Damn it! Now I have to answer you. You can use your approach and you will get 72 combinations and guess what? 72*3,760=~270,725. My approach is to consider each pair of aces separately.

The chances of the first pair of aces is 52*51/6/2=1/221, but since there are 9 players who can get the first pair of aces, you have to divide that by 9 which gives you 1/24.55 (rounded to 2 decimal places).

Now consider the chances of the second pair of aces. That is 2/50*1/49=1/1,225, but since there are now only 8 players who can get the second pair of aces, you have to divide that by 8 which gives you 153.12 (rounded to two decimal places). So, you simply multiply 1/24.55*1/153.12 and you get the chances of any 2 of the 9 players getting AA in the same hand as ~1/3,760, (I'm sure you've noticed that divide by 9*divide by 8=divide by 72, so to speak).

When considering only 2 players (either 2 specific players at the 9 handed game or 2 players heads up), you simply multiply 4/52*3/51*2/50*1/49 which does give you 1/270,725. There is no need to divide by either 9 or 8 because there are only 2 players who can get the pairs of aces, so there is only one player who can get each pair.

Seriously though, you have to quit doing this. Otherwise, God forbid, there is actually a chance (a really small chance), that we might end up friends. Permit me to make a suggestion. Go find that Mason guy and insult him. He stuck me as kind of arrogant.

lol i figured out why our answers are differing. You are using permutations in your calculations whereas I'm using combinations. Thats why i get 36 and you get 72. 9c2=36 whereas 9p2=72.

Using permutations in this case is wrong though. Let's say paying attention to order utg gets aces. then utg +1 gets aces. That's one instance where two players get aces. Then in another hand utg+1 gets dealt aces first, then utg gets dealt aces. You see that the order of their getting dealt aces is different and if that mattered your answer would be correct. But it doesnt matter because poker hands are dealt in a certain order so utg + 1 wiill never have aces before utg has aces.

I have to admit, you're right about this one. In my calculation, I forgot to take into consideration that the cards must be distributed in such a way that cuts the number of possible AA,AA combinations in half. As you said, it's impossible for utg+1 to get AA before utg.

"The probability of heads is 1/2, so mathematically, the probability of heads is 1.0 every two flips."

0.o

Also, even if the cards were dealt in a different order every hand, there would still only be 36 possible AA,AA combinations, since the order in which the pairs of aces were received doesn't matter. Oh well, they can't all be gems. So, the correct calculation for the chances of any 2 players getting AA in the same hand in a 9 handed game is 52*51*50*49/4/3/2/9/8*2=1/7,520.13888888889. Or, more simply put, 36/270,725=1/7,520.13888888889. And, in the words of Lou Gump, that's all I have to say about that.

The problem here is that of definition. You MUST define the system from which we are observing. If it's a question of a two outcome system, it is 48k to 1. If they are 1 outcome systems looked at seperately, it's 221 to 1. If it's 50,000 outcomes, the odds are more favorable. But defined as a two outcome system, the odds are 48k to 1.

Probability is not REALITY, it is only a predictive measure that attempts to reflect reality... not unlike palm reading. Although highly improbable, AA could lose to KK 4000 flops in a row. Is it likely? Of course not... but it is still possible.

But again, without specific definition of the system, there is no correct answer.

Even here, we're not talking about 3 times in a row, we're really talking about twice in a row. You're not starting before a 3 card sequence, you're always on the second hand of a 3 card sequence. And, if we go from an observer's perspective, the observer isn't aware of anything until afterr he has already seen the same cards twice in a row. So really, its just 1/(52*51).

It was interesting reading all the responses to this question. I think it says something about ones ability to think a bit more abstractly. After thinking about it, "of course it is much less than 40k:1" is probably the best answer because this poorly defined question cannot have a well defined answer, but the mere assumption that the woman isn't going to play a 2 hand session is enough to kick it under 40k.

Thanks, I was hoping someone would answer my question seriously. I apologize for the poorly constructed question. Would you please infer what I meant and ask the question in a more accurate and specific manner?

Ditto! I thought I tried to answer your question seriously back on post #114 when I said: The odds of getting dealt exactly the same two cards (same rank and suit), but ignoring order, twice in a row are 1/(2/52*1/51*2/52*1/51)=1/1,758,276 or 1,758,275 to 1. That's about 1.7 million to 1. But remember, Full Tilt has dealt about 11.7 billion hands to date, so what happened to you has happened about 6,900 times so far at Full Tilt (11.7 billion/1.7 million=~6,900). I apologize for being so far off base, John. Still, you gave me the chance to post one last time and end it in a way I've wanted to end one since I got involved with this thread.

Best wishes,
Mason

SERIOUSLY!??!?

lol mason coach me please. I think I've proved I can become winning player and have something of a brain.

Not so fast there buddy. The second part of the question is, assuming the odds are 1.7 million to 1: what are the odds of this happening 3 times in say 100,000 hands?

I don't understand what knowledge you hope to gain by asking this question. If the chances of this event happening are 1/1.7mll (which is already a small number) then the chances of it happening 3 times in a significantly less amount of trials than 1.7mill is obviously going to be a ridiculously smaller number. I would guess any calculator would approximate it to zero.

If you are interested in the method behind finding it and not just the "how rare is this" factor, google binomial distribution and input the numbers. That should give you the actual answer.

That's a tricky one and I've actually come up with 2 answers. The first, which I'm not sure of, is 1/(1,700,000/100,000/3)=1/5.666666667 or about 17.6% or odds of ~4.6:1 against, or a bit less than once every five 100,000 hand sessions you play. I have to disagree with junwagh (big surprise) since if something has a probability of 1/1,700,000 but you do it 100,000 times, I say the probability shrinks to 1/17.
However, since the principle of Lou's Razor says that when you have two solutions to a problem you should always choose the simpler one, I'm gonna go with answer number two: slim and none.

shrinks to 1/17.That should have been increases to 1/17. Also, the reason I'm not sure of answer #1 is that I don't even know what a binomial distribution is since I have no formal training in mathematics beyond high school algebra. I just think junwagh has his numerator and denominator reversed (no offense).

Are you a product of the California public school system?

I am a product of the Lousterville school system. Btw, who is the girl in your avatar, very cute, in a Miley Cyrus kind of way.

My point is that the chances of this happening are so small, that it shouldn't be happening.

That's assuming that online sites are truly random. Based on what I've seen, that's open to question. I can't prove it, but I suspect the game software is written to "manufacture action" by giving out more good hands, permit more suck outs and so on, which leads to more betting and bigger pots, which leads to more rake income. It's only a poker game to us. It's a business for profit to them. That was the significance of the Absolute Poker cheating scandal; it proved that online poker sites can, if they want to, rig the software.

Sorry, but I can't resist.

Best wishes,
MasonLouster

"since if something has a probability of 1/1,700,000 but you do it 100,000 times, I say the probability shrinks to 1/17."

You have done this twice in this thread, and you seem to have a genuine interest so maybe you can look into it and get it right.

You are confusing expected value calculations (means) for Bernoulli random variables for proability calculations. As an example the probability of getting 1 or more (a bigger probability than what you calculated) is very close to 1/170 and not at all close to 1/17....

The expected value for the 1s and 0s is 1/17 after 100k hands, but the probability of getting 1 or more in 100k times is approximately 1/170. Two different numbers.

You seem to have a preference for reasoning in expected value so by all means keep doing it. But it just looks silly to call it probability when you do.

THISSSS!!!

I think my question should have been, "If the odds of this happening are 1 in 1.7M, then how unusual is it for it to happen 3 out of 100K hands". You math geniuses should be able to calculate the Z, M, P and find how far from the standard deviation it is.

My wife did this calculation last night and the answer was pretty much zero chance of this happening.

I'm gonna have to take your word for this since I don't know what Bernoulli random variables for proability calculations are. Also, after re-reading my post, I realized that even if my approach in answer #1 had been right, I made an arithmetic error (that happens when you hang out with junwagh). Should have been 1,700,000/100,000*3 or a probability of it happening 3 times in 100,000 hands of 1/51.

However, I'd still like to point out 3 things. First, In answer #1, I never said the probability of it happening 3 times in a 100,000 hand session was 1/17; I said, or intended to say, 1/51. Second, you said As an example the probability of getting 1 or more (a bigger probability than what you calculated) is very close to 1/170 and not at all close to 1/17.... How can a probability of 1/170 be bigger than a probability of either 1/17 or 1/51? Third, I still stand by my answer #2. I think even you would agree that's a tough one to argue with.

So, thanks for the corrections.

I'd say your wife's answer agrees with my original answer #2.....slim and none.

Which begs the question: "What's up with that?" I will start saving HH when this happens.