Running It Twice: who does it favor?

[captZEEbo1]

Let's say two people go allin on flop, and in order to reduce variance (perhaps b/c they are playing out of their bankroll), they run it twice, meaning they deal the turn and river and whoever wins that gets 1/2 the pot, then they put those 2 cards to the side, and put out a new turn and river and the winner of the second turn/river wins the other 1/2 of the pot.

Outside of meta-game reasons (it'd be good to not let someone playing outside their comfort zone to run it twice, b/c you'd want them to fear going broke or whatever), my intuition says that this should favor someone, but I don't know for sure. Obviously if the turn and river went back into the deck and it was reshuffled, it'd be neutral ev.

I read about it here, and it sounded like it took doyle some time before he agreed to run it twice. " 'Oh, I don't know; I don't see how I could win it twice!' Doyle quickly understood that despite the fact that he had the best hand at the moment, my draw was a significant favorite to win the pot." That seems like some bad logic by Doyle, unless he was just holding out for metagame purposes.

[pzhon]

Running it twice favors no one. It just reduces the variance.

E(A+B) = E(A) + E(B), even when A and B are not independent.

[captZEEbo1]

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Running it twice favors no one. It just reduces the variance.

E(A+B) = E(A) + E(B), even when A and B are not independent.

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but they don't reshuffle the turn/river cards back into the deck, shouldn't that affect something?

[pzhon]

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Running it twice favors no one. It just reduces the variance.

E(A+B) = E(A) + E(B), even when A and B are not independent.

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but they don't reshuffle the turn/river cards back into the deck, shouldn't that affect something?

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It means the deals are not independent. It decreases the variance slightly. It does not change the EV.

[RiverTheNuts]

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Running it twice favors no one. It just reduces the variance.

E(A+B) = E(A) + E(B), even when A and B are not independent.

[/ QUOTE ]

but they don't reshuffle the turn/river cards back into the deck, shouldn't that affect something?

[/ QUOTE ]
It means the deals are not independent. It decreases the variance slightly. It does not change the EV.

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I think ya gotta be wrong here, with random cards, if someone wins, they most likely hit a pair or more, without the turn and river cards that probably had an impact on them winning shuffled back into the deck, its gotta favor the first hand loser...

[TomCollins]

Search the archives, we proved its exactly the same even if you don't reshuffle.

[Mountainhawk]

Take an extreme example. Imagine their are 10 cards in the deck. Red 1,2,3,4,5 and Black 1,2,3,4,5

You have 5,4 Red He has 1,2 Black.

The board is 5B, 4B, 2R, 1R.

3B gives him the straight flush and he wins, and 3R gives you the straight flush and you win.

You are 50%, he is 50%.

If you run it twice without replacement, You each win 50% of the pot with no variance.

EV the same, variance reduced.

[pzhon]

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It means the deals are not independent. It decreases the variance slightly. It does not change the EV.

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without the turn and river cards that probably had an impact on them winning shuffled back into the deck, its gotta favor the first hand loser...

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That the first hand loser is favored means the variance is reduced (relative to reshuffling). It does not chance the EV, as you don't know who wins the first time.

Consider the extreme example that the rest of the deck has two cards, one which gives you a win, and one which causes you to lose. Dealing once or twice gives you an average of 50% of the pot. Dealing it twice helps whoever loses the first deal, but you always get 50% of the pot.

Let me make this very explicit:

Suppose the probability that you win is p.

The value from dealing it once is p*pot.

The value from dealing it twice is probability(win 1st time)*(.5*pot)+probability(win second time)*(.5*pot). Both of those probabilities are p, though the events are not independent. (Winning the second time is like winning after burning a few extra cards. Burn cards do not change the probablity of winning.) So, the value is p*(.5 pot)+p*(.5 pot) = p*pot, the same as if you deal it once.

Dealing it twice favors no one. It just decreases the variance.

[yellowjack]

This question reminds me of a question in class:

Suppose there are 5 participants, and each one gets a coin. There are 5 coins, 3 of which are gold. Each person randomly selects a coin, one by one. Are the chances of getting a gold coin even for all participants? Why or why not?

Answer: Surprisingly, against basic intuition, each participant has an equal chance of getting a gold coin. [img]/images/graemlins/shocked.gif[/img]

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I do see what Doyle means by not being able to see himself winning it twice. Suppose you're a 75% favorite to win, and you run it twice. There is only a 56% chance of winning it both times. Granted the chances of you losing both times are now just over 6%, but that's besides the point. I've read every reply that states the EV is the same, and realize this. I just want to point out that Doyle does have a point, and that it's not a misconception.

[meow_meow]

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Running it twice favors no one. It just reduces the variance.

E(A+B) = E(A) + E(B), even when A and B are not independent.

[/ QUOTE ]

but they don't reshuffle the turn/river cards back into the deck, shouldn't that affect something?

[/ QUOTE ]
It means the deals are not independent. It decreases the variance slightly. It does not change the EV.

[/ QUOTE ]

I think ya gotta be wrong here, with random cards, if someone wins, they most likely hit a pair or more, without the turn and river cards that probably had an impact on them winning shuffled back into the deck, its gotta favor the first hand loser...

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Arg.
It's been discussed ad nauseum. As long as you don't reshuffle, it is EV neutral.