HH? In the spirit of Regis Philbin, this is my final answer. If the odds of it happening are 1,700,000:1 against, the odds of it happening 3 times in 100,000 hands are 1/(1/1,700,000*1/17*1/3)=86,700,000:1 against. Even if this turns out to be wrong, notice how I cleverly avoided using the word probability.

Stop grunching. Been there done that. Check out 4L.

im new! WOOO! Aces two times in a row? hmm...id say only 220 squared...whats that? lol...divided by...PIE
hah

well i can see the logic of using the odds of gettin aces ONE time and SQUARING it for two times, but it just doesnt seem that realistic...id say the odds are drastically LESS than 40,000. id say youd have to sqaure it, then multiply by 25 (50 cards in poker deck divided by two)

OK all you math geniuses, what are the odds of this exact circumstance happening? http://forumserver.twoplustwo.com/54...ce-row-472669/

lol donkaments.

The probability is easy, the hard part is explain why it doesn't mean anything. You see this sort of reasoning all the time around games with random numbers, but it doesn't imply anything because of two effects.

1) a couple million people play poker of those who starts forum post about a run of luck? The people that didn't see anything conspicious happening or the person that ran into something unusual? We calculate the probabilities of one person playing X hands viewing Y events, but your one of 100,000s of people that play a lot of poker and you were filtered out of those 100,000s.

2) Poker is a game where there are 1000s of different things you can measure in a hand history. For anyone one person you almost certain to be able to find one thing that is peculiar and extremely unlikely in their hand history. In stats people recognize the term "fishing expedition"; it is where you find unusual ways to look at your data over and over until you find something inconsistent with how it was generated. As the number of ways to measure your data increases towards infinity you should expect something inconsistent with a true null hypothesis to occur.

#1 and #2 combine to make this whole thread not meaningful. If anyone of 100,000s of people were going to post about wierd things, it would be the few of 100,000s that had wierd things happen to them. And everyone can look at their hand history and find patterns that are lucky looking or unlucky looking.

I just want to reiterate the point that LargeLouster has been incorrect in probably half of his posts (so that means if he posts twice, he'll have a 100 percent probability of being wrong once, right? )

Just to start, as someone already pointed out, this is wrong. You've calculated the probability of getting an EXACT hand (disregarding order) twice in a row...but you're completely wrong about how many times it has approximately happened on FullTilt. The probability of it happening on ANY GIVEN HAND, that is, you receive the exact same hand that you just had, is 1/1,326. Your calculation is correct in that 1,326*1,326 are the odds of your NEXT TWO hands being exactly the same. But, your odds of getting the exact same hand you just had is NOT that...because you've already received it once. If you going to argue semantics and that you were indeed trying to calculate for the NEXT two hands, then I'll just leave it at the fact that you're way off on how often it happens. If you think it happens 6,900 times every 11.7 billion hands, then you're wrong.

Among many other things, you are also wrong about the probability of two people receiving pocket aces at a 9 handed table. I'm not 100 percent sure myself, but I'd venture to say that it's ((4/52)*(3/51)) * ((2/50)*(1/49)) * 8)

I added some extra parentheses and spaced it out so that it's easier to see how I arrived at the answer. Again, not so sure myself, but I'm willing to admit my mistakes and reassess if someone proves it wrong.

By the way, the notion that your probability of having gotten AA twice in a row by your 48,841st hand is 100 percent is absurd. It's just as absurd as the notion of you getting tails on a fair coin by your second toss is 100 percent.

binomdist(1,48841,1/221^2,0) = 36.8%

I'm pretty sure it's actually

C(9,2) * 1/C(52,4) = 1/7520

You can also use
C(9,2) * 6/C(52,2) * 1/C(50,2) = 1/7520

Your formula gives 1/33841

LargeLouster gave 1/3760 because he missed a factor of 2 in his calculation.

Every single hand has a 1/1326 chance to be the same as the hand that came before it. So in 11.7 billion hands, we have 11.7 billion (less 1) opportunities for this to happen. They are independent, so we just add probabilities. So 11.7 billion * 1/1326 is about 9 million times, multiplied by the number of players dealt cards in each hand. If that averages 6, then we have about 54 million times of back-to-back identical cards. Amirite?

54 million or 6900, big difference there.

Whats the probability of winning a state lottery jackpot (matching all numbers) such as power ball and mega whatever etc twice *in a row*?

Answer tiny. Other answer it happened once already. And multiple people have won it twice but not in a row. A harvard statistician computed we were over 50/50 on someone winning twice in their life sometime in the late 80s.

There is a problem in statistics called the Birthday problem. And students often have to calculate it but rarely do instructors tell you the intuition it represents. The birthday problem is like this: how many people do you have to have in a room before you expect two to share the same birthday without respect to year (ignore leap years assume all birthdays are equally likely). And the answer is about 24 before you expect a pair to share the same one of 365 days as a birthday. And the intuition behind it is when you don't restrict what can have a collision ... you should expect collisions much more frequently (comes up in math of hashing tables too). Contrast for example with how many people do you need in a room before you should expect one shares your birthday.

The moral of the lottery story is this:
things that appear unlikely with respect to a particular person happen all the time, but you don't know who they will happen too or what they will look like prior to seeing them. When you take into account that information then what appears unlikely is quite ordinairy.

oops, I see what I did. I left out the 9 in the first part. there are 9 chances to get AA the first time, and then 8 chances the second time.

(4/52)*(3/51)*9 * (2/50)*(1/49)*8

So my calculation actually does come out to be the same as Louster's. But like I said, I wasn't too sure about that one, so I could be off...care to explain how you arrived at twice as unlikely as 1/3760?

Yep, I'd say this is exactly right, assuming that 11.7 billion hands dealt means that 11.7 billion times, a table has been dealt as many hands of poker as there are people sitting down. So like you said, you would multiply 11.7 times the average number of people at a table at any given time.

The only alternative is if 11.7 billion hands means exactly 11.7 billion, and already takes into account the number of people at a table.

The way they count them is deals to a table, regardless of players seated. That is the displayed counter on the site.

OK Einstein, be happy to explain this one. First, go read my post #178 where I corrected my error and post #177 which shows why my original calculation was wrong. (4/52)*(3/51)*9
this is correct. (2/50)*(1/49)*8 this is wrong. There are only 36 AA combinations in a 9 handed game. Using my original calculation, you have to divide the 8 by 2 to get the correct answer of ~1/7,520.

As jungwah correctly stated the solution, a simpler approach is to start with heads up and reason from there. Heads up is 4/52*3/51/*2/50*1/50=1/270,725. Since there are 36, not 72 combinations in an 9 handed game, then 1/(270,725/36)=~1/7,520.

Next time you decide to be full of yourself, make sure you have a) done your homework and b) have the competency to back up your arrogance.

Competency? I brought up three issues. Two of which I (and others) knew you were completely 100 percent wrong about. I guess you might call it arrogance, but if you're wrong about an objective fact, I think it makes sense to bring it to light - if not for your sake, then for the sake of everyone else reading it.

The third issue, I myself was unsure of. I was pretty sure your original calculation was wrong, but I wasn't too sure how to do it myself. I missed your corrected post.

So as far as competency goes, I guess I'm 2/3, and you're 1/3?

And that's assuming your AA calculation is correct. I'll be honest, I can't quite wrap my head around it.

Why do I need to divide 8 by 2? And that's not some sarcastic, snide question. I'm just trying to follow the logic behind this example.

skopivy think about the porblem this way. 9*8 represents 72 combos, but there are only 36 combos. if u write out the combos on a piece of paper (or start to) you'll see that you have to include utg, button and button,utg as separate events when in actually they are the same thing since we don't care about order in this instance. button will never be dealt aces before utg, and even if it were the case order doesnt matter in the context of the problem.

Also I remember someone showed me that birthday thing and it freaked me out, specially when the guy was putting large sums a maneys on the line.

As usual, I made a typo. 4/52*3/51/*2/50*1/50=1/270,725 should be 4/52*3/51*2/50*1/49=1/270,725. In my original calculation, which was wrong, I went back and divided the 8 by 2 just so I would end up multiplying the two pairs of aces result by 36 and not 72. Bottom line, it was a quick fix to an incorrect calculation.

More precisely, you should be using C(9,2) for that factor, which equals 36, rather than using 9*8 and remembering to half it and why. It's also logically easier to follow.

yea thats what i did but i was trying more to show that 9p2 (what he did) was factoring in order and 9c2 wasn't.

Ah I see, putting it in terms of combination vs. permutation helps. The way I had been looking at it was:

(1/221 * 9 players) * (1/1225 * 8 players)

Seems logical, but i guess it takes the order of AA received into consideration when it is in fact irrelevant.

Does anyone know how to calculate this without C or P? Or would it be far too complicated?

Forget permutations when dealing with poker hands, and always use combinations. I don't know any way to count possible combinations other than counting possible combinations. I'm not trying to be cute, but it's sort of an irreducible operation as far as I know. Many calculators can do the C function for you, I usually just plug it in Excel since I'm at the computer. With small sets you can do them in your head or easily on scratch paper, but I don't think there is any way to avoid counting combinations to figure out poker probabilities.

The formula to do it longhand is:
nCr = n!/ (n-r)!r!
so now you have to calculate factorials.

Originally Posted by skopivy
oops, I see what I did. I left out the 9 in the first part. there are 9 chances to get AA the first time, and then 8 chances the second time.

You made the same mistake I did, which was starting from a faulty assumption and not taking into account a restriction, that the cards must be dealt in a specific order and not at random. Its not true there are 9 players who can get the first pair of aces, there are only eight. Focus on the first nine cards dealt. Unless one of the first 8 players gets an ace, it's impossible to meet our requirement that two players end up with a pair of aces. Also, whoever gets the first ace determines the number of players who can get the second ace, which can be from 8 down to 1. If you count the number of players who can combine with player #1, then player #2 and so on, you end up with 36 (8+7+6+5+4+3+2+1).

However, even if you could deal in a completely random order, there would still only be 36 meaningful combinations of AA,AA since order doesn't matter and we end up back with 9*8/2=36 combinations of AA,AA again.

4/52*3/51*2/50*1/49=1/270,725 is still valid since only 4 specific cards, aces, meet our requirements. So, that gets us back to 1/270,725*36=~1/7,520 as the correct answer. The incorrect answer was 1/270,725*72=~1/3,760.

Btw, don't take the trash talk seriously, I just think it's fun sometimes, nothing personal.

I'm sure no one will be upset when I say this will be my alsoLOUte last post on this thread. It's been fun, interesting and I've learned a lot about probabilities I didn't know before. I wasn't kidding about a lack of formal training, I ended up counting up the combinations in my last post #224 on my fingers. Also, I was deliberately provocative in the way I presented my posts because I figured that someone who actually knew the correct answer would get just pissed off enough to take the time to point out my errors and show me the corrections; between junwagh and spadebidder, I exceeded my expectations. What can I say? Socrates had his method, I have mine. So, gl e1 & tc.

Best wishes,
MasonLouster