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Suppose on the flop in no limit hold'em two players go all-in. They both have 2 hole cards and there's 3 community cards on the board at present. There will be 2 more cards drawn and each player will make the best 5 card hand out of the 5 community cards and their own 2 hole cards.

Sometimes people like to "run it out" more than once, and deal the 2 final cards 2 times, and either split the pot, or one person wins twice and wins the whole pot. There's also variations where they'll run it out more than 2 times.

Does this benefit one player more than the other? Figuring out the percentage chance of 1 player winning is no easy feat for running it out once only, and I have no idea how to approach this problem for running it out more than once.

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  • $\begingroup$ It is just to reduce variance. If you are a 3:1 dog and run it 4 times then statistically you would win once. In a tournament a big stack would typically only run it once as in the payout structure putting the other player out is worth more than the pot. No tournament I know of lets players run the deck more than once. $\endgroup$
    – paparazzo
    Jul 16, 2016 at 16:00

2 Answers 2

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Expected value is linear, even over dependent random variables. The chance to win each run (not conditional on particular outcomes of the previous runs) is the same as the chance to win the original, so there is no advantage in expected winnings to running the hand $n$ times instead of once. Although some people find it counterintuitive, running the hand multiple times doesn't favor either the drawing hand or the hand which is ahead, whether the draw is likely to hit or not.

The variance is reduced significantly when you run the hand several times. It can greatly affect the probability that the player is ahead or behind for the day, or the probability of reaching or keeping other thresholds of psychological importance.

Some people feel that allowing your opponents to run a hand multiple times decreases their risk aversion, and mistakes due to risk aversion, earlier in the hand. This is one reason some players such as Barry Greenstein might have the policy of not letting people run the hand multiple times.

Another effect of running the hand multiple times is to decrease the probability that players bust out, and have to decide whether to rebuy or leave. Leaving may bring in other players you would prefer to play against or may leave the table shorthanded, which may suit you if you are relatively strong at shorthanded play. If a player wins the whole pot, he may also reach an uncomfortable stack depth where he might make more errors. Running the hand multiple times also takes time which could be spent playing poker instead.

On the other hand, if someone asks to run the hand multiple times, and you refuse, this may look unfriendly, and a casual player may decide to spend his money elsewhere.

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    $\begingroup$ From my understanding, the two runs typically occur without replacement. It's true that the EV happens to be the same regardless, but I don't think your first line is likely to be very convincing to a skeptic. For instance, if a player only has one "out" and hits it the first time, he cannot win the second time. Your answer is right, but I wonder if maybe you could improve it by more explicitly dealing with cases like this? $\endgroup$ Sep 29, 2012 at 22:52
  • $\begingroup$ What if there's 2 outs for the behind player? Doesn't running it twice risk both outs coming on one run and being completely dead on the other run? I did the calculations for 1 out and you're definitely 100% correct. $\endgroup$
    – user14281
    Sep 30, 2012 at 0:32
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    $\begingroup$ The chance that the first run is favorable exactly balances the chance the that the first run is unfavorable. The chance that you win with cards $3$ and $4$ is the same as the chance that you win with cards $1$ and $2$, by symmetry. $\endgroup$ Sep 30, 2012 at 0:42
  • $\begingroup$ (+1) For some reason this (1) is very obvious from linearity and (2) somehow seems miraculous, like there is a conspiracy making things cancel out exactly right. $\endgroup$
    – guy
    Jan 10, 2014 at 18:27
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"Figuring out the percentage chance of 1 player winning is no easy feat for running it out once only"

  • On the contrary, approximations are easy on the flop and turn.

From the flop, with 2 cards to come. Percent to win = (# outs) x 4. Example, if you have 9 clean flush outs on the flop, then you are about 36% (actually, slightly less than 36%).

On the turn, with 1 card to come, percent to win = (# outs) x 2 - 1. Example, you missed the the flush on the turn but still have 9 clean outs, then it's (9 x 2) - 1 or about 17%.


Re: running it twice.

E.g., Let's say that I have a flush draw (9 outs and thus, about 36% on the flop to win) against a made hand on the flop. We agree to run it twice.

At this point, I am behind and will lose 36% of the time if we run it once. But if we run it twice, my chances of winning or splitting the pot is greater than 36%. Because at this point, I'd be happy with a split pot and all I'm looking to do is win at least one of the runs.

So run it once, I am 36%.

Run it twice, given that I have some extra chance (an additional) 36% chance, my chance to chop OR win is greater than the 36% to win in a single run. Ergo, there is a benefit to run it twice for the person getting the chips in bad. It doesn't increase the chance of winning the pot but provided there is now a better than 36% chance of winning OR chopping the pot, this is surely a benefit.

Thus, to answer the OP's question. There is a benefit to running it twice for the person getting the chips in bad. It doesn't increase their chances of winning the pot. But there is the added value/possibility of a chopped pot, which should be clearly welcome for anyone getting their chips in bad.

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