But think about what someone means when they say in amazement: "I got dealt AA twice in a row today". Do they mean "at some point in the session I wondered about the next 2 hands and then they were aces" or do they mean "at some point in the session two of them in a row were aces"? This is what has been solved in the last few posts, so therefore the longer the session, the higher the odds of seeing AA twice. If you play a 2-hand session, the number in the OP (40k:1 I think it was) is correct. But the more you play the further it is away from 40k:1 and the closer it is to 1.

the odds of being dealt AA twice in a row are equal to the chances of someone wishing you the best twice in a row.

If the odds against having AA on any given hand are 220:1 or probability = 1/221, how could the probablity that you would have aces two hands in a row be anything else but 1/48841 or 48840:1?

Ok. I see the semantics of it all. In my lifetime I have been dealt aces twice in a row, so the chance of me getting aces twice in a row in my life is 1, and even if I had never been dealt aces twice in a row, the probability is very close to 1 that it will happen sometime if I play a reasonable number of hands. And I also see that the odds against it happening sometime during a session are much better than 40000:1.

And if I get aces, then the chance that I will get aces on the very next hand is 1/221.

And the probability I will get aces on the next two hands is 1/48841. Which I think the question implied.

And I sorta get the Monte Hall problem, but only vaguely. I don't think it has much to do with this.

I think that is an error, and the source of confusion. Her question isn't in a vacuum about any two hands of poker. I interpret her question as, "After holding AA on my last hand, what are the odds that my next hand will be AA also?"

Yes if I asked what the odds of a flipping a fair coin twice and it landing on heads both times the answer is clearly 1:2.

Because clearly the question assumes that the first flip was already heads and is only inquiring as to the second flip also being heads. WTF is so confusing about this really?

I assume you are joking. Otherwise, the odds against flipping a fair coin twice and it landing heads twice is 3:1.

And do you mean 'odds' as 'probability', or 'odds' as 'odds against/for'?

Classic Mason.

Mathematically speaking, because you have no idea what question the woman was actually asking, you have no idea if the answer is right or wrong.

In real life she was given exactly the answer she wanted - the most improbable and amazing, although correct under the right assumptions, one.

Not sure how you're defining the problem, but do you mean the odds of someone saying it after they've already said it once or just at random 2x in a row. During good weather or bad weather (etc)?


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Cards, Coins, Dice, Large Spinning Wheels with red and black spaces, etc all have no memory which is key to the discussion.

And yes it is symantics in defining the problem.

The chance of getting dealt A's after having just had A's is 1 in 221, same as it is for being dealt A's after getting 96o. The issue here is that you are defining the past which has already happened and thus has absolutely no outcome on the future. Therefore the Chances of being dealt A's AFTER already being dealt A's is 1 in 221. This is the same "logic" that people use to bet roulette in the wait for 3 reds in a row then bet Black and double your bet beacuse the odds of 4 red in a row is so low and 5 is downright crazy, and 6 never happens. Well we all know (or I hope we do) that the past has no impact and the chance of black is the same as it always is every single spin of the wheel (or flip of the coin, or toss of the dice). Same as saying I have rolled a 6 twice, what's the odds I will roll another 6. Obviously 1 in 6.

Now absent of the logical test of Current Hand=AA being already true, the chance of AA being dealt twice in row is 40k+ to 1, same as the chance of the next three dice rolls being a six are about 1 in 216.

So it is not trivial, not a game, just a matter of determining what has already happened and thus is known, versus what is yet to happen and does not care what has happened in the past.

Stungar's calculations are absoLOUtely correct! 1/221*1/221 = 1/48,841 Anyone who thinks differently doesn't know much about probabilities, algebra, or both. A more interesting question is what are the odds of any 2 players in a 9 handed game of NL Hold Em being dealt AA in the same hand? I saw this 2 nights in a row online and got curious. I like algebra so I figured it out. If you come up with an answer, you should show the math behind it. If no one gets it, I'll put the answer up tomorrow. In both cases, the odds apply to any 2 pocket pairs. Of course, there is also the question of the odds of 2 players being dealt AA in the same hand in a heads up game.

I think it would be the odds of one person getting dealt aces and the odds of the second person getting dealt aces factoring in removal of the first two aces (hypergeometric i think). so (1/221) * (1/how many hand combinations are left after removing two aces from the deck). Sorry I'm to lazy to do the actual math.

Disregard my last post. I found another way to look at it.

This is a hypergeometric. The probability of the first person getting dealt aces is (4/52)*(3/51). The proability of the second person getting aces is then (2/50)*(1/49). Multiplying these two will give you the probability of those two getting aces. However you have to multiply that by 9 choose 2 to account for the different combos of how the two aces could be dealt around the table, since the first product would simply give you the probability of two specific people getting dealt aces.

My final answer was .00026 or .026%. I'm pretty confident in the method though my actual arithmetic skills suck so that number may or may not be off.

[ ] delt

[x] dealt

mathematical breakdown would be:
1/{[4/52*3/51]^2} = 48,841
ie. the odds of any ace as the first card * the odds of getting an ace the second time. squared to say that the odds of it happening back to back

Wow. That merits a s-l-o-w golf clap.

Let's quote exactly what you said:

It's pretty clear to me, and I wager to most. "Get dealt aces two hands in a row". She wasn't wondering about the odds of getting aces, taking into account she already had aces and that you should discount these odds. Aces. Two. Hands. In. A. Row. Easy. Obvious. Ignoring the first pair of aces is the same as redefining the problem to say "get dealt aces after any random hand". That's not what she was asking about, obviously. Of course, if you don't understand what she's asking, or you redefine it under your own terms, you're bound to get it wrong.

Greetings, sports fans! LargeLouster here again (but plz, call me Lou, e1 does ^__^). The odds of any 2 players being dealt AA in the same hand of a 9 player NL Hold Em game are 52*51/2/6/9*50*49/2/8=3,760.06 and is usually expressed as 3,759 to 1. In a heads up game, the odds of both players being dealt AA in the same hand are, as junwagh correctly pointed out yesterday, 4/52*3/51*2/50*1/49=1/270,725 and is usually expressed as 270,724 to 1. Btw, at bodog, where I usually play, there are 2 chances of AA actually holding up and winning the hand, slim and none; ask anyone who plays there.

There are two seperate issues here.

1. What is the porbability of getting a pair of aces GIVEN THAT I was just dealt a pair of aces in the previous hand?

and

2. What is the probability of getting dealt aces on hand A AND THEN hand B.
They yield different answers.
Since the cards have no memory, I am likely to fall on the side of option 2 as the answer.

Exactly. When someone asks you the odds of hitting red 5 times in a row at roulette, who ever thinks of it as ~50%, because we are ignoring the first 4 times? No one, that's who.

I think he was trying to say that two hands never exist in isolation, and the size of the hand series determines the answer. But the answer was not wrong. The only reasonable way to answer that question is with the frequency it will occur on average.

To illustrate this, let's look at some progressions.

If only two hands are played, the odds they will both be AA is ~48840:1

If 100 hands are played, the odds that two consecutive ones are AA is ~493:1

If 34000 hands are played, the odds that two consecutive ones are AA is ~1:1 (50% chance).

If 230,000 hands are played, the probability that two consecutive ones are AA is ~99%.

But the key question is, how often will this event occur on average? And the answer is exactly the same as the odds of it occuring on the next two hands. It will happen on average once for every ~48,840 hands played.

So I say the answer was precisely right. It is in fact the average frequency that this event will occur.

Over a larger sample the probability doesn't change, the expected frequency changes.

I have concluded this is all a big joke.

So then the odds of getting dealt exactly the same two cards (same rank and suit) twice in a row must be much greater than 40K to 1. Interestingly this has happened to me 3 times online and never in a live game. I have played many more hands live than online.

the odds of calling twice in a row aces for the next two hands are 40k+:1.

the odds of getting aces immediately after getting aces is 221:1.

221:1 is the more realistic representation of what is going on here. No one is surprised by getting aces once. Aces once is basically a precondition for being surprised by aces at all.

You figure that you get aces regularly, so you are taking the set of all the times you get aces, and trying to figure out how often you get it again. Its 221:1. The perception problem comes in because of the frequency.

Now, lets say winning the lotto has odds like 100,000,000:1. Now I can say that winning the lotto twice genuinely has odds around 1E16:1. Winning the lotto even once is an event that will most likely never happen and is rarely attempted. I might only even attempt to hit the lotto twice in my life, so I personally really do have a 1E16:1 shot at getting it twice. I will be very surprised both times I hit.

But with aces twice, you figure you play thousands and thousands of hands... you get aces with high frequency (221:1), so when you say, "what are the odds of me getting aces twice in a row", you really, really, really mean "what are the odds that I just got dealt aces, given that I had just previously been dealt aces"

and that figure is 221:1

You're implying a dependent event when you say "given that..." and there is no dependency at all. That terminology has a specific meaning in statistics, and you've misused it and changed the question. Obviously there is always a 1/221 chance (which is 220:1 btw not 221:1) of any deal being AA, the previous deal is totally irrelevant to that. So describing the question that way isn't valid. The question was about the chance of two consecutive deals being AA and it doesn't make sense interpreted any other way, so I don't agree with your explanation. The event being asked about is specifically: [two consecutive deals of AA], not two separate events as you implied. The right answer is the one already given, i.e. ~48K:1. That is the average frequency the event will occur.

"Being dealt AA" is one event.
"Being dealt two consecutive AA" is a different event.

The question was about the latter.