Let's say it's KK vs AA, players run it three times
Is the formula for scooping with KK 0,18*0,18*0,18*100?

If yes what about a spot where it gets like 99vsAA on K82? 99 has only two outs and should have 0 chance to win all three times instead of 0,09*0,09*0,09

Thanks for replies

Assuming all possible outs are known...

First run is normal...

If win the first board, then second board has the first board odds minus the dead out(s) from the first run...

If win first two, then the third board has the odds from the second run minus additional dead out(s)

Can be run for up to N boards where N=the number of outs, or the number of boards the dealer can produce with the remaining cards in the deck.

So a hand that must hit one of two outs to win can only have odds to scoop two boards.

Possible straights and flushes must be blocked by the hands somehow, or one of them is likely to be a slight favorite with hidden outs that might not be obvious.

Excluding straights and flushes, the chance kings win all three runs is zero. To win a run, a king would have to fall with no ace falling. Since there are only two kings remaining, there is no way kings win three times. This assumes, of course, that there is no replacement after each run.

Why would we exclude straights and flushes?

I am a little confused to be honest, I thought I am asking a relatively simple math question regarding equities.

I just needed a clarification whether the formula is exactly x to the power of n (where x is equity and n the number of times we run) or if the "discounting ours after each run" is really a thing.

The first poster seem to claim that it, indeed, is.

No, it is precisely not x^n for the reasons that I posted.

Outs are a thing.

I see. So...

1. if we wanted to find out what were the chances of me winning with my AJs vs 89s 3 our of 3 times, there would be no simple formula to do it. Correct?

2. x^n would give as a pretty decent approxmation (right?), but not the exact odds. Is it possible to quantify how far is x^n off from the real odds?

Also, one more thing:

3. If what you are saying was truly the case, wouldn't that mean that the number of times we run it then actually "changes the equity" a bit in some situations? Like, the 99 vs AA on K82 example. Wouldn't it be objectively better for the 99 guy to run it less than three times in comparison to three times or more? (I am assuming that the answer is "no" but I'd like to see a lil more elaborate explanation)

Because the probability of kings winning with a straight or a flush, especially if a suit is shared is negligible. The main point of my response was to point out that the 0.18^3 answer is outside the realm of possibility.

By the way, you left out flushes in your second question.

As for how many runs, it can be shown mathematically, that the expected value is constant but the more runs the lower the variance.

Barry Greenstein said he never runs it more than once because players who like to run it twice know that and therefore the chance they fold increases to avoid a one-time showdown.

Yes, no matter how many times you run out the board, each player's equity in the pot ("EV") is the same.

It turns out mathematically that the "card removal" effects exactly offset from board to board.

That's kind a a neat result, though it is "obvious" that it must be true.

P.S. Of course, the scoop prob is affected by card removal effects.

I do, of course, know that. I did not ask the question to find out the exact odds of KK vs AA winning three times, but to get the definitive answer regarding the overall formula of calculating the probabilities of scooping.

The 99 vs AA on K82? I thought it is evident that by omitting suits I meant this to be on a rainbow board.

Right.

Can you please rephrase? Not 100% sure what do you mean by these two statements.

Let me ask you the question this way:

If I wanted to find out the odds of hand X winning three out of three times against hand Y before any of the boards are dealt. What would be the best way to do it?

Thanks very much.

This was just confirming the result that everybody seems to agree with in the thread that the OVERALL equity for each player does not change whether you run the board once, twice, thrice, etc.

Regarding scoop probs, I don't think there is an easy way to derive this. Unless, I guess, you construct a weird and extreme example in which you can reason in your head that it is logically impossible for one person to win all the boards.

Since I have a NLHE simulator already programmed, if I have some time later I'll kick off a simulation of KK vs AA (three boards) and see what we find.

I chose to simulate running-it-thrice for KK vs. AA with one suit overlap. ProPokerTools tells us that the overall equities are 18.05% vs. 81.95% in this case.

The table below presents the results of a simulation of 100,000 deals where three complete boards are run out for each deal. Logically, taking into account the possibility of a board resulting in a split pot, there are a total of 10 possible cases for each deal.

Ks Kc wins	As Ah wins	Split Pot	Tally of Breakdown
0	3	0	51,544
1	2	0	41,041
2	1	0	5,825
3	0	0	106
.
0	2	1	963
1	1	1	491
2	0	1	23
.
0	1	2	6
1	0	2	1
.
0	0	3	0

As a sanity check, the above detailed results "roll up" to overall equities of 17.94% for KsKc vs. 82.06% for AsAh which is consistent with the overall equities (via ProPokerTools running it once).

The 3-scoop probabilities are easy, I hope, to read from the above table.

- KsKc scoops all three boards 106 out of the 100,000 deals (0.106%)

- AsAh scoops all three boards 51,544 out of the 100,000 deals (51.544%).

I am not sure how long it will take, but since I already have everything programmed, tonight I'll kickoff a simulation of 1,000,000 trials in an attempt to get even more precision on these numbers even though I realize that OP was essentially asking for ways to derive these figures, not necessarily the figures themselves.

There is no such formula, for the same reason that there is no formula to calculate the chance of AA winning against KK when only running it once. The only way to determine that equity is to count all possible outcomes and then see what portion of those are winners (plus half the ties too).

All equity calculators work off brute force enumeration or pre-counted results tables, not by using any formula. There is no arithmetic that is made for poker hand rankings. Poker hands follow poker rules, not arithmetic rules.

If you run as badly as Scarmaker, it happens every time.

Thanks for doing the calcs. I wouldn't have liked to guess at the numbers, but I'm kind of surprised the aces scoop more than 50% of the time in this instance, but then I haven't won an 80/20 since 2011. Interesting stuff.

I run bad?

Thanks for all the replies guys, especially the calcs.

Seems to me that while there is no formula, for all the practical purposes it is pretty alright to use: x^n - approx 0,1 - 0,5%

If we take whosnext equity for KK vs AA, the 0,1759^3 gives us 0,577% for scooping while the sim shows 0,106%, a 0,471% diff.

That's mostly coincidence, and other equities and matchups will be off by various amounts. There are many cases where only one win is possible and your formula will ignore that and multiply it anyway. Same when only two are possible. There really isn't any rule of thumb to estimate the scoop chance.

Updating for the simulation results for 1,000,000 deals:

Ks Kc wins	As Ah wins	Split Pot	Tally of Breakdown
0	3	0	516,353
1	2	0	412,125
2	1	0	56,698
3	0	0	1,000
.
0	2	1	8,809
1	1	1	4,630
2	0	1	338
.
0	1	2	38
1	0	2	9
.
0	0	3	0

The 3-scoop probabilities are easy, I hope, to read from the above table.

- KsKc scoops all three boards 1,000 out of the 1,000,000 deals (0.1000%)

- AsAh scoops all three boards 516,353 out of the 1,000,000 deals (51.6353%).

To echo what others have said above, there is no easy formulaic way to derive any of the numbers in the table above. Poker equities are not amenable to math in that way.

Also, I would not take the above results to suggest that E^N is a good approximation of scooping N runouts when E is your overall equity.

In the hypothetical world when you know nothing else except for your overall equity, then E^N is basically all you have, but in all real-world cases you have more information about how many "outs" you have and can do better than resorting to a simple assumption that all runouts are "independent" and "card removal" effects are insignificant.

Perhaps it will be of interest to look at the case of KK vs. AA with 2 suits of overlap. Of course, in this case (where KK has 17.36% overall equity) it will be even rarer for KK to scoop three boards as KK cannot win with a flush.

Ks Kh wins	As Ah wins	Split Pot	Tally of Breakdown
0	3	0	527,151
1	2	0	406,252
2	1	0	49,907
3	0	0	448
.
0	2	1	10,668
1	1	1	5,166
2	0	1	325
.
0	1	2	66
1	0	2	17
.
0	0	3	0

The 3-scoop probabilities are easy, I hope, to read from the above table.

- KsKh scoops all three boards 448 out of the 1,000,000 deals (0.0448%)

- AsAh scoops all three boards 527,151 out of the 1,000,000 deals (52.7151%).

KsKh, as mentioned above, has 17.36% overall equity. So the E^N approach would suggest that KsKh has a (.1736^3) = 0.523% chance of scooping all three boards.

But we see above that the actual chance of KsKh scooping all three boards, via the simulation of 1,000,000 deals, is 0.0448%, more than an order of magnitude less than the guesstimate formed via an unwarranted assumption of "independence" across boards.

One more for the road? Here is one where I hoped that the "independence" assumption underlying the E^N formula might yield a decent approximation.

6s 6h vs. Jd Td

JdTd, the trailing hand (actually having more equity) has so many different ways to win the hand that perhaps the card removal effects will be minimal.

According to ProPokerTools, 6s6h has 47.80% overall equity and JdTd has 52.20% overall equity.

Here are the detailed results of a simulation of 1,000,000 deals, each deal having three separate boards.

6s 6h wins	Jd Td wins	Split Pot	Tally of Breakdown
0	3	0	119,703
1	2	0	406,363
2	1	0	368,070
3	0	0	88,749
.
0	2	1	4,298
1	1	1	8,897
2	0	1	3,831
.
0	1	2	52
1	0	2	37
.
0	0	3	0

The 3-scoop probabilities are easy, I hope, to read from the above table.

- 6s6h scoops all three boards 88,749 out of the 1,000,000 deals (8.8749%)

- JdTd scoops all three boards 119,703 out of the 1,000,000 deals (11.9703%).

6s6h, as mentioned above, has overall equity of 47.80%. So the E^N approach would suggest that 6s6h has a (.4780^3) = 10.92% chance of scooping all three boards.

We see above that the actual chance of 6s6h scooping all three boards, via the simulation of 1,000,000 deals, is 8.8749%.

Similarly, JdTd has overall equity of 52.20%. So the E^N approach would suggest that JdTd has a (.5220^3) = 14.22% chance of scooping all three boards.

We see above that the actual chance of JdTd scooping all three boards, via the simulation of 1,000,000 deals, is 11.9703%.

These "actual" scoop probs are still significantly lower than the figures based upon the E^N calculation. Even in this scenario where we chose the match-up to give the E^N calculation the best chance, the true scoop probs are between 80-85% of the E^N figures due to card removal effects.

If we only are concerned with the most obvious outs, and not the tiny backdoor runner-runners, then a much much more accurate estimation is the method of card removal.

If x number of outs appear on one board, the chance of winning the next board are readily available with regards to those specific outs.

The only error here would be said runner-runners to backdoors.