about "running it twice"

[Jezco]

Hello!

My question is fairly simple. I did not find the answer by using search quickly so I hope this isn't a post you get every day.
Is it always mathematically correct (more +EV) to run the turn and river twice when you are holding, let's say, top set against over pair with 2 outs, compared to running it only once? please if you can calculate quickly it would help.
also if you include meta-game (barry greenstein theory), does it make your opponents more agressive when they know
you'll "do business" when ever they semibluff into you?
And how does all this affect all together? please I need some help with this. All of this for big bet poker (nl/pl).

[AKQJ10]

Equal EV, less variance -> "better" to most people (but not the most visceral gamblers, I suppose)

[Louie Landale]

Running it twice has no effect on your EV, it just reduces your varience. That's generally a good thing if you want to stay in this game.

Its not such a good thing when you have the other tight player well covered and you prefer that he gets busted out, giving a chance for a bad player to take his seat.

[Jezco]

the guy who I talked to about this said it would affect EV when you're running this twice. In his own words "It's harder for him to scoop the whole pot, even if he takes half of the pot because he has fewer outs" this just didn't make sense to me and I considered meta-game aspect to be much more important anyway. This is why I'd love to see some calculations.
Is he really correct about the EV?

[Jezco]

My friend couldn't help but calculate this and it was equal EV. thanks for the first replies.

[AKQJ10]

[ QUOTE ]
"It's harder for him to scoop the whole pot, even if he takes half of the pot because he has fewer outs"

[/ QUOTE ]

Sure. It's harder for him to get scooped, too. When your opponents draw comes in and you've run it once, you don't get a push -- you lose the whole pot.

I'm glad your friend eventually came to the right conclusion. If you really want a generalized solution, it's not hard. For running it N times where your probability of winning each time is p and pot size = P, your expectation is:

sum (i = 1 to N) of 1/N * p * P

=>

1/N * p1 * P + 1/N * p2 * P + 1/N * p3 * P ... + 1/N * pN * P
where
p1 = p2 = p3 = ... = pN = p

collect the ps and the Ps

p * P * (1/N + 1/N + 1/N + ... + 1/N) for N terms

=>

p * P * 1

Which is the same as the EV of running it once.

[TheBronzer]

use search button

[PokerZombi]

Running twice is exactly the same. Even though some outs might be drawn on one particular drawing (changing odd of subsequently scooping), the number of ways you can run the two cards twice is still in the same proportion of long run wins / loses.

However if I'm way behind I like to do it once because a bad beat might put the guy on tilt.

[binions]

[ QUOTE ]
My friend couldn't help but calculate this and it was equal EV. thanks for the first replies.

[/ QUOTE ]

The thing I like about running it twice when I am ahead is that only 1 out of 4 outcomes is bad.

Win both = same as running it once
Lose both = same as running it once
Lose top, win bottom = better than runnning it once
Win top, lose bottom = worse than running it once.

Plus, you give up no EV by running it twice. Compare insurance.

[Dangeresque]

Yeah, it's no fun when we're in a big pot and they won't let us insure. AKo vs AKo and they won't let us chop in a $400 pot... grr at my province's gaming regulations.