total pot

X1= 1 if a player wins on the first run; else X1 = 0. The same is true for X2, the random variable representing the second run result. X1 and X2 are binary random variables.

Then the expected value of X1 is E(X1) = W*1 + (1-W)*0 = W and the same for X2, since we are defining these variables prior to any action.

For each run, if the player wins, he wins P/2 according to the standard Payout for running it twice.

Therefore, we can write the amount won in 2 runs as

Amount Won = X1*P/2 + X2*P/2 = (Sum Xi)*P/2

Since the expected value of a sum is equal to the sum of the expected values,

EV =Sum E(Xi) * P/2

But, E(Xi) = W for all i; therefore

EV = 2W*P/2 = W*P

But, W*P is the EV for the player if it is run only once, proving that EV does not change with running it twice. The above proof easily extends to running it r times.

To show that the variance for RIT is lower than it is for running it once (RIO), we note that Var(cX) = c^2Var(X). Since the variance of a Bernoulli variable with mean W is W(1-W), we have,

Var(RIT) = 2*(P/2)^2*W(1-W) =1/2 * P^2*W*(1-W) <= P^2*W(1-W) = Var(RIO)

Oh duh, thanks statmanhal… yes, expected value of a sum is equal to sum of expected values really says it all. And the proof of linearity of expectation here is succinct and intuitive: https://math.stackexchange.com/quest...-are-dependent
Basically if we sum up the individual expectations weighted by probabilities of occurrence, it’s easy to see that’s the same as the expectation of the sum weighted by the same probabilities of occurrence. And when all of the probabilities of occurrence are added together, we get “the expectation of the sum is the sum of the expected values.”

Bonus question… Is it possible to show that variance of running it multiple times without replacement (Ie normal definition of running it) is less than or equal to variance of running it the same number of times *with replacement* (ie put cards back in deck and shuffle after each run)? I expect this to be true. I’d also expect the variance to be *strictly* less without replacement for two or more runs, except for the trivial case of all runouts being equal - ie 1 player having 100% equity, or all runs being chops.

I had a long debate (and ultimately won a bet) on this topic some time ago in a live poker room.

The bet was something like "On the river, running out the entire deck delivers exactly the expected value every time."

My bet was that the above statement was untrue.

i tend to think it doesn't affect EV much.

the one thing i wonder about is if you need a miracle for a suckout. but even then i don't think it matters. i don't think variance has value, but not certain.

one thing people might focus on is "turn/river #1" you wasted one of the cards you needed, so the "next turn/river" has smaller win %... BUT if you didn't hit a wasted good card on "turn/river #1" then your chances on the "next turn/river" are higher.

the big thing is whether somehow variance has a positive $EV i don't think it does... obviously, it may have a higher "maximize utility" EV (i.e. lower variance, which most people like)

As far as the utility of reducing variance goes, it really depends on the amount risked for most people. Consider a toy example - you’re on “Let’s Make a Deal” and Monty Hall hands you a quarter. If you flip the coin twice and it comes up heads both times, he will give you $1000. He offers you $100 to give the coin back and quit. Do you flip or take the sure payout?

The EV of flipping is obviously $250. Many people would be willing to risk the sure $100 to gamble for $1000, especially given that gambling has a higher EV. Change it to $10 million if you flip and $1 million to quit. I’d bet a lot more people take the sure million.

In real poker, this goes to the notion of proper bankroll management and not playing with scared money. You should not be playing at stakes where you are tempted to take lower EV in order to reduce variance.

Well, what is EV? It's is already the convergence limit of running the hand an infinite amount of times.

So why should the EV of running it once suddenly be different from that of running it twice and then the EV of running it an infinite number of times be suddenly again the same as running it once?

There's no reason why The EV of n runs (for n > 1) should deviate.

I would say running it multiple times does affect EV.

In my opinion, it should be somewhat intuitive:

Lets say we have an flush draw against an overpair on the flop without blocking the flush cards for simplicity so all 9 outs are clean.

if we catch multiple of our outs on the first runout, then our EV for the second runout must decrease

We have roughly 36% to hit. (9x4)
after two flush cards we are now 28% (7x4) on the second board

To take this to the extreme, lets say we have the option to runout the entire deck. once there are no flush cards left, our EV should be zero

Yes, but you did not consider the following. If we do not get any of our outs on one run, that increases the probability you will hit on the next run. The math shows this to result in the same EV as running it once taking into account that for n runs, the amount won on any run is 1/n *Pot,.

Consider a very simple example. You have 3 cards, 2 red and 1 black. Selecting the black wins. Win amount is 12

Run It Once; P(win) = 1/3 : EV = (1/3)*12= 4

Run It Twice: P(Win) = P(win on 1st run) + P(lose on 1st run)*P(win on 2nd run|lose on 1st run)

EV = (1/3)*6 + (2/3)*(1/2) *6 = 2+2 = 4

Run it Three Times (third run win= lose on 1st and on second and win on third)

EV = (1/3) * 4 + (2/3)* (1/2)* 4 + (2/3) * (1/2) * 1 * 4 = 4

Only after you know what the first board is.


Are you also going to look through all the folded cards to see how your outs have decreased?

I wonder if this will help it make sense. Think of the cards as if they were face down. One entire board is dealt face down, and a second entire board is dealt face down. Intuitively (I hope) it seems like at this point your chances of winning on one board are the same as on the other.

If we decide before looking at the cards that we are only going to usee board 1, or only going to use board 2, or we are going to use both boards and split the pot, how can that possibly change the odds? We haven't learned anything new, we can't change our actions once we see the cards, etc. Of course it is always possible that your gin cards are already mucked, or in the stub of the deck, but that is always the case when we talk about poker hand odds.

Eh, maybe that doesn't help at all.

Your equity stays the same if you run it out twice.
(I am copying this answer from another forum)(Credit to Haizemburg)

Just(roughly math from top of my head) for run it twice KKvAA
-64%of times you'll win whole pot
-4% of time you lose everything
-32%of time you split the pot

EV=0.64*100-0.0.4*100=0.6*100
For run it once
EV=0.8*100-0.2*100=0.6*100

-------------------------------------
So, If Hero refuses to run it twice and Hero holds AA v KK,
then Hero wins villains 100bb 80x out of 100 (+8000)
Hero loses his 100bb 20x (-2000)
Total EV (+6000)

If Hero accepts run it twice,
the Hero wins villain's 100bb 64x (+6400)
Hero loses his 100bb 4x (-400)
hero and Villain break even 32x = 0
Total EV is the same (+6000)

Only as it affects the play of the hands, which it does. I don't (or only very rarely do) run multiples. Strongly prefer not to whether a good spot or bad. I make exceptions if someone is really asking hard and all-in or if it seems to make you the enemy of the table not to. I prefer not to play long hours nowadays and that is one reason to just let the volatility happen.