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The practice of “running it twice” (or even more times) has become a common practice in some poker games. What it usually means is that, when a heads-up hand allows no more betting, because one player is all-in, the opponents can agree to see the outcome dealt twice, instead of once.

A player winning both times claims the entire pot. Otherwise, the pot is split.

Should you agree to run it twice? Does it matter whether you’re the favorite or the underdog? Here’s a technical explanation, with the answer at the bottom.

A $200 pot

Imagine that it’s a $200 pot. How the money got there doesn’t change the decision. Also imagine that you’re a 3-to-1 favorite to win. If you only see one outcome, you’ll win three-quarters of time, assuming you replay the result forever. That means you’ll have a gross win of $200 three times, on average, for a $600 total win and lose $200 once. The advantage is $400 every four pots or $100 per pot.

For a single pot the math is (.75 times $200 = $150) – (.25 times $200 = $50) or $100.

If you play it twice for each of the four instances, you’re likelihood of winning both “runs” is .75 times .75, or .5625. So, 56.25 percent of the time, you’ll take $200. You’re likelihood of losing both hands is .25 times .25 or .0625. So, 6.25 percent of the time you’ll take nothing.

More math

Those two possibilities account for 62.5 percent of the outcomes, when added together. The remaining 37.5 percent of the time, you’ll split the pot, meaning no gain either way. So, the math is (.5625 times $200 = $112.50) – (.0625 times $200 = $12.50) per two-run events. That’s still a $100 advantage per hand, because on the remaining 37.5 percent, you split and take $100.

The full math for a double run would be:

((.5625×200)+(.375×100))-((.0625×200)+(.375×100)) = 100

The simple answer

You can see that it doesn’t make any difference in the long run whether you take the deal or not. And it doesn’t matter to the underdog, either. The process reduces the risk and makes it more likely that the outcome will be closer to expectation. But that’s all it does. **— MC**

Does the fact that the second run is dealing off of a smaller deck that has had the previous run(s) cards eliminated from it, change my +EV odds?

Said another way, if I run it three times, but each time pick the runouts back up, shuffle them into the deck, and then deal the next runout, aren’t those odds going to be different from multiple runs where the runouts (correctly) aren’t shuffled back in? Or would it not matter what method I used because the variance works either way (eliminated cards are sometimes good/bad for me, in a way essentially randomly determined)?

Hi, Paul —

The odds don’t change, no matter how many times you agree to deal. Whether you shuffle or not doesn’t change the odds, either. However, if you shuffle each time, there is more variance and the luck factor is greater. To use an extreme example, suppose there are only two cards left in the deck, one red and one black. Black wins for you. Red loses. So, if you run it twice after shuffling each time, you can get lucky and win both times, or you can get unlucky and lose both. But if you run it two more times without shuffling, you’re sure to get a “fair” outcome and win half the time.

Thanks for your reply. It shows that you’re on the right track without my explanation.

— Mike Caro

Wait, .75 * 200 – .25 * 100 = 125, not 100, right?

Yes, you’re right. I should have read “.25 * 200.” The original article used a $100 pot, but when I changed it to 200, one instance of 100 remained.

Thanks. I fixed it.

If you run it twice isn’t your odds of winning the second run less than 0.75? (Assuming you won the first and the villain would have same number of outs but there are 2 less cards in the remaining deck?

Whether or not cards are shuffled between runs, the odds don’t change overall. Without a shuffle, the first run cards certainly do affect the odds for the second run, but so what? The cards aren’t known unless you look at them, so their influence is mathematically unknown before the run-twice process begins. Suppose there were just two cards left in the deck, an ace and a king. If you only deal one card, the chance of an ace is 50 percent. If you deal both cards, that 50 percent is absolute, regardless of whether the ace comes first or second.