High Stakes Poker - run it twice

[maurile]

[ QUOTE ]
...So I guess I have the same EV with my one-outer running it twice without shuffling?

You have proven nothing.

[/ QUOTE ]
Yes, you have the exact same EV. Cooker showed why.

[maurile]

[ QUOTE ]
Barry Greenstein on a Live at the Bike episode talked about how he doesn't like running it multiple times as he wants to punish people by making them have to play for the whole pot.

Thought that was an interesting perspective when I heard it.

[/ QUOTE ]
I suspect the reason is that losing a big pot causes most players to tilt. Barry figures he doesn't tilt as badly as his opponents do, so he wants to increase variance. If he loses, his EV is similar going forward. If he wins, his EV is increased going forward due to the other player's propensity to tilt.

Just my guess.

[NegativeZero]

Simple example:

$1-2 NL, I hold T-T in small blind, big blind holds A-A. Everyone else folds, I complete, big blind checks. We each have $1,000 behind.

Flop: A-T-5
Running once:

I win 4.545%, lose 95.455%, EV = .04545*1004 - .95455*1000 = -908.91.

Running twice:

I SPLIT the pot 8.971%, lose 91.029%, EV = .08971*2 - .91029*1000 = -910.29

I have $1.38 the worst of it by running it twice.

You are ignoring the possibility that A and T can both come up on the same turn/river, which happens 2 * (1/45) (1/44) = 0.1%. This accounts for the reason you should not run them out twice in this case.

[graydot]

[ QUOTE ]
Simple example:

$1-2 NL, I hold T-T in small blind, big blind holds A-A. Everyone else folds, I complete, big blind checks. We each have $1,000 behind.

Flop: A-T-5
Running once:

I win 4.545%, lose 95.455%, EV = .04545*1004 - .95455*1000 = -908.91.


[/ QUOTE ]


Your numbers and math are off, recheck them

[maurile]

[ QUOTE ]
$1-2 NL, I hold T-T in small blind, big blind holds A-A. Everyone else folds, I complete, big blind checks. We each have $1,000 behind.

Flop: A-T-5
Running once:

I win 4.545%, lose 95.455%, EV = .04545*1004 - .95455*1000 = -908.91.

[/ QUOTE ]
Huh?

I don't know where you're getting that, but you're going to win 43 out of 990 times (4.3434343%) and lose 947 times (95.6565656%).

Those numbers are true for both the first iteration and the second iteration.

The numbers for the second iteration are different given a certain result of the first iteration, but that doesn't effect the overall EV.

[MrDannimal]

[ QUOTE ]
[ QUOTE ]
The more you run it the more times you come out loser.

[/ QUOTE ]

that's some strange math you've got going

[/ QUOTE ]

No, he's right. What he leaves out is that it also means more times you come out the winner.

If you run it 10 times, you lose 7 and win 3. 20, you lose 14 (more) and win 6 (more).

He's right, but not how he wanted to be right. [img]/images/graemlins/smile.gif[/img]

[Cooker]

[ QUOTE ]
...So I guess I have the same EV with my one-outer running it twice without shuffling?

You have proven nothing.

[/ QUOTE ]

I am not going to post on this stupid topic again. I showed in detail that the EV for a 4 outer is exactly the same whether you run it once or twice with no reshuffle. Due to the clumsiness of forum formatting, I didn't show it holds for the general case of running it N times with M outs and N killer cards and no reshuffle, but rest assured the technique and results are the same. The EV doesn't change and is absolutely identical. I can lead you to water, but I can't make you drink.

I then gave an intuitive and correct arguement based on burning cards that the EV of each run is identical. If you don't see why after reading that, then you have gaping holes in your understanding of probability that I cannot hope to patch in this setting. What if your 1 out is a burn card in a live casino? Did your EV change because they burned cards? What you are suggesting is just as absurd as saying the dealer burning cards changes your EV. Of course it doesn't and everyone here should know that.

As to your question, yes you have the same EV running it twice with your 1 outer and I will prove it. When you run it once your

EV = 1/45 +44*1/45/44 = 2/45.

When you run it twice with no reshuffling you have

EV = 1/2*(probability hit first run + probability hit second run)

= (1/45 +44*1/45/44 + 44*43*1/45/44/43 + 44*43*42*1/45/44/43/42)/2

= (1/45 + 1/45 + 1/45 + 1/45)/2 = 2/45.

BIG SURPRISE, EXACTLY THE SAME AS THE 1 RUN CASE. Not close, not within rounding or some such, but EXACT. Also, that was much easier than the example I actually did, since you don't have to worry about using up outs when you hit twice in the first run, winning the whole pot versus winning half the pot,etc.

[BarronVangorToth]

Cooker, you're right and its EV neutral running it more than once, which then leads to the question regarding my quote in the other thread from Barry Greenstein as to why he said on Live at the Bike that he doesn't (in general) run it multiple times.

After all, if BG said something, invariably that means that is the EV move, yes...?

No.

I think his sole motivation for not running it more than once is to make people less likely to call him knowing that they will not have the option of running it multiple times to smooth out their variance.

"You want to call me -- you're playing for your stack, and not 1/3rd of your stack three times on a 40% shot."

Imagine this scenario:

You're going to wager $1,000,000 against someone heads-up.

You're flipping a coin where the coin will come up 51% of the time in your favor.

It's a EV move but can you call? Can you take one crack risking a million?

OR

Imagine you know that you're going to run 1,000,000 flips each for $1. Or 1,000,000,000 -- now that 51% looks like an immediate call, *EVEN THOUGH THE EV IS THE SAME*.

[Cooker]

Certainly, the decision on whether or not to run it multiple times is pure metagame and may influence the overall EV of your play.

If you feel the people you play with are more likely to gamble with you in -EV situations if you are known to run it twice then running it twice would be +EV for you from a metagame point of view.

Likewise, if you think people are on scared money and are likely to fold in close situations where they would have the correct pot odds to draw out on you if you are known to never run it twice, then it would be +EV to never run it twice.

I think it depends mostly on the types of games you frequent. In a soft game, I think you want to be known to run it twice, while in a tougher game maybe that extra fear factor will be more useful to you.

[NegativeZero]

You are correct, my numbers and math were both wrong.

EV in my example = -912.96 in both cases.