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When players agree to "run it twice" (or more times), should any additional cards be burned before each replayed step?

Also, shouldn't the cards turned in the first run be shuffled back into the deck before flipping the next one? I understand it wastes additional time, but the odds do actually differ for each of the runs, which they should not, right?

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I think that part of the intent in running it twice is often to have the odds differ, reducing or increasing the likelihood of a draw hitting the second time based on whether it did or did not hit the first time around. –  Jeffrey Blake Jan 25 '12 at 21:15

2 Answers 2

up vote 4 down vote accepted

There is no reason to deviate from standard procedure of burning a card before the flop, turn, and river when dealing with an all-in situation - it saves nothing and can only cause confusion.

Running it twice isn't really part of any official poker rules, so it will always follow the house rule. I would prefer that known cards not be reshuffled between - if the point is variance reduction, which it usually is, you have a better chance of realizing your true equity if the cards are not reshuffled.

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I have a hard time comprehending the first part: does that mean you would always burn the cards or not? –  Mikulas Dite Jan 25 '12 at 21:47
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Yes - the standard procedure is to burn cards. I'll update the answer. –  Chris Marasti-Georg Jan 26 '12 at 0:28
    
Besides if the cards have to be reshuffled to run them again that takes more time and introduces other risks (card damage, separate burn piles, etc.) I doubt you would get a casino to allow multiple runs if the dealer had to shuffle. –  Chad Jan 27 '12 at 20:31

There's no need to burn a card after the first run because the players are all-in and have no other decisions to make. The point of the burn card is to protect against marked cards.

Cards turned in the first run are not shuffled back into the deck. There is no point because it doesn't change the equity. This may seem counter-intuitive but if you do the math you will see that it works out like that.

Here's an example. Obviously not a general proof, but hopefully this illustration suffices. Say you are headsup on the turn and all-in, needing a one-outer to win and you run it twice. There are 8 cards out (2 per player and 4 on the board).

Not Reshuffling

  • Win first, lose second (1/44 * 43/43) = 0.022727
  • Lose first, win second (43/44 * 1/43) = 0.022727
  • Lose first, lose second (43/44 * 42/43) = 0.95455

In the cases where you win, you only win half the pot, so your equity is 0.5 * (0.022727 + 0.022727) = 0.022727 = 1/44.

Reshuffling

  • Win first, win second (1/44 * 1/44) = 0.0005165
  • Win first, lose second (1/44 * 43/44) = 0.0222107
  • Lose first, win second (43/44 * 1/44) = 0.0222107
  • Lose first, lose second (43/44 * 43/44) = 0.95506

Your equity is 1 * (0.0005165) + 0.5 * (0.0222107 + 0.0222107) = 0.022727 = 1/44.

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Equity is the same, but the chance that you come away with some portion of the pot is not. –  Chris Marasti-Georg Jan 26 '12 at 0:33

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