Running It Twice
 

for those who don't know what it means i think it's easiest to show an example:

two players playing NL holdem get it all in on the flop, player A has K[img]/images/graemlins/heart.gif[/img]K[img]/images/graemlins/spade.gif[/img], player B has
A[img]/images/graemlins/diamond.gif[/img]J[img]/images/graemlins/diamond.gif[/img], and the flop is 9[img]/images/graemlins/diamond.gif[/img]4[img]/images/graemlins/diamond.gif[/img]2[img]/images/graemlins/spade.gif[/img]

after seeing each other's hand and realizing it is about a coin flip, they agree to 'run it twice'

the dealer now deals the turn and river, and a second turn and river:

first turn and river: 8[img]/images/graemlins/diamond.gif[/img]5[img]/images/graemlins/diamond.gif[/img]
second turn and river: T[img]/images/graemlins/spade.gif[/img]J[img]/images/graemlins/club.gif[/img]

now, because they won one each, they split the pot

had one of the players won both boards, he would take the whole pot

running it twice reduces variance, which many players are happy to do, but my question is can running it twice change the EV for the players?

Re: Running It Twice
 

Assuming the turn and river cards are replaced and reshuffled, then EV remains the same. (Variance is halved)

Re: Running It Twice
 

[ QUOTE ]
Assuming the turn and river cards are replaced and reshuffled, then EV remains the same. (Variance is halved)

[/ QUOTE ]

They aren't.

Re: Running It Twice
 

EV is still the same even if the cards are not shuffled. Let me think about it for a bit and try to come up with a proof.

Re: Running It Twice
 

Id like to see the proof, since this is counterintuitive to me.

Re: Running It Twice
 

Call the players A and B. The key observation is this: the probability that player A wins the second run is the same as the probability that he wins the first run. What may confuse some people is this: the conditional probability that he wins the second run, given that he won the first or given that he lost the first, is something different. But before the runs take place, the probability he wins the second run is exactly the same as the probability he wins the first.

So let P be the size of the pot, let X=1 if player A wins the first run and 0 otherwise, and let Y=1 if player A wins the second run and 0 otherwise. Then the amount that player A wins is

X*P/2 + Y*P/2

and his EV is

E[X*P/2] + E[Y*P/2] = [P(X=1)+P(Y=1)]*P/2
= [2*P(X=1)]*P/2
= P(X=1)*P.

On the other hand, if they only run it once, he will win X*P and his EV will also be P(X=1)*P.

Re: Running It Twice
 

[ QUOTE ]
The key observation is this: the probability that player A wins the second run is the same as the probability that he wins the first run.

[/ QUOTE ]
That's one of two basic facts that may be helpful. The other is that E(X+Y) = E(X) + E(Y), even when X and Y are not independent.

Re: Running It Twice
 

Here is an example showing how EV is the same even without reshuffling. This is not a proof that this always works. I'll try to get that in another post.

Suppose you have KK, opponent has 55. Board is K952.

He wins 1/44 times. EV = 1/44*pot for opponent.

Suppose you run this three times without reshuffling:
He has 2 possibilities:
Lose all 3:
(43/44)*(42/43)*(41/42) = 41/44. EV = 0

Win 1, lose 2:
3/44 (1- 41/44).
Also: (1/44)+(43/44*1/43)+(43/44*42/43*1/42) = 1/44+1/44+1/44 = 3/44.
His EV = 3/44 * 1/3 pot = 1/44*pot.

Total EV for taking 3 cards is 1/44+0 = 1/44. Exactly the same as taking 1 card. But there is less variance since he wins 1/3 of the pot more often.

Re: Running It Twice
 

So instead of playing say one $10 pot with a particular flop of XYZ , they're essentially just playing two $5 pots with XYZ flops, and asking the dealer to expedite the process? Makes sense.

Re: Running It Twice
 

Your original example might hint at why I think it does change the EV.

Basically, AJs needs an A or a diamond to win (lets exclude other more obscure cases, like JJ or some kind of diamond combo that gives KK a boat).

AJs needs only ONE diamond to win, but in the first turn and river, he used TWO of his outs. When running this once, that's irrelevant, and won't change anything. When running it twice, it slightly reduces his chances of winning the second run. Same if he got a diamond AND an A in the first run.

As I see it, the idea of running it twice is that doing so would maintain an "average" probability of each hand winning over the two runs equal to its chance of winning the first run. But I would think that using two outs in the first run would violatet this assumption. I'm going to take some time soon and do some math on this, it really interests me.