Running it Twice.....what are the dis/advantages???

[TheWorstPlayer]

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Why is this such a difficult concept? Instead of being in one pot for, say, $100 you are just saying let's play the exact same hand twice for two $50 pots. So if you have 70% equity in the $100 pot ($70) then you have 70% in both $50 pots ($35 each = $70).

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Absolutely false.

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No change in (EV).

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Wow. Magically, the correct end result.

Your explanation is typical of the idiots who think it is "obvious" the EVs are the same. If you initially think it's obvious, you are probably an idiot. If you don't think it's obvious, think it over, and then realize it was obvious, you are probably fairly intelligent.

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WTF? If you're implying that once you draw the card for the first half pot you no longer have the same equity in the second half pot, that is correct, but irrelevant since before you draw the first card you still have 70% equity in BOTH halves. If you're talking about something else, explain. And don't call people idiots.

[somapopper]

I don't know why this was hard for me to figure out, but I thought shuffling mattered to. Here's some math to prove that it doesn't (and that the ev doesn't change running it twice the standard way).

To simplify things let's say you have one out.

running it once: 1/44

run it twice, no shuffle: 1/44*.5 + (0/43*1/44 + 1/43*43/44)*.5

the second term = 43/1892 which reduces to, surprise surprise, 1/44.

run it twice, shuffle: 1/44*.5 + 1/44*.5

The thing that makes it clear that shuffling doesn't matter is once you understand that putting down the river, shuffling the cards, and putting down the river again, is exactly like playing the exact same hand on two different occasions. Obviously your EV for these two different hands won't change.

[Wintermute]

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Question: when you run it twice, do you shuffle the first turn/river back into the deck before running it the second time? Without replacement, the EV will change.

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Incorrect Sir, the EV *over* however many times you run it will never change, although the EV from one particular run will change, over the full amount of times you run it, the average is the EV of just running it once.

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I think this is right. I tried to come up with a counterexample and couldn't.

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An obvious way to see that Ribbo is correct is to simply imagine running it as many times as there are cards left in the deck (of course then you would also want to know all the dead cards to calculate the percentages correctly). Then you would always win exactly the EV of dealing it once but with zero variance. I think this is a fairly clear way to show that the variance goes down while the EV stays the same as you increase the number of cards dealt out without shuffling.

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You've got a set over set situation in hold'em. Say the board is such that no other draws can win the hand for the player who trails. Now, you claim that we deal out boards with the rest of the entire deck and the EV gets distributed as the percentages dictate. Not always: if one of the boards gives both players quads, then the underset will never win, yet we know he does not have zero EV. QED. Ergo, I am smarter than you, per se, as it were.


<font color="white">Of course this is wrong, it's just that the granularity of the number of boards the deck can deal out is not fine enough to catch the small EV that the underset has every time you run this. If you dealt the remainder of the deck out a few hundred times, you'd get the right EV. So I agree w/ you and Ribbo, the EV does not change. </font>

[somapopper]

Yeah, once you start adding the turn card in to the analysis things do get a lot trickier. If you want to take this "run it every way" approach, you actually have to run every river for every single turn card. That would be fun.

[Cooker]

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Question: when you run it twice, do you shuffle the first turn/river back into the deck before running it the second time? Without replacement, the EV will change.

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Incorrect Sir, the EV *over* however many times you run it will never change, although the EV from one particular run will change, over the full amount of times you run it, the average is the EV of just running it once.

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I think this is right. I tried to come up with a counterexample and couldn't.

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An obvious way to see that Ribbo is correct is to simply imagine running it as many times as there are cards left in the deck (of course then you would also want to know all the dead cards to calculate the percentages correctly). Then you would always win exactly the EV of dealing it once but with zero variance. I think this is a fairly clear way to show that the variance goes down while the EV stays the same as you increase the number of cards dealt out without shuffling.

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You've got a set over set situation in hold'em. Say the board is such that no other draws can win the hand for the player who trails. Now, you claim that we deal out boards with the rest of the entire deck and the EV gets distributed as the percentages dictate. Not always: if one of the boards gives both players quads, then the underset will never win, yet we know he does not have zero EV. QED. Ergo, I am smarter than you, per se, as it were.


<font color="white">Of course this is wrong, it's just that the granularity of the number of boards the deck can deal out is not fine enough to catch the small EV that the underset has every time you run this. If you dealt the remainder of the deck out a few hundred times, you'd get the right EV. So I agree w/ you and Ribbo, the EV does not change. </font>

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I was assuming there was only one card left to come. I thought that was obvious. [img]/images/graemlins/tongue.gif[/img]

Post deleted by FeliciaLee

[gergery]

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Jesus, I guess I should expect this thread would have the largest number of replies. It's like this entire site is designed for a bunch of retards.

You shouldn't give a [censored] about this as long as there is an automatic shuffler. If it's shuffled by hand, the player with the lowest EV gets an advantage from running it twice. If it's run at least five times, it goes back to neutral.

So there you have it. Now can we lock this stupid fking thread?

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Are you and RawTurkey cousins or brothers?

-g

[Ribbo]

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but if I only have 1 buyin and have to make it last,

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Then that doesn’t meet the criteria I noted of being able to handle variance

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It meets the criteria of being able to play poker, yet again stop just quoting other people and think for yourself. If some guy wants to play poker and has money, he will play poker.

[gergery]

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stop just quoting other people and think for yourself.

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Much better. I feel properly insulted now.

-g

[KJ o]

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Let's say Wintermute would rather ramble

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FYP