What are the odds and math of two players all in preflop, running 3 full boards, AsKd vs AdQs, AQ wins all three.

Buddy and I have a bet and have disagreement on the odds.

Thank you!

Something less than 25.3%^3~1.6% if you do not reshuffle after each run. Probably rather significantly less. AdQs vs AsKd has 25.3% but if you run it multiple times, although the EV doesnt change, the chance to win all 3 is not the cube of this chance because of removal effects that become conditions on every new run made after a win affecting often substantially the new win probability.

For example if it were to win the first 2 times by getting Q and in one of them you had 2 Qs then the 3rd time it cannot really win with a Q as there is none left and it can only win by very rare in comparison ways like flush or straight.
(Even if it didnt win one with 2 Qs but won twice with Q and so there are now say only 1 Q and 3 K left the chance is on the 3rd run far less than 25% say around 15%)


It may be easier to see this by an example with a river all in that is ran many times. Say

2s2d vs 7c7h with the board 2cAd7s8h

Now 22 can only win with the last 2 left. So clearly it cant win 3 times if you run it 3 times. If it wins 1 time then the others must be losses. So the probability to win all 3 is 0.


This is why what you ask is rather complicated to be answered exactly.

It can be simulated and then this will probably yield an answer smaller than 1.6%. But its a good question how smaller it proves.

I imagine the concept of card-removal effects is understood by OP and his buddy to some degree. Or else this would not be much of a problem and they would likely not have a disagreement.

Over all possible 5-card boards, Ad Qs beats As Kd 392,345 times out of 1,712,304. This is 22.913%. With no card-removal effects and perfect independence, cubing would give the prob of winning all three runouts. 22.913% cubed gives 1.203%. Card-removal effects will bring down the actual answer in reality.

There are two general approaches in these types of problems. Seek exact answers or do a simulation. Unfortunately there are just too many 3-board runouts to run through them all to get an exact answer that way. I think there are 10^17 3-board runouts in heads-up NLHE.

Let me muse a bit to see if there may be other ways to get an exact answer. Theoretically, we could construct the set of 392,345 boards in which AQ beats AK and then loop over each one in turn to see how many 3-board sets it would be part of. It would be a triple loop over the 392,345 items and in each loop you identify which other items are still possible in the runout by taking into account card removal. The problem with this is that a triple loop over 392,345 items seems infeasible on first blush.

Alternatively, we could group AQ-winning boards into categories based upon the poker hand that AQ wins with. I think for our purposes the relevant categories might be:

A. Pair of queens
B. Two pair aces and queens
C. Two pair queens and a lower pair
D. Three queens
E. Ace high straight
F. Queen high straight
G. Diamond flush
H. Full house aces full of queens
I. Full house lower rank full of queens
J. Full house queens full of aces
K. Full house queens full of lower rank
L. Four queens
M. Straight flush in diamonds
N. Straight flush in spades

Maybe this isn't the ideal categorization but you get the idea. Then, we come up with a way to tally how many 3-board runouts are possible by choosing one runout from each category. The triple-categories would be:

AAA
AAB
AAC
AAD
.
.
.
NNN

Ideally, the tally for each triple-combination could be done via combinatorics. Of course, many of the tri-category cases will be impossible (e.g., there are only so many queens left in the deck).

However, I am not sure that this alternative is much of an improvement to the brute-force counting of the number of possible 3-board runouts starting with the 392,345 winning boards. Since, in the absence of approximations, to get the exact answer we'd still have to manually take into account card removal effects when forming possible 3-board triple runouts from the different categories. Maybe a clever categorization could be devised to diminish the importance of card-removal effects.

Of course, if an exact answer is too difficult to derive, a simulation could be performed to give an approximate answer. But I don't know how many people have simulators that can handle 3-board runouts. I could do it but it would take a day or so and I don't have the time right now to devote to this.

Yes i should have said 0.229%^3~1.2% as the upper limit not 0.253^3 because i forgot about the ties that generally are not significant between top 10% range all ins but they are over 4% here when they share an ace.


I did a rough estimate on a lower bound using wins by 3 queens one at a time no kings or 2 times with Queens and 1 with straight or flush and it came at about 0.31%.

So my guess is the correction is significantly below 1.2%.

So lets say for now the answer ought to be between 0.31% and 1.2% and probably closer to 0.5-1% range with 0.5% a not so terrible guess.

Okay, I kicked off a job before I went to bed last night to compile the list of the 392,345 boards that win for AQ.

I verified that it is beyond infeasible to do a triple loop over all 392,345 items to tally how many triple-combos are possible taking into account card removal effects.

So what else can be done?

I had a brainstorm of sorts (a very minor brainstorm). I could randomly sample triples from this list in order to estimate the pct of of winning-triples that would win all three runouts, meaning that all three triples are possible (no duplicate cards). Then based upon that information we can back into the overall winning-triple pct.

I ran 10 million trials of randomly selecting triples from the list of 392,345 winning boards. Of these triples 8.025% were possible (no duplicate cards among the three five-card boards).

So my current estimate of how often Ad Qs beats As Kd in all three of three runouts is the following:

C(392345,3) * 3! * (8.025%) / [C(48,15)*C(15,5)*C(10,5)*C(5,5)] = 0.586%

where the 3! comes from there being that many ways of ordering the triples I selected (in order to match the approach used for the denominator in the formula above). Equivalently, you can divide the denominator by 3! to arrive at the number of "non-degenerate" triple runouts.

Comments welcome.

It would be hilarious if OP just comes back and announces that yeah btw we shuffle the deck after each board

Unless the shuffling is not done deep enough at which point it will be an even more complex problem lol.

Yeah, wouldn't it be even more complex if they reshuffle? Since the same unknown dead cards (folds and burn) would be omitted from the pool.

It would be easy if they reshuffle always at 22.9% for a win or so. If they do it poorly it becomes extremely hard problem depending on their method to even crudely estimate. Indeed the way humans reshuffle if not done deep enough or worse if its done by cutting decks and changing their position alternating one up one down etc and few times and cutting again etc you may get significant biases still (like see next video in the first minute to be shocked at the claim made about the shuffles some people do and think its ok lol).

I have a game i play with cards for fun to kill time say when waiting or something that is basically a player (fictitious opponent ) that pushes all in every hand and i call or fold (hero trying to guess proper range) but always the cost is 1bb and we start from 10bb and its a freezeout. Proper theory how to play this to maximize win probability of the event is like 53% edge at only 10bb starting stacks or so for the careful caller. So i once did a cut like that multiple times (because its hard to always shuffle properly in such unimportant single game) and dealt myself the same 9 cards all in called sequence to my complete shock.

I think the proper shuffling is where you put all cards at the table and move them around for over 20 seconds at reasonable speed mixing them from all sides (smoothing) (say one rotation every half a second or so). He claims that even this needs to be for close to 60 seconds, yikes.





I always imagine a good test of any shuffling is to consider you have 4 aces on the board in some rare deal and then you shuffle. In principle the chance the new board has also 4 aces should be like 4C4*48/52C5=1/54145. Now you go ahead and tell me that the way people shuffle you will break those 4 aces apart and then mix them again to come again together with that probability and not end up basically either almost never getting it that way all 4 again on the next deal or getting 2 of those 4 aces stack as pair together dealt one after the other etc but the other 2 very far typically.

The very easy instant test of any shuffling done is to put the cards in order of suits and ranks and then do the standard shuffling you would perform and notice once you deal it again face up to see the new order how many same suits and connected cards you have in the deck. What should be the correct avg number of such 2 card streaks of suit and rank (not very big lets do it as exercise)? Compare that to what you observe. I bet you will still see plenty of same suit connected cards in the deck left the way most people shuffle at home.


Ps: He is wrong about the number of particles in the universe though. Its just the protons/neutrons of milky way galaxy.