Mason Malmuth said that being dealt aces two hands in a row isn't over 40,000-to-1, however I can't understand why.

"The other night when playing limit hold ’em at The Bellagio, a young lady managed to get dealt aces two hands in a row. She couldn’t believe her good fortune and wanted to know what the odds of that happening were. Another player spoke up and told her that it was over 40,000-to-1. Of course this is wrong."

I'm probably wrong, but here are my calculations.

Number of possible poker starting hands:
(52*51)/2 = 1326

Combinations of aces: 6

Odds of being dealt aces:
6/1326 = 1/221

Odds of being dealt aces twice in a row:
(1/221)(1/221) = 1/48,841

That's definitely over 40,000-to-1. Are my calculations wrong?

Yes, you will get aces 1/221 times the hand after you first get aces.

Those are the odds if you're dealt exactly two hands.

However, generally when you ask this question, you're probably actually thinking of the odds of getting dealt aces twice in a row in any two hands in a given session, or at least over a range of hands.

Those odds are much higher.

Came to this forum just to post about this. Glad somebody else was as confused as I was since I would have assumed the odds of any two given hands were 40k:1 as well. Don't feel like digging up my copy of Super System but I thought it specifically said "odds of getting AA on consecutive hands = 48k:1" (or whatever it is).

Anyway, I do not know why it isn't 40k+:1 and I don't appreciate the weird condescension of the statement "of course this is wrong" without providing any explanation for the reader. It makes me feel stupid and pretty much turns me off to the magazine as a whole.

Why would you get them any other number of times?

Hi Bob:

It depends when you start counting. If you don't start counting until you have been dealt a pair of aces, then the odds are 220-to-1. If you start counting before your are dealt a pair of aces, such as "What are the chances I will be dealt aces on each of the next two hands? then the larger number becomes correct.

Best wishes,
Mason

Oh okay.

Mason is playing a parlor trick, or a lawyer's game.

Kind of like the logical fallacy about carrying a bomb onto the airplane in order to dramatically reduce the odds of anyone else also having a bomb on the plane. That old saw has probably gone out of fashion sometime in the last decade, I expect.

I don't know about that, I agree it really does depend on when you start counting.

If you were just dealt aces and you start trying to do the calculation at that point, it's the 221-1 because these are supposed to be (allegedly) approximately independent trials.

There has been somewhat similar debate in the past regarding the Monty Hall problem. As I see it, the contrast relates to "ex post" vs. "a priori" estimations.

Forgive me for citing wiki but it is pretty thorough when it comes to explanation of the problem, which I believe is somewhat related to this question:
http://en.wikipedia.org/wiki/Monty_Hall_problem

Mason - Thanks for the explanation. But I think you know what the other person means by 40k+ to 1 and this is as much a semantics issue as anything else. So there's the part about it not being true but the other part where you know that getting AA on the very next two hands you play indeed are 40k+ to 1.

Well, not exactly; I think getting a hand like AKo occurs twice as likely as getting aces. So that'd be (1/110)^2, or 1 in ~12k.

No. It's not a semantics issue. In the world of mathematics, there is something known as a problem being well defined. Specifically, if the problem at hand is not well defined, incorrect conclusions can be reached. And in terms of poker, inaccurate strategies can be applied.

Here's a similar example. Suppose you are playing no-limit hold 'em in a full ring game. If you are dealt a pair of kings, what are the chances a pair of aces is also out. If you think it's approximately 5,400-to-1 as opposed to approximately 25-to-1, you may play your hand very differently.

best wishes,
Mason

Best wishes,
Mason

lefty - When I said, "I would have assumed of any two given hands" I was referring to AA since that was the subject. That on any two given hands you choose the odds of both of them being AA were 40k+.

Mason - Not quite sure I understand your example. Are the odds of KK vs. AA in a given hand of heads-up roughly 5400 perhaps? While the odds of one out of 8 players having AA (if you are holding KK) somewhere in the 25:1 range?

I do believe I'm getting what you're saying about the well-defined bit for mathematical purposes and obviously there are many poker players who think they know the math but apply it so horribly that it gets close to making no sense.

Not really related but it reminds me of the player who was telling me how strong a hand T5 is because it makes all the straights (confusing the idea that every straight contains a T or a 5).

What Mason is saying is basically just about the different parameters. For the aa kk problem the odds of kk being out aren't the same as the odds of aa and kk being out. AA being out has already been defined so the probability of that is 1. Mason, coach me in teh limit hold thems?

Also if you want to talk about the problem being well defined isnt the 1/48k answer correct. She simply wanted to know what the probability of getting dealt aces two times in a row was, not the probability of getting dealt aces assuming she had jus been dealt aces.

The very first post in this thread actually quotes Mason's definition of the problem verbatim. As written, I would very much interpret the problem as:

'What are the odds of getting Aces two hands in a row'.

If this sentence alone is the only thing we have, then we can assume that only two hands are played and in that sense probability to happen is 1/221*1/221 = 1/40k.

But this is not how one would practically look at such question, or statement. And normally when one asks the above, he would imply an event that occured during a session, or something like this. When a fish says "lol he got aces 2 times in a row", and we try to calculate the probabily for that, we wouldn't end at 1/40k, but the answer will more lean toward 1/221.

Think also about the more radical statement, "wow, he got aces 3 times in a row". Most often, the mathematical event that triggered this "wow" statement would be with probabilty 1/40k or less, instead of 1/4million.

i like what mm said about the 48k:1 odds being right for getting dealt aces twice in a row "on the next two hands". But then, if you played 100 hands that night, that's 99 opportunities for 'two specific hands in a row'. So the odds of getting aces twice in a row, anytime that night, would be 99 times that, or like 4.8k:1.

furthermore, if you played for longer and longer time intervals, then your probability of getting dealt any pre-specified string of starting hands at any point in that interval, would converge to 1.

so umm, what ulrichw said.

im sorry but what are u talking about??? As you play more hands the probability of getting dealt any pre specified string of starting hands does not change.

No, but you've got many more individual opportunities for the specific string to occur. In this case, there are 99 discrete two-hand sequences, thus 99 chances for this 1/48K probability to occur. That's like 99/48K, I hope.

Or maybe I'm just wrong, and if spoken with authority, stating that I am wrong is sufficient unto itself.

Yeah, that's it. And it's correct that the more hands you play, the probability to have "dealt aces two times in a row during all these hands" will converge to 1. It starts at 1/40k if you look at 2 hands only, and converge to 1 if you increase the number of hands.

by my approximation the chances of being dealt aces on consecutive hands in a 50 hand session are of the order of 1088 to 1
With 100 hands it is 605 to 1
with 200 hands it is 369 to 1
with 500 hands it is 245 to 1

As a very rough approximation if you said a session was about 100 hands, and that with ten players at a table the chances of it (someone being dealt AA twice in a row) is about 10 times the above, you'd expect it to happen about once every 37 sessions you play, i.e. rare but hardly newsworthy

I didn't realize you had introduced a different probability distribution. But you are right. The probability of getting dealt aces twice in a row at least once would converge to one as more and more hands are played. However, I feel you need to include the at least part in order for this to be accurate, which goes back to what Mason was saying about the problem being well defined.

And if you wait until after you've already been dealt AA twice in a row than the probability is 100%. We can come up with all sorts of numbers if we completely ignore the original question that the lady asked.

Next time i'm wrong about something obvious, then i state that my incorrect answer is obviously correct, i'm gonna remember that poorly defined question excuse.