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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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1  
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
up vote 6 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
1  
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
    
If you are playing 9 handed, there are 18 cards out pre-flop, then counting the burn card there are 22 cards out after the flop, leaving 30 cards left in the deck. If you agree to run it some ridiculous number of times, say, more than 7, then you don't physically have enough cards left in the deck to burn a card before every turn and river. – Michael Feb 4 at 23:08

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

Tried to post this as a comment to chris's reply to User93, but i lack reputation.

Chris is correct. The odds of both winning the whole pot and losing the whole pot are increased very slightly with a reshuffle, while the odds of a chopped pot are decreased very slightly. The extent of this increase to all or nothing outcomes depends on the situation.

Examples:

For someone drawing to a 7 out situation with 1 card to come, no reshuffle, running twice, there is a 2.2199% chance of total win on the draw, and 70.4017% chance of total loss on the draw.

For someone drawing to a 7 out situation with 1 card to come, with reshuffle, running twice, there is a 2.531% chance of total win on the draw, and 70.7128% chance of total loss on the draw.

So .3111% more chance for total win, .3111% more chance for total loss, .6221% less chance for chopped pot.

Example 2:

For someone drawing to a 7 out situation with 1 card to come, no reshuffle, running 3 times, there is a .2643% chance of total win on the draw, and 58.6681% chance of total loss on the draw.

For someone drawing to a 7 out situation with 1 card to come, with reshuffle, running 3 times, there is a .4027% chance of total win on the draw, and 59.463% chance of total loss on the draw.

So, .1384% more chance for total win, and .795% more chance for total loss. Interestingly, there is also an increase of .518% chance to win 2/3 pot, with a decrease of 1.45% chance to win only 1/3 pot. (Need to check my math on this one, since the results look odd, but the result is interesting because it illustrates that while the overally equity stays the same, the changes to the variance may not equally favor each outcome when running the pot multiple times)

Example 3:

For someone looking for a running flush to win (2 cards to come), no reshuffle, running twice, there is a .1409% chance of total win on the draw, and 91.05% chance of total loss on the draw.

For someone looking for a running flush to win (2 cards to come), with reshuffle, running twice, there is a .2066% chance of total win on the draw, and 91.1157% chance of total loss on the draw.

So .0657% more chance for total win, .0657% more chance for total loss, .1313% less chance for chopped pot.

Basically, running multiple times with reshuffle slightly increases the hand variance relative to running multiple times with no reshuffle. If you're playing to not lose everything, run it multiple times, but you tradeoff a chance at a bigger payday.

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