The concept of “running it twice” is well known especially among the high-rollers. The concept is simple: Say two people are ALL-IN on the flop. Now, instead of “running” the turn and the river once, to determine a single winner, they run them twice. And each time the winner gets half the pot. Similarly, they can run them 4 times, each time for a quarter of the pot.
I am often asked the question if “running it twice” affects your odds of success. Especially, if the same deck is used and the cards are not reshuffled (which is usually the case). The answer is of course: It doesn’t! The only thing it does, is reducing your overall variance while maintaining your Expected Value (EV) intact. In other words, by “running it twice” (or more times) not only you will make the same amount of money in the long run, but also your bankroll will fluctuate less.
But WHY is it that our EV stays intact? Well let’s look at a simplified example, where we have a Flush Draw (FD) with only 8 cards left in the deck, only 2 of which will complete our draw.
The assumptions:
- Hero has a spade draw on the flop and he will win if and only if he hits his FD
- The deck has only 8 cards, 2 of which are spades.
Question 1: What are the chances of Hero winning the hand?
Probably the easiest way to compute this is by computing the 3 disjoint winning scenarios for Hero. Namely:
- He hits the turn and misses the river: (2/8)*(6/7) = 0.214 or 21.4%
- He misses the turn but hits the river: (6/8)*(2/7) = 0.214 or 21.4% (no surprise there: We switched the “2” with the “6”, plus the two scenarios are symmetrical)
- He hits both turn and river: (2/8)*(1/7) = 0.036 or 3.6%
Total chance of winning the hand is: 21.4 + 21.4 + 3.6 = 46.4% (not too shabby! – This result is skewed from the typical
FD because 9/44 is closer to 1/5 than to 1/4 – but who cares? :-D)
Question 2: What happens if we run the entire deck? (a total of 4 runs)
Well, since there are 2 spades there are two cases:
- Case 1: Hero wins 2 out of the 4 runs, or 50% of the pot (if the spades are spread)
- Case 2: Hero wins 1 out of the 4 runs, or 25% of the pot (if both spades come in a single run)
Final Question: What is the EV of hero if we run it 4 times?
In order to answer this we need to know how frequently each case happens.
It is easier to compute the frequency of the second case. The chance is simply 4 times 3.6% = 14.4%, where 3.6 is the chance of hitting both turn and river. (This is because, the chance of hitting it on the first run is the same as hitting it on any other run)
Therefore, we have that case 2 happens 14.4% (which nets Hero 0.25 of the pot) while case 1 happens the remaining 85.6% of the time (which nets Hero 0.5 of the pot). Now a simple EV calc shows that:
EV(hero, run it 4 times) = 0.856*(0.5) + 0.144(0.25) = 0.464 = 46.4% which is EXACTLY what we found above!
Conclusion: Hero’s and Villain’s EV ware not affected by the multiple runs!
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A few things to notice:
- Hero runs below EV only 14.4% of the time but his realized losses are huge (46.4 – 25 = 21.4%)
- Hero runs above EV a whooping 85.6% of the time but his realized gains are tiny (50 – 46.4 = 3.6%) (This is where a lot of people think that when they have a draw, they should run it twice. They think that because they have a higher chance to chop the pot, they are golden. They don’t realize how much value they lose, when they don’t. As a matter of fact, if anything they help villain who now has nothing to lose – see next point)
- This is essentially a risk-free situation for villain who in reality they CAN never lose more than half the pot. Sure they do sacrifice a bit of EV the majority of the time (85.6%), but they get to keep 3/4 of the pot once in a while (14.4%) making a huge profit! And all that, without having to risk a single penny!
- The above point also shows the benefits of running-it-multiple-times in the form of reduced variance and consistent earnings. In other words, villain is perfectly hedged!