# Can “run it twice” be a +EV in a long run?

The concept of "run it twice" is that when players go all in, the rest of the board will be dealt two times with the cards left at the deck. The pot is then by splitted into two parts; one for both "runs" where winner of one "run" gets 50 % share or splits ot with the other. This significantly reduces variance in some games.

Imagine that you are playing omaha 4/5 hi and you flop 223. You have 33xx and the opponent has 22AK (you have 1 out).

If you go all in at the flop, you have approximaely 4,88 %. Could it be possible, that the 33xx holding player will get +EV from the run it twice in long run (let's suppose 1 000 000 times 100 € pot)?

Note that if hero hits his 1 outer, the second "run" will be 0 outer, so 0 % 50 € pot.

Running it multiple times does not move EV an inch. It only reduces the variance.

I think the example from kiota is spot on (+1). On the river the number of down cards is 44. Even after you see 2 cards the bet was placed before.

If you hit on the first then you are less likely to hit on the second. If you miss the first you are more likely to hit on the second. Think about it this way. The last 2 card in the deck should have an equal chance of improving your hand as the first 2 cards. Heads up the most you can run is 22 times.

Math on 1 card left run it twice

``````say take m outs with n unknown cards

Run it once your chance is m / n

k is number of cards left = 1

run it twice

You either win the first or not

If you win then take away 1 out

m / n x (m - k) / (n - k)

If you lose the first then

(n - m) / n x m / (n - k)

net

(m(m - k) + m(n - m)) / n(n - k)

m(m - k + n - m) / n(n - k)

m(n - k) / n(n - k)

m / n
``````

if k = 2 it gets more complex and not going to try and prove that

• Yes, this is the existing concensus, but can it be proven? Perhaps this is a mathematics stack exchange thing. – Kasperi Koski May 14 '18 at 12:59
• How do you need to prove a definition? The chance of winning is not dependent on number of tries. If I flip a coin it is heads 50% and tails 50% if I flip it 1 or 1 million times. – paparazzo May 14 '18 at 13:55
• @paparazzo the odds of winning each run are not necessarily the same – Clarko May 14 '18 at 18:25
• @Clarko OK from that can you prove any different EV? – paparazzo May 14 '18 at 18:55
• @Clarko Actually I think they are the same. See example from kiota. You see the first two cards but at the time you made the bet they were unknown so you treat each run as same number of unknown. E.G. for the river 44 unknown. – paparazzo May 16 '18 at 20:22

EV does not depend on how many times you run it, only variance does. I will try to illustrate it with a simple example:

Assume heads-up play. You play all-in on the turn and you have x outs to win the hand.

# Scenario 1 - Run it once

Cards_left = 52 - 4 (deck) - 4 (in your hands) = 44
P[win] = outs / cards_left = x/44 (you have x/44 equity to win the whole pot!)
EV_TOTAL = (x/44) * pot

# Scenario 2 - Run it twice

i) P[win_first] = x/44 (as previously, but now for half the pot)
EV_first = (x/44) * 0.5 * pot

ii) P[win_second] = P[won_first] * P[win_second] + P[lost_first] * P[win_second] =
(x/44) * {(x-1)/43} + {1-(x/44)} * (x/43) = x/44 (for the other half pot)
EV_second = (x/44) * 0.5 * pot

So, EV_TOTAL = EV_first + EV_second = (x/44) * 0.5 *pot + (x/44) * 0.5 *pot = (x/44) * pot

--

As shown, the EVs of the two scenarios are identical, regardless of the number of outs. Same can be shown for more complex scenarios (run it 3 times from the flop), but I will leave it as an exercise to all the math geeks out there :)

• I agree some people think there are only 42 down cards after the first two are shown but at the time of the bet there were 44 down cards. – paparazzo May 16 '18 at 20:30