Running hands twice

[icewizard008]

Suppose you are all in on the flop and can only catch one out to win (for example, set over set). Is it true that running it twice decreases your expected value?

Pot size = P
Your chance of winning once = x
Expected value of running it once = Px

Your chance of winning twice = 0 (x^2 in most cases, but 0 now because you only have one out)
Your chance of winning once and losing once to split the pot = 2(1-x)x
Expected value of running it twice = 0*P + 2(1-x)x*(P/2) = Px - Px^2.

Is the analysis correct?

[zeppo]

Unless I'm missing something obvious....

Using your notation, the EV of running it twice is
0*P (chance of hitting the card twice is nil)+
2 (x * 0.5P) (two chances of winning half the pot)
+ 0 (EV of not hitting)
= Px which is the EV of running it once

[btmagnetw]

you're twice as likely to hit that one out for half the pot. same EV, less variance.

[jay_shark]

Lets say villain contemplates and shoves all in otf for an amount $P . You think for a while and make the call . The probability you win is B .

EV(running it once ) = P*B -P*(1-B)=2*P*B - P

EV(running it twice) = P*B^2 - P*(1-B)^2=2*P*B - P .

Same as before .

Var(running it once) = P^2*B + (-P)^2*(1-B)-(2PB-P)^2
= 4*P^2*B*(1-B)

Var(running it twice) = P^2*B^2 + (-P)^2*(1-B)^2 -(2PB-P)^2
=2*P^2*B*(1-B)

So we can conclude that your variance is reduced by a factor of 1/2 when you run it twice . Your EV is exactly the same .

[JocK]

[ QUOTE ]
Pot size = P
Your chance of winning once = x
Expected value of running it once = Px

Your chance of winning twice = 0 (x^2 in most cases, but 0 now because you only have one out)
Your chance of winning once and losing once to split the pot = 2(1-x)x
Expected value of running it twice = 0*P + 2(1-x)x*(P/2) = Px - Px^2.

Is the analysis correct?

[/ QUOTE ]
No, it is not. If you take card elimination effects into account for the chance of hitting twice, you should do the same for the chance of hitting once and losing once:

Chance of winning first and then losing is x

Chance of losing first and then winning is (1-x) times x/(1-x) = x

[icewizard008]

But, when calculating the EV of running twice, it doesn't make sense to include P*B^2 because it's not possible to win it twice with only one out.

[GeniusToad]

[ QUOTE ]
same EV, less variance.

[/ QUOTE ]

[TrvChBoy]

I updated my running it twice spreadsheet to calculate the EV with a single out. It still makes no difference to the EV whether you run it once or twice with only one out. Yes, it is impossible to win twice with only one out, but it does not change the EV.

For a $100 bet:

Running it once with one out: -$95.45
Running it twice with one out: -$95.45

To check my math, see the Excel spreadsheet at
http://www.saltlakepoker.com/poker/runningittwice.xls

And see my post on the subject at:
http://forumserver.twoplustwo.com/showfl...part=1&vc=1

[Explicit65]

[ QUOTE ]
you're twice as likely to hit that one out for half the pot. same EV, less variance.

[/ QUOTE ]

[binions]

Running it twice or three times or 4 times or once or 10 times etc is the same EV.

Multiple threads on this. Search for them if you still don't get it.