Nate,

Frogs falling from the sky on two days days are probably not much more surprising than falling on one... on day two I'm still surprised it's happening at all!

But back to cards, the fact that she was just dealt aces on the previous hand seems the most obvious reason that she was prompted her to ask the odds of getting them again the very next hand. Had she not been dealt aces on the previous hand and inquired the odds of getting them twice in a row I would feel differently.

She probably didn't know there was a difference, may very well have been satisfied with the answer, and perhaps cares very little for the reasoning, but the answer was, imo, not the best answer for the circumstances as they seem to have played out.

For two people who weren't there (and don't read minds), as you said, it is only speculation, but this rings of "OOH I just got em again, what are the odds?" to me. Operative word being "again", as in, it already happened once.



Absolutely? Even if you were there, how do you know absolutely to the point of all other views being nonsensical?

There is more than one answer because aces on the next two hands is much different than aces on the last two hands. When referencing the last two hands, the second pair is only noteworthy in the context of having already been dealt them on the previous hand. Ignoring that possible interpretation of the frame of reference for the question tosses out a perfectly reasonable interpretation.

Additionally, for aces twice in a row to have a precise probability with only one answer possible you need to be dealt a specific number of hands. If you are dealt two hands, you get one shot to get them back to back, if you are dealt four hands you can get them on the first two hands, the second two, or the third two, etc.

The odds of someone being dealt pocket aces in hold em on two consecutive hands is 1 in 48,841.

Of course the odds of being dealt aces after having gotten them the previous hand is 1 in 221.

1/221 * 1/221

vs

1/1 * 1/221

It's really not that complicated. Mason shouldn't have been so glib to dismiss the other person as "of course wrong" because to me it sounds like he was correctly answering the question the lady was intending to ask.

played a 45 man recently on Stars. Very first hand 2 guys get it all in and both flip over KK. Very next 3 (repeat 3!) hands in a row I get dealt KK.

you can have your 48k:1 but what about those apples??!!

ubercuber--

Suppose that, simultaneously, the Mets beat the Yankees and the White Sox beat the Cubs. You're watching both games on TV. Now, during the postgame celebration in Chicago, someone sneezes, setting off a complicated physical chain reaction that ultimately affects the atoms in your brain in a way that causes you to wonder what the odds of the parlay are. Here, again, the cause of your asking the question is pretty obvious, but would any reasonable person say that the right answer to your question is the probability of the Mets beating the Yankees (because you wouldn't have asked the question if the White Sox hadn't won).

Of course this is an absurd example, but the point should be simple and obvious: the cause and the content of a person's question are totally different things. Not keeping these things straight for oneself (as most people, myself included, do not always do well) can lead to fallacies. But saying that the woman wasn't really asking about aces two hands in a row is just a different kind of sloppy thinking.

All my best,

--Nate

You're right, I wasn't there, but it's pretty clear the event being questioned was getting Aces twice in a row.

The probability of Aces two hands in a row is always 1/(221)^2. Framing that in context of how many hands were dealt, or when you start counting, just means there are multiple opportunities for the event to occur, so you multiply the probability by the number of opportunities. It doesn't change the probability of the event itself at all.

This topic is being beat to death. Everyone is saying the exact same thing or remarking about how they got dealt aces twicce at bellagio or w/e. Please lock the thread already.

Odds of AA three times in a row are 1/221*1/221*1/221=1/10,793,861. At Poker Stars they have dealt about 10,000,000,000 hands, so it's probably happened about 1,000 times there.

The odds of 2 players getting KK in the same hand depend on how many players there are at the table and whether it has to be two specific players or it can be any two players. Nine handed, where it can be any two players, the odds are 52*51/2/6/9*50*49/2/8=1/3,760.06 or about 3,759 to 1. If it has to be two specific players (as it would be in heads up), the odds are 4/52*3/51*2/50*1/49=1/270,725 or 270,724 to 1. The larger number is exactly 9*8=72 times larger than the smaller number.

The odds of getting KK three times in row are 1/221*1/221*1/221=1/10,793,861 or 10,793,861 to 1. As I just said in the post above this one, at Poker Stars they have dealt about 10,000,000,000 hands so it's probably happened about 1,000 times.

Sorry, but there are actually three separate issues. The two you mention plus will Mason ever make a post agreeing with anyone? Just because somebody doesn't ask a perfect question doesn't mean the vast majority of us realise what the person is actually asking. My grammar may not be perfect but I'm fairly certain the majority of people understand my post.

thank you. please advise odds of being dealt KK 3 hands in a row straight after two players held KK in first 4 hands of a tournament.

Regards, Sqwerty

....and while we're at it.....can someone please confirm the odds of the following happening because I experienced it in a LIVE MTT and it was absolutely freaky to all that witnessed it.

2 players go head to head in a pot. One player has KK the other has AQ. Can't remember the suits and the cards exactly because it happened a while back but the flop comes (say) KhJs3d, turn is 4c.

Anyway.....the result of the hand is unimportant.

BUT.......VERY NEXT HAND......cards have been washed and shuffled....same 2 players are at it again but this time the hole cards are reversed...ie the guy who had KK holds AQ and vice versa.....THEN....the flop comes in EXACTLY the same order, EXACTLY the same cards (including suits) ...ie. KhJs3d.....AND the turn is also identical ie. 4c, as the previous hand.

Keeping in mind we were 8 handed here, now what on earth are the odds of THAT happening ???.....must be over 1000:1 !

ps: please no responses saying 'hh or never happened'....because it did.

Imagine someone buys a lottery ticket only once in their entire life. What do you know, it's a winner! Amazing! Now imagine someone buys millions of lottery tickets. Are you amazed that one of the tickets is a winner?

can anyone tell me how i could be like mason but not get punched out everyday? this looks too fun if possible.

rogged

OBViously

hobokes--

How about we tell 25 statisticians, 25 analytic philosophers with central interests in rationality and statistical reasoning, 25 mathematicians, and 25 expert poker players all the information that's in this thread, ask them whether "over 40k-1" is "obviously wrong," and bet $10K on who gets the majority?

Yes, this would be hard to pull off, but it might be possible, and I'm entirely willing to bet on anything similar that would be logistically feasible.

All my best,

--Nate

I just want to go on Let's Make A Deal and win a God damn goat. I have some vines growing out of control at the top of my property and goats are the answer.

i've gotten aa 3 x in a row before.. whats the odds of that?

mike

1/221*1/221*1/221=1/10,793,861

Sqwerty,

This is a little tricky because ur not specific enuff about the two players who held KK in the 1st 4 hands and the number of players at the table. So, I'm gonna give u the answer if it could be any two of nine players on each of the four hands. Odds are 52*51/2/6/9*50*49/2/8=1/3,760.06^4=1/199,884,491,835,615.00. Odds of u getting KK 3 times in a row are 1/221*1/221*1/221=1/10,793,861. This is where ur question gets a bit murky again. Did u mean u got KK 3 times in a row on the 5th, 6th and 7th hands of the tournament and are therefore asking about the odds of 7 specific hands in a row or did u mean u got KK 3 times in a row sometime later in the tournament. In the first case, the odds are 1/199,884,491,835,615.00*1/10,793,861=1/2,157,525,420,929,260,000,000.00. In the second case odds are 1/221*1/221*1/221=1/10,793,861.

Since I know my typing sucks, I'm not gonna bother to proof read this for errors since I never find them all. I'll just accept what my girlfriend says. "There are no bad orgasms, only bad typists."

Regards, LargeLouster

Odds of KK are 1/(52*51/2/6/8)=1/27.625. Odds of AQ are 1/(50*49/2/6/7)=29.16666667. Since u don't seem concerned with the order of the 1st flop odds are 1/(3/47*2/46*1/45)=1/56,752.5. Odds of turn card are 1/(1/44)=44.

Product is 1/2,011,994,359.60494.

Odds of guy who had KK getting AQ on 2nd hand are same as odds of his KK on 1st hand cuz ur just figuring odds of two cards of a certain rank. Same applies to odds of guy who got AQ getting KK on 2nd hand. The flop odds change cuz u said they have to come in a specific order and become 1/(1/47*1/46*1/45)=1/97,290. Turn odds remain the same since there is still only one card that meets ur requirements and are 1/(1/44)=44.

Product is 3,449,133,187.50

So, the overall odds are 1/2,011,994,359.60494*1/3,449,133,187.50=1/6,939,636,518,776,210,000. I'll be honest, I'm not 100% certain this is right cuz it's easy to make an error in this many calculations, but for sure it's greater than 1,000:1 !

Regards, LargeLouster

oops, made an error there, should have been 48,47,46 and then 45..sorry, ill recalculate it tomorrow unless u wanna do it

what are the "odds" of on a single hand having 1 person dealt AA other KK QQ JJ 1010... as seen on my screen on this site

1/221 * 1/221 = Odds of getting back to back aces

If you have an infinite number of opportunities at the first pair of aces, the first 1/221 in that equation can't be correct.

lol @ admitting the problem is ambiguous and then authoritatively dismissing a proposed (and valid, depending on what the real problem is) solution as "Wrong."