Assume you choose to run it four times after a flop all in, can you just use binomial distribution to determine the odds of winning 3 out of 4 boards? Do the dead cards matter in this instance?

The hand in question was AA and KQ on a J62 flop and villain couldn't believe he only got one heart out of eight turn & river cards.

When you run it more than once you are choosing extra cards without replacement from the stub of the deck.

The binomial will be a reasonable approximation, but you really want a hypergeometric distribution (in the case of flushes, a slight tweak of one because you have to add up the cases of 1 heart and 2 hearts on the same 2-card board) :

Choosing 8 cards out of 45 (36 non-hearts and 9 hearts):

All 8 are non-hearts 36C8/45C8 = 30 260 340 / 215 553 195 = 14% of the time.
1 heart and 7 non-hearts 9*36C7/45C8 = 75 129 120 / 215 553 195 = 35% of the time
2 hearts and 6 non-hearts 9C2*36C6/45C8 = 70 120 512 / 215 553 195 = 32% of the time. One-seventh of these make only one flush, six-sevenths of these make two flushes.

That means that you will make no flush 14% of the time and only one flush 39% of the time. (And a small handful of those times you'll still lose to a full house.)

Thanks alot siegmund. Probability is by far one of the most (if not the most) helpful and least trollish boards on all of twoplustwo.

Yeah, this is true. I think it's possibly because so many probability questions have True Answers, and most argument comes from Opinions.

A common myth of running multiples is that the person ahead in the hand is doing a favor for the person behind in the hand by running multiples. Actually, no one is doing anyone a favor, unless both players are playing above their heads, in which case the reduction of variance helps both. Equity remains constant regardless of the number of runs.

In the hand above, running it 4 times insures to a 97.96% degree of certainty that you will not lose the entire pot. The "price" you pay for this insurance is that you drop from 62.22% to win the entire pot all the way down to 14.99%, and you are 48.62% to chop or lose money.

In the breakdown below, notice the steadily decreasing chance of ruin. Contrast this with the dissimilar pattern of chance of profit. If booking some kind of a profit is more important to you than the amount of that profit, while minimizing your chance of losing the entire pot, then running it three times is best for you. If your only goal is to avoid losing the entire pot, then run it four times. Of course, your opponent could always just say "rip it!"

Run it once (62.22% chance of profit; 37.78% chance of ruin)
62.22% to scoop
37.78% to lose it all

Run it twice (38.71% chance of profit; 14.27% chance of ruin)
47.02% to chop
38.71% to scoop
14.27% to lose it all

Run it three times (67.95% chance of profit; 5.39% chance of ruin)
43.86% to win 2/3
26.64% to lose 2/3
24.09% to scoop
5.39% to lose it all

Run it four times (51.38% chance of profit; 2.04% chance of ruin)
36.39% to win 3/4
33.14% to chop
14.99% to scoop
13.41% to lose 3/4
2.04% to lose it all

These numbers would be correct if running it multiple times went as follows:
1. deal the turn and river, allocate 1/n of the pot to the winner(s)
2. shuffle the turn and river back into the stub, along with their respective burn cards
3. repeat

However, this isn't how it works. (This is exactly the point that Siegmund was making earlier.)

Consider 7T vs AA on A A 6 9; the two hands' equities are precisely 1/44 and 43/44. But it is not true that T7 scoops (1/44)² of the time. In fact, it is trivial to see that running it twice never leads to T7 winning twice.

Yes, but you aren't including the increased chance of T7 hitting the second run when it misses the first run. The disparity between my numbers and the actual numbers is statistically insignificant because of the vastly larger number of combinations by which OP could win or lose. As you approach "one out" this disparity becomes significant. It is functionally correct to say that your probability of scooping = probability of winning one run to the X power where X is the number of runs.

I do get your point, though. The reason my numbers aren't going to be much different than the exactly correct numbers is because when you hit the first run your chances go down (to zero in your T7 example, but much less significantly in my example), but when you miss the first run your chances go up (infinitely larger than zero in your T7 example, but much less significantly in mine).

From T7's perspective:
WW = (1/44)(0/43)=0 (IMPOSSIBLE)
WL = (1/44)(43/43)= 2.2727272% (chop)
LW = (43/44)(1/43)= 2.2727272% (chop)
LL = (43/44)(42/43)= 95.454545% (lose)

Running it twice DOUBLES T7's chances to win half the pot. Because there is only one way to win on one card to come, it was easy to figure exactly what the actual odds are in this case. Multiplying odds together to get the numbers I got is a functional approximation when there are many ways to win (villain could get two hearts, villain could get running two pair, etc.)

I'd be interested in seeing how you DO actually figure it out without it being too complicated to be functional, and see how far off my numbers were, if at all.

Huh? What in my post are you responding to here? I see no significant error or even plausibly misleading omission in my original post.

You have no evidence to support your claim of statistically insignificance here. That said, you are probably right, but a priori this may not be the case.

(Also, just FYI, I think you want something like "practical," "workable," or "usable" where you used "functional" in your posts.)

I have several problems with this posting and your question of whether the results might not be exact. One, an approximation is just that- it is not exact and in this case I would say your results are off by at least several percentage points. For example, you assumed that the chance villain wins on the last hand is the same as the first hand win percentage after he has won three times. In winning the first three times, he likely did so each time with one or two hearts for a flush so the ratio of hearts to the deck size on the fourth run is much lower. For the first hand, it is 9 out of 45 =20%. For the fourth hand, assuming 3 previous flushes (chances for winning without a flush are less than 2%), it is at best 6 out of 39 = 15.4%. I agree that coming up with the exact answer here is complicated and your assumption of independence makes for easy computation and could get you in the ball park but is not by any means exact.

The second problem is your definition of profit. You assume a profit only if you win more than half the pot. That is not in keeping with accepted EV theory. The money in the pot includes what you, your opponent and others put in but is now considered up for grabs -- any winnings is “profit.” So, how should you choose the number of runs? Well, you can calculate expected profit. We know that expected profit is the same for all cases. That leaves the choice to depend on variance and theory tells us that the more times you run it, the lower the variance as your approximate results suggest.

Judging from the responses about the exactness of the numbers, and an exact solution being complicated, would a simulation make more sense here?

I really like this. Thrice it is!

There are other ways to win besides flushing, and please prove I was off by that significant an amount or gtfo.

Dude he acknowledged that:

Still waiting for proof that I was significantly off.

Well, at least you said ‘please.’

Look, you are using an approximation which it appears you were not aware of until two other posters pointed it out. However, you still seem to think that the probability results you got could be exact or at least so close to being exact that it won’t matter. You asked for proof that you might be off by more than just a smidgen, “if at all”. Well, doing that for the hand being discussed would be extremely tedious so let me simplify the problem a bit

Assume the turn card was dealt and it was an innocuous 3. That card helps neither player improve but gives hero a higher winning chance for he is currently in the lead. So, we have the following:

Hero: A A Villain: K Q Board: J 6 2 3

Now, the only way villain with two hearts can win is if a heart falls on the river. There are 9 hearts in the remaining deck of 44. Hero wins the first run if a heart doesn’t fall and the probability of that is 1 - 9/44 = 0.795; villain wins with probability 0.205.

Here are the results for the villain after the fourth run using your approximate method (binomial distribution assuming independence) versus the exact method (hypergeometric distribution accounting for cards dealt on previous runs).

Formulas for P(villian wins x hands in 4 runs ), x = 0,1,2,3,4:

Approximate method: C(4,x)p^x * (1-p)^(4-x), where p= 9/44 = 0.205
Exact method: C(9,x)*C(35,4-x)/C(44,4)

Wins Approx... Exact .....Abs. Error
0..... 0.4004..... 0.3857..... 0.0147
1..... 0.4118..... 0.4339..... 0.0221
2..... 0.1588..... 0.1578..... 0.0011
3..... 0.0272..... 0.0217..... 0.0056
4..... 0.0018..... 0.0009..... 0.0008
............Total Abs. Error..... 0.0442

The total of the absolute errors for each possible outcome in using the approximation is 4.42%. Whether the total error will be larger or smaller than this for the case of having both a turn and river card dealt I cannot say but certainly this result should indicate that the assumption of independence in the approximate method can lead to more than just an infinitesimal error in the win probabilities. Villain’s expected value for both methods is 0.205, the same as it is for just one run.

Let me also state that for some hard problems (at least for me), I have used the same independence assumption – but I always note this in the posting and I appreciate it if someone was able to provide a better solution.

I don't understand the concept of absolute error.

Approximate method: C(4,x)p^x * (1-p)^(4-x), where p= 9/44 = 0.205
Exact method: C(9,x)*C(35,4-x)/C(44,4).

Why are you making the approximate method harder than it was? I don't understand what C stands for or what the heck the (4,x) is in there. The approximate method is just adding probabilities of each possible outcome. All you have to know is the probability on running it once. This method is not erroneous because for every time the villain wins the first run and loses a card he needs, he sometimes wins the second run with the extra card instead, so they wash out. Sure, its complicated and time consuming when you run it four times, but it works pretty well running it twice, especially when you have multiple cards to come.

I could not follow your "exact method." Maybe if you explain in plain English what is all going on in your formula. I would love to know the magic formula for figuring this stuff out explained to me in English that someone with a 19th grade education can understand.

C(n,r) is the number of combinations of n things taken r at a time = n!/(r! *(n-r)!) where k! = k * (k-1) * (k-2) * ... *3*2*1 and 0! =1.

The approximate formula I showed is just a mathematical way of calculating exactly what you did. It actually is the binomial distribution, which assumes that for each run, the probability of winning a hand is the same as that for the previous hand. This is obviously not true unless the board cards of the previous hand are put back into the deck and reshuffled.

Your argument that the runs "wash out" is not valid when you are talking about results of a series of runs such as consecutive wins -- that was clearly demonstrated in a previous posting by wobuffet? when a player had only one out.

In my example, for four runs, villain wins four times only if a river heart appears on each run. There is no wash out effect. The probability of a heart on the first run is 9/44. Given that happens, the probability of a heart on the second run is 8/43; third run 7/42; fourth run 6/41. Therefore total probability is

(9*8*7*6)/(44*43*42*41)=0.000928.

By your method, you have

(9*9*9*9)/(44*44*44*44)=0.00175.

Your method gives a result that is almost twice the actual value although here the arithmetic difference is small.

Absolute error is simply the difference between the approximate and true value, ignoring sign.

Thanks...but don't understand "taken r at a time" And you don't define what k stands for.

The C(n,r) notation is simply the binomial coefficient: http://en.wikipedia.org/wiki/Binomial_coefficient ; it is quite standard in any discussion of probability.

starrazz, have you tried being a little less aggressive in tone? There's no need to assert your correctness repeatedly or, conversely, get really defensive when someone suggests you might be wrong. It'd go a long way.

Combinations = the number of different ways you can select r distinguishable things out of a total of n = C(n,r)

I used the symbol k only to define what the math term k! means, which is called k factorial.

If you have 9 basketball players, the number of starting teams without regard to position is

C(9,5) = 9!/(5! * 4!) = 9*8*7*6/(4*3*2*1) = 126

The number of starting teams considering position, equivalent to permutations, is
P(9,5) = 9!/4!= 120*126 = 15120.

Note, for each of the 126 combintions of 5 players, there are 5 players that can be chosen to play position 1, then 4 players to choose from for position 2, etc. That is, for each combination there are 5*4*3*2*1 = 120 possibilities.

If you really want to learn about these things in detail, I don't think a forum like this is the way. You ought to get a basic book on probability or applied mathematics. I even think there may be one that focuses on poker and probability.

Hmmm, maybe I'll write one.