One of the problems here is trying to answer a mathematical question with language. Language may make it easier to understand a mathematical equation, but words seldom give you a precise answer since words seldom mean exactly the same thing to different people. That's one of the reasons humans invented mathematics. The question being debated is the chance of being dealt pocket aces twice in a row. If you let A represent the correct answer, the only precise way to describe it is the equation:

A=1/(52*51/4*3)*1/(52*51/4*3)=1/221*1/221=1/44,841

Except you mistyped the result, 48,841.

Oops! I stand corrected, it is 48,841. Sorry about that, my typing sucks!

What are the probabilities that at a nine-handed table, I and the same opponent get dealt JJ and 99 on two consecutive hands, with me getting the JJ twice? Happened to me a while back in a Pokerstars EPT Round 1 qualifier, with both of us all-in on both hands.

I think, but am not certain, it's 1/(52*51/4/3)*1/(50*49/4/3)^2=1/2,035,889,600.69444

I've decided my 1st answer is wrong because I assumed you meant a specific opponent you had selected before any cards were dealt. Now I think a more realistic reading of your question is that you don't really care which of the other 8 players got the 1st pair of 9's, only that the same player got them twice in a row. This changes the way in which you calculate the odds of you getting JJ in the 1st hand and your "opponent" getting a pair of 9's. Since there are 8 other players it's 8 times as likely that someone will get a pair of 9's. So, the first part of the equation changes to 1/(52*51/12)*1/(50*49/12/8). The second part of the equation remains the same since there are now only 2 players involved and the overall answer is still the product of the two calculations. So, my new answer is 1/(52*51/12)*1/(50*49/12/8)*1/(52*51/12)*1/(50*49/12)=1/254,486,200.086806. I think this is a good example of what i mentioned earlier in this thread, that language is imprecise and can create confusion when discussing things that can really only be expressed with mathematics. Either that or I'm just lousy at reading questions correctly.

Mason--

So, when I get dealt aces, and when I think to myself that I just got lucky at the 1/221 event that just happened to me, I guess I'm making some kind of terrible cognitive mistake?

--Nate

MicroBob--

This is pretty much exactly right, and I'll put my math credentials up against anyone's ITT.

Thanks for your consistently good posting, & all my best,

--Nate

Hey Nate, since you actually have math credentials (I don't, just like algebra) I'd really appreciate it if you would check the math in my answers and tell me if it's correct. Thx in advance.

--LargeLouster

Guys...it's like this: If someone asks you what is the probability of flipping heads 4 times "in a row" it is 1/2*1/2*1/2*1/2= 1/16, or 15:1 against.

On the other hand, if someone has "already" flipped a coin three times and gotten heads all 3 times, what is the probability of getting heads on the "next flip"? 1/2, or 1:1.

Similarly, if someone has flipped a coin once and gotten heads, the probability of getting heads the next three flips will be 1/8, odds against being 7:1.

But if someone asks me the question of what are the chances of getting 4 heads in a row, I'll assume that no flip has yet been made, and you will now start the 4 flip sequence with the next flip. In this case, the chances that your next 4 flips will all be heads is 1/16, or odds against of 15:1.

Is that an easy to understand explanation?

The odds of being dealt aces 'twice in a row' or flipping heads 'four times in a row' is undefined without reference to a timeframe, which is why people are offering the known odds after the first event. What you want is the odds of being dealt aces on the next two hands.

/thread

rigged

The odds are much greater against being dealt 2 specific cards of the same rank twice in a row, ignoring order, because there are 6 ways to get any 2 cards of the same rank, ignoring order, and only 1 way to get 2 specific cards of the same rank, ignoring order. Any two cards of the same rank, twice in a row, ignoring order, is 1/(4/52*3/51)*1/(4/52*3/51)=1/221*1/221=1/48,841. Two specific cards of the same rank, twice in a row, ignoring order, is 1/(2/52*1/51)*1/(2/52*1/51)=1/1,326*1/1,326=1/1,758,276. If you compare the 2 final numbers, you'll notice that the larger number is exactly 6*6=36 times larger than the smaller number. So, getting two specific cards of the same rank, twice in a row, ignoring order, is 36 times less likely than getting any 2 cards of the same rank, twice in a row, ignoring order. I said "ignoring order" because in your post you didn't specify in which order you received the 2 cards.

Every 221st time you get aces, you will get them on the next hand as well on average.

[x] Aces twice in a row twice, once at Bella, Once at Jokerstars....A10 3rd hand at bella boooo

What we're talking about here is the difference between a priori and a posteriori probability. Expectation is a priori, so being dealt AA the next hand is always 1/221 regardless of what the last hand was. The a posteriori probability, the chance of two sequentially dealt hands being aces, is 1/48,841.

A fair coin has an expectation of heads 1/2 of the time. After flipping heads once, a second flip has the same expectation as the first. -After you have flipped heads twice-, however, you must revise the probability a posteriori.

Say you ask a hold 'em player to select at random a single hand from her entire history, ad infinitum. If that single hand is aces, what are the chances the next is aces?

Now say you ask that player to select at random groups of two sequentially dealt hands from her entire history, ad infinitum. If she tells you -at least one- hand is aces, what are the chances the other is aces?

Are the answers the same?

hmm...interesting, i was dealt aces 3 times in a row in a live cash gme once

the way that mason describes it when he says its much less than 40k to one is just him saying that the odds of getting aces the next hand when you currently have them is the same. i dont think it takes a mathematical genius to figure this out; when people ask the question "what are the odds of AA twice in a row" i think the parameters are pretty well defined already, that is, what are the odds of AA once then AA again, which is why OP and microbob had issues originally.

The question should be assumed to be framed within her circumstances, which is either that she was dealt exactly two hands or that she was dealt any other number of hands >2 as a part of a session. The most improbable and amazing scenario is the former, therefore most likely she was asking "Given that I was just dealt aces on the last hand, what are the odds I would be dealt them again on this hand?"

Additionally, the other relevant possible question is much more specific and most people would, imo, instinctively phrase the question to include the fact that only x number of hands were dealt if that number were any small number, especially if that number were two.

The low probability of someone being dealt aces on their first two hands AND excluding that additional piece of information from the question make Mason's assumption so overwhelmingly unlikely to be incorrect that it is hardly worth pointing out.

Finally, since Mason's assumption is correct so frequently, and the math is so frequently misunderstood, pointing out the common error is still relevant even if wrong about the circumstances, imo.

Ubercuber--

That strikes me as very improbable. If I see a frog fall from the sky and think "what are the odds of that?", it's pretty obvious what's going on. The event I'm surprised by is the frog falling from the sky. Now, if I see a frog falling from the sky today and tomorrow, and again I ask myself what the odds are, it's still pretty clear what I'm surprised about: frogs falling from the sky twice in two days. Perhaps it's some variant: frogs on two consecutive days or something like that. It's even conceivable that what I do care about is the conditional probability of a frog tomorrow given a frog today.

But just because event E is a necessary condition for E', or even a necessary condition to make you care about E', doesn't mean that when E' happens, and you wonder what the probability of "that" is, that you aren't referring to E'.

Of course there's a complex causal process leading to the woman asking what the probability of "that" is. And we can argue about where the clearest "joints in nature" here are. But this is, at the bottom, largely a psychological question. The woman got aces twice in a row, and I'd bet a whole lot of money that the "that" she asked about was the aces twice in a row. Mason's taken plenty of courses in statistics, but that doesn't mean he gets to choose what other people's pronouns refer to.

All my best,

--Nate

This.^^
"Aces twice in a row" is a specific event (one), with a precise probability, and only one answer to the question is possible. The woman absolutely was not asking about the probability of "being dealt a pair of Aces". Twisting the question to imply that is nonsensical, and in the circumstances it would require a belief in the gambler's fallacy to even ask it.

It might be helpful to look at the way in which a mathematical probability is determined. Mathematically, you simply divide the number of outcomes that meet your requirements by the total number of possible outcomes. This applies to coin flipping as well as getting AA. First, however, it's necessary to distinguish between dependent events and statistacally independent events. In both the case of the coin and the cards, the 2nd flip and the 2nd deal depend on the outcome of the first event, but only in the sense of whether it happened or not. If it happened (the coin came up heads or you were dealt AA) then the second event becomes possible. If it didn't happen, the second event becomes meaningless since something can't happen twice in a row until it has happened once. However, mathematically, these two events remain statistically independent of each other because the probability of the outcome of one has no effect on the probability of the outcome of the other. If the coin came up tails, you could still flip it a 2nd time and the probability of it coming up heads would remain the same.

Flip the coin and there is 1 way to meet your requirement of heads, but 2 total possible outcomes, heads or tails (leaners don't count). So the probability of it being heads is expressed as 1/2 or 50%. If you want to determine the probability of it coming up heads twice in a row, you apply the same method. The 2nd flip, by itself, has the same probability as the 1st flip, for the same reason. However, when you look at the 2 flips as a single event, things change. There is still only 1 outcome that meets your requirements..heads, heads. But now there are 4 total possible outcomes..heads,heads..heads,tails..tails, heads..tails, tails. So the probability of heads twice in a row becomes 1/4 or 25%.

The same principle applies to the cards. On the 1st hand, there are 6 two card combinations of AA which meet your requirements (we can ignore order since we don't care in what order we received the cards) and 1,326 possible two card combinations (again ignoring order). So, the probability of the 1st pair of aces is 6/1,326=1/221. Just as with the coin flip, the probability of the 2nd AA is also 1/221 since, like the coin flips, the probabilty of each seperate event has no effect on the probability of the other event. However, just as with the coin, when you are determining the probability of AA twice in a row, things change. There is still only 1 outcome that meets your requirements, AA,AA. However, just as with the coin, the number of total possible outcomes changes. They are too numerous to list, but fortunately you can just multiply 221*221 and get the total number of possible outcomes as 48,441. So, applying the same basic principle, divide the number of outcomes that meet your requrements by the total number of possible outcomes and you get 1/48,841.

There are two other things worth mentioning. First, it's not necessary for the girl this whole debate is about to have the hands dealt to her at the same time. She could play one hand of Hold Em, be dealt AA and not play another hand of poker for 10 years. If, when she sat down to play again, she was dealt AA, that would be considered twice in a row in her frame of reference, which is the only one that counts in answering her question. Second, the answer 1/48,841 says nothing about when the "two in a row" will occur. It could be the first 2 hands a person is dealt or you could go 48,841 deals without it happening at all. The probability only says that the more hands you deal, the closer and closer your actual results will get to 1/48,841. The only exception, ironically, is online poker. You could program the game software to make sure that each player gets AA twice in a row exactly once out of every 44,841 deals. But we all know that can't happen ( ) because the sites assure us that they use something called a "random number generator" which accurately reproduces the random distribution of cards you would expect from a properly shuffled deck of 52 cards.

Needless to say, the probability that my typing still sucks remains at 100%. On line 6 of paragraph 1, I incorrectly spelled statistically as statistacally and on line 15 of paragraph 3, I incorrectly typed 48,841 as 48,441.

I was delt AA 3 hands in a row in a cash game 2.50 5.00 Nlh won all three thet wernt big pots though