Let's say someone flopped top two pair and I have the naked ace high flush draw. Assume we get it all in on the flop and I have 33% equity. If we run it twice, what percentage of the time will I win both boards running it twice, what percentage will I get scooped and what percentage of the time will we chop? Can someone please show me how to break this down? Thank you.

Are you running flop,turn and river twice? Or just the turn and the river? Because I'd think that makes a big difference also .

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To clarify, I'm running only turn and river twice. Flop is the same for both boards.

No mathematical change in odds running it twice . Does lessen variance and luck . You'll have the same 33% the second run out as the first.

But anyone please correct me if I'm wrong


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Alfred, you are correct, but that wasn't one of the questions I asked.

Will not have the same eq when won first run. But who cares, win once is more than can expect as "average". Will have higher eq often in the 2nd run when lost 1st.
Very likely EV of run it once/two is close if not exactly the same. Not able to prove it, maybe there is a chance EV could change after several runs when the deck is almost exhausted, in the end run it several times is not the same as Repeat game with a fresh deck.

Does anyone know the exact math involved and the percentages?

Yeah it seems like people are missing the point here.

The chance of A happening AND B happening is prob(A) * prob(B)

The chance you win is 33%, the chance you lose is 67%.

win both: P(win)*P(win) = .33*.33 = .1089
win first, lose 2nd = .33*.67 = .2211
lose first, win 2nd = .67*.33 = .2211
lose both = .67*.67 = .4489

So you'll win both 10.89%
win one 22.11 + 22.11 = 44.22%
lose both: 44.89%

Yes, perfect, thank you!

I should add, for clarity:

This is only true if event A and event B are independent. Which, before the cards are drawn for the first flop, is true.

This is not correct.

In order to see why this is not correct, let's simplify the example.

Let's say you get all-in on the turn and you have 11 outs out of 44 cards (now that you've seen your opponent's cards) and you're going to run it twice.

If you ran it once, you'd have a 25% chance to win.

But if you run it twice, your chances to scoop are not 1/16, as they would be if the two rivers were independent. Instead the probability that you will scoop is

(1/4)*(10/43) = 5/86

which is about 5.8%, less than the 6.25% that 1/16 is equal to.

The fact that you can't hit the same out on both runs means you're less likely to win the second run after you've already run the first one.

At the same time it also makes it MORE likely that you will win the second run after losing the first run.

So while the EV of running it twice is the same as the EV of running it once, the math is much more complicated than just "it's independent events". They're not independent but the EV still works out as we think it should.

Your chop percentage also just has a lot to do with what your hands are. A2 vs A3 will obviously chop a lot more than AK vs QQ on a single runout which means there's a higher chance someone is getting quartered on a double runout.

Here's another factor to consider.

If the game is short and busting a player, can result in the game breaking, it is an argument for running it twice.

If the game is full and has a waitlist (must-move), then running it once will result in overall deeper stacks (assuming you have money to reload) and therefore a bigger edge to an otherwise winning player.

It is true when prob(A) does not influence prob(B) and vice versa. Im affraid this is not the case.

E. pardon you already mentioned it I missed it.

Yeah and I kind of forgot that they're not independent. But it's... close. It is worst when your chances of winning are very large or very small I think.

The EV of the guy who bets only on the first deal is exactly the same as the EV of the guy who bets only on the second deal.

Actually doing all the math correctly in the case you describe can be quite tedious since you need to keep track of each possible runout (or at least categories of runouts).

To give some concreteness, suppose Villain has Kh 9d and Hero has Ac 7c. Flop comes down Kc 9c 2h. Players get it all in on the flop. If the turn and river cards are only dealt once, Villain will win 647 possible runouts and Hero will win 343 runouts of the possible 990 runouts.

If I tallied these correctly, here is how Hero can win the first or only runout:
3 AA
9 A7
3 77
36 two clubs
286 exactly one club
6 A2 (no clubs)
-----
343 TOTAL

Suppose players agree to run it twice. Now, to determine the exact probs of each player winning both runouts and the prob that they chop, you need to keep track of what comes on the first runout (at least in which above category) since that affects the prob of winning the second runout.

You can see immediately that how Hero wins the first runout would affect the prob of winning a second runout. Of course, the converse is also true. How Villain wins the first runout (compare first turn/river of Qc/Kd to 4s/3h) affects the prob of Hero winning the second runout.

Due to "card removal" the prob of either player winning the second runout is not independent of the prob of either player winning the first runout. In theory, this can be handled by a huge "runout tree" or computer program.

Of course, in the simplest case of Hero having exactly one out, the tree is easily constructed and the respective probs can be easily calculated. But things are not so easy in more complicated cases such as getting top two pair vs. ace-high flush draw all-in on the flop.

Yeah, this is why I used a simplified example to demonstrate my point earlier. The OP's question is a lot tougher than they probably thought it was.

I ran a computer program last night to do these tallies in this specific case. Recall from above that:

0.34646 = 343/990 = Overall prob of Player 2.

The computer program tallies show:

0.32353 = Prob(Player 2 wins 2nd runout | Player 2 wins 1st runout)
0.35862 = Prob(Player 2 wins 2nd runout | Player 1 wins 1st runout)

So we have the probs of the possible outcomes of running it twice:

0.41916 = Prob(Player 1 wins both runouts)
0.46875 = Prob(Players chop)
0.11209 = Prob(Player 2 wins both runouts)

You can verify that (0.34646)*(0.32353)+(1-0.34646)*(0.35862) = 0.34646 so that Player 2's overall equity is the same whether or not they run it twice.

Note also that the prob of Player 2 winning both runouts (0.11209) is less than .34646*.34646 (= 0.12004) due to the "card removal" effects discussed above.

The card removal effects can vary from situation to situation, so there is no easy way to come up with a simple rule of thumb to derive the exact probs of each player winning both runouts in all run-it-twice situations.