I know it reduces variance, but does it save or cost you money in the long run? does it have no effect at all ? Lets say you got ak vs qq 10000 times, and you did it again 10000 times running it twice, would the qq or ak hand receive greater benefit from running it twice? would it have no effect on the outcome?

TLDR: From an EV standpoint lets say I played 50000000 hands, would running it twice be positive and make me more money, or less, or no effect?

no effect.

so its totally up to personal preferance to do it or not? you never receive an edge at all?

You never receive an edge at all

Furthermore, you can't even gain an edge by selectively running it twice. There is no scenario where you gain EV.

Tangentially, Barry Greenstein talks about a pseudo-edge that can exist if one player has a much bigger bankroll than the other player (can better handle the variance) and makes it known that he only runs it once.

In that case Barry can put pressure on under-rolled players by getting them to fold spots that are probably flips so that he picks up a lot of dead money this way. Whether Barry has the hand that he is representing or is bluffing.

No mathematical/absolute/true ev edge is gained or lost. Arguments can be made both pro/agains depending on many things such as:

I think there is some edge to one player due to card removal effects when there are multiple cards to come. But I haven't worked out the math yet.

You are mistaken, so don't waste your time. This issue has been discussed and proven to death on these forums.

Links? I've searched the forums for multiple threads but haven't found any mathematical proof of equal equity when there is more than one card to come. Which I think it would be too complicated to prove without computer simulation to work out numerous combinations. For all practical purposes, though, the difference is probably less than 1%.

You certainly don't need a computer simulation, and the proof requires virtually no math at all. You only need to think about it. With n cards to come, it should be obvious that your probability of winning with the second set of n cards will be exactly the same as with the first set of n cards BEFORE ANY CARDS ARE TURNED which is when you compute the EV. No set of n cards is any better than the other. If you can't see this, then stop reading right now, and never place another bet again. Now running it twice is the same as running it for half the pot each time, so if E is your EV of running it once, then the EV for each run will be 0.5*E, and your EV for running it twice will be the sum of the EVs for the 2 runs or
0.5*E + 0.5*E = E, the same as running it once.

The difference is after the first set of n cards gone, there is n less cards in the deck which would affect the probability of the second set. If n=1 it is fairly obvious the times where one player's equity goes up balances out the times one's equity goes down. For n>1 I don't think it is obvious.

Go back and read what I wrote in bold. I wrote it in bold for a reason. The card removal effect is irrelevant to the probability that you win the second hand BEFORE the first hand is dealt. This is fundamental.

Suppose you had 100 ping pong balls numbered 1 to 100. If you draw a single random ball, the average number on the ball will be 50.5. If you draw a ball and then draw a second ball without replacing the first ball, the number you draw on the first ball will effect the average number on the second ball, but the average for the 2 balls is still 50.5.

No. The decision to run it twice must be made before the first run. At that point your equity is exactly the same whether you choose to run it once or multiple times (any number). There is nothing to balance out.

After the first run, recalculating equity is meaningless as there is no new betting. Your equity is for all runs combined and is the same no matter how many runs you choose.




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I realize the following is completely unnecessary but there are some who just refuse to accept the "non-math" proof. So,here is a math proof that says exactly what others have said using basic logic, exemplified by the bolded statement below. I also know that we can simplify the math proof considerably but I think it useful to write out all the nitty gritty so as not to leave any doubt, but I'm not too optimistic about that.

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 r times, then the player’s expectation can be found in the following way:

Let
P = total pot

r = number of runs

Xi = 1 if player wins on the i-th run; else Xi = 0. Xi is therefore a binary random variable

Then the expected value of Xi is E(Xi) = W*1 + (1-W)*0 = W

For each run, if the player wins, he wins P/r according to the standard payout for running it r times.

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

Amount Won = X1*P/r + X2*P/r + . . . +Xr*P/r

= (Sum Xi)*P/r (the sum goes from 1 to r)

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

EV =Sum E(Xi) * P/r

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

EV = rW*P/r = 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 more than once.

I suggest Charlie runs all the computer simulations he wants, or considers all the funky situations that come to mind (e.g., one outers, two outers, etc.), until he is convinced one way or the other.

I'm convinced that running it twice has no effect on EV now, after calculating some cases, then reread his message.

There is an interesting case when it is three handed and two players agree to run it twice against each other though. It has been discussed here before. I don't remember all the details (e.g., it might require that the game be hi/lo split), but it is definitely possible for players to gain a significant equity edge by running it more than once while another player only runs it once.

http://forumserver.twoplustwo.com/25...ities-1410105/

Yeah but that wasn't really running "it" twice. That was running a completely different matchup the second time, two-way instead of three-way. Of course the equities change when you remove the third hand, and depending on what that hand contained, some advantage shifts to one or the other of the remaining players.

That's the one. Thanks.