Running it twice, math question

Obviously, this didn't happen online.

Player 1 has As Kd
Player 2 has Ah Jh

Flop is Ks 9h 6h

For simplicity sake, say the pot is $100 (both players are all-in).

Running it once, Player 1 has $63.33 pot equity and Player 2 has $36.67. How does this change if you run it twice?

It would depend what came out the first time if you want to get exact. If you're trying to figure out how much the chance of the flush coming (on board 2) goes down after making it the first time, you would subtract one out, or two if the first turn and river are hearts.

Forgive me if Im wrong, but I dont think EV changes no matter how many times you run it

Originally Posted by andy-akb

Forgive me if Im wrong, but I dont think EV changes no matter how many times you run it

It doesn't. The odds change slightly though the second time the hand is run.

Running it twice has the following effect on the results:

If you run it once, top pair has 2/3 to win, 1/3 to lose.
Run it twice top pair has 4/9 to scoop, 4/9 to push, 1/9 to lose.

Essentially, running it twice kind of ressembles chopping- Player 1 agrees to minimize odds to get sucked out (1/9) and Player 2 agrees to make it likely he'll chop or better (5/9).

However, Player 2 gives up any likelihood of winning, and Player 1 basically gives up his edge (reduces his shot at the pot). Don't run twice. Split pots n chops are no fun.

Primarily it reduces vairencea little, but it does change the EV.

By running it twice, it is like saying ok, this is two serperate instances where I'm in this exact position. Instead of being in one coinflop right now, and if you miss you lose a decent chuck, and where next time you might win since "on average" you expect to win 1 out of 2 coin flips. But by running it twice, you get to be involved in 2 coin flips right now, instead of maybe having to wait hours or days to be involved in the same situation.

However, running it twice increase slightly EV for a drawing hand (or which ever hand has more outs to take/keep the lead). For example:

Over Pair vs Top Pair. $100 All-In Pot. Top Pair has 5 clean outs with only the river card to come.

Running it once:
TP has 5 outs and a 10.86% chance to win the pot, 89.14% to loss.
TP = $10.86 Value
Over Pair = $89.14 Value

Running it twice:
TP has 5 outs with 46 cards left in the deck the first time around. So TP is 10.86% to win & 89.14% to lose, same as running it once. But on the second time around, if he doesn't win the first one, he still has 5 outs with only 45 cards left in the deck. So now he is 11.11% to win and 88.89% to lose. So, what does this mean?

Winning Both = (5/46) 10.86% * (4/45) 8.89% = 0.96%
Winning One = [(5/45.5) * 2 chances] 21.97%
Winning None = 77.07%

So as a result,
Winning Both = $100 * 0.96% = $0.96
Winning One = $50(split pot) * 21.97% = $10.98
TP running it twice = $11.94 Value

The cost for TP to draw did not change just because he is running it twice, but the value of his draw went up by almost 10%! And as a result, the Overpairs value decreased by almost 10%. Overpair is still a winner and significant favorite, but his EV did drop a decent amount.

(and if you are running it twice from the flop, the difference is even a little more)

So, of course when you are behind, but have more outs, you always want to run it more than once, the more the marrier as it does increase your EV a little.

But why would the hand favorites agree to it? Well, I can't speak for them (since I don't play in a game that does this), but I would say it is primarily to help with variance and as a show of sportsmanship and kindness to the weaker player that continues to get his money in behind, whom you want to continue to come back (I would rather give up some EV and be a 3:1 favorite twice, than a 4:1 favorite once! ). But I don't know....

Doesn't the dealer shuffle between the runs?

Originally Posted by sandstorm

Doesn't the dealer shuffle between the runs?

No

Originally Posted by NWNewell

Over Pair vs Top Pair. $100 All-In Pot. Top Pair has 5 clean outs with only the river card to come.

Running it once:
TP has 5 outs and a 10.86% chance to win the pot, 89.14% to loss.
TP = $10.86 Value
Over Pair = $89.14 Value

Running it twice:
TP has 5 outs with 46 cards left in the deck the first time around. So TP is 10.86% to win & 89.14% to lose, same as running it once. But on the second time around, if he doesn't win the first one, he still has 5 outs with only 45 cards left in the deck. So now he is 11.11% to win and 88.89% to lose. So, what does this mean?

Winning Both = (5/46) 10.86% * (4/45) 8.89% = 0.96%
Winning One = [(5/45.5) * 2 chances] 21.97%
Winning None = 77.07%

So as a result,
Winning Both = $100 * 0.96% = $0.96
Winning One = $50(split pot) * 21.97% = $10.98
TP running it twice = $11.94 Value

Your math is not correct. Under the assumptions that you made the overpair wins (41/46)*(40/45) = 79.23% of the time. The EV doesn't change at all if you run it twice.

Originally Posted by mcatdog

Your math is not correct. Under the assumptions that you made the overpair wins (41/46)*(40/45) = 79.23% of the time. The EV doesn't change at all if you run it twice.

Yeah, I see where my math went wrong .... Thanks pointing that out.

But I'm not sure I've got my mind wrapped around the logic of how odds to hit don't get better the second time around, and since it is not costing you any extra the second time around, your EV should go up a little (drawing odds get better, cost stays the same = greater EV... I would think).

I've got to think about this.....

Originally Posted by NWNewell

Originally Posted by mcatdog

Your math is not correct. Under the assumptions that you made the overpair wins (41/46)*(40/45) = 79.23% of the time. The EV doesn't change at all if you run it twice.

Yeah, I see where my math went wrong .... Thanks pointing that out.

But I'm not sure I've got my mind wrapped around the logic of how odds to hit don't get better the second time around, and since it is not costing you any extra the second time around, your EV should go up a little (drawing odds get better, cost stays the same = greater EV... I would think).

I've got to think about this.....

EV should be the same, by its definition - if you run the same situation over and over again EV represents your EV oin average from those inifnite runs.
The fact that you did not shuffle the deck between runs should not really affect EV, odds to hit the flush improve the 2nd time only in the case of two non suit cards coming out the first time.

On High Stakes Poker, Daniel Negraneau kept saying how running it twice did not affect win percentages at all, it just reduces variance.

Originally Posted by bigslikk

Don't run twice. Split pots n chops are no fun.

yes variance is lots of fun

It doesn't change the EV.

If you need peace of mind, realize that the cards that come on the 2nd run don't know what the cards where that came the first time.

If you need proof, run it for a 1-outer, the calculations are much simpler.

Edit: 1-outer calculations with 1-outer redraw for completeness, like in the case of set over set.

There's 43 possible cards to come, and we'll say it's AA v KK on AK8.

Running it once

Chance to make K on turn: 0.0232
Chance to make K on river: 0.0238
Total chance to make K: 0.047
Chance to make A: 0.047
Chance to make both: 0.002
Chance to make K without A: 0.045

Running it twice

(A) Chance to make K without A on first run: 0.045

Chance to make A without K on first run: 0.045
Second run after above, chance to make K on turn: 0.0243
Second run after above, chance to make K on river: 0.0250
Second run after above, total chance to make K: 0.0493
(B) Total chance for A on first run, K on second run: 0.0022

Chance to miss both on first run: 0.9081
Second run after above, chance to make K on turn: 0.0243
Second run after above, chance to make K on river: 0.0250
Second run after above, chance to make both: 0.0012
(C) Second run after above, chance to make K with no A: 0.0481

I rounded all of these wayyyyy too much, and I'm too lazy to go back and redo it, but if you add A+B+C and divide by 2 (as those are the only cases where KK wins half the pot, KK can never win both halves) you'll have the total equity which will be the same as running it once.