Been trying to work out the odds of this but it happened the other night.

The answer depends on what assumptions are made to answer it. Because I'm feeling lazy, I'll make the assumptions that make answering your question the easiest: It's virtually 100% guaranteed to happen sooner or later to someone somewhere.

Same table. One after another.

That's not enough to change my answer. In theory, your session at that table could anywhere from 3 hands long to a zillion. Are just 2 players playing or 10? Is this hold'em? Is it the same player getting dealt 99 or different players at the same table? All these variables alter the answer. After my 1st answer, this should not have to be explained to you.

I understand probability. Unfortunately for you I am not stupid and do not take this kind of tone well. I should have said this was hold'em on a 9 handed table, myself being the player. Just because sometimes people may post things that lack clarity does not give you the right to talk to people this way.

Your "intelligence" is clearly fogged over by epic issues with condescension.

Do not reply to my posts ever again. You are ignored.

1/221 chance of being dealt 9's, plus a 1/8 chance of making a set is 0.0565% I think.

Three times in a row would be 0.0565^3, so 0.0001803%.

Not sure if that's right!

OP, I think you're getting misled by his tone. The problem here is that there are different answers depending on how you look at it. The 2 most common points of view might be

1. how often would this happen starting now, over the next 3 deals. This is obviously quite rare. It's not that hard to compute but there's not much point because of how meaningless it is.

2. what percentage of players who play X hands would experience this? The answer obviously varies with X and as X increases, it becomes more and more likely. And in fact, the liklihood increases more than people think, so that even rare events are relatively certain for high volume players.

Added to this, would you be less shocked if you got a set of 2s 3 hands in a row, or 6s? What if it was 222 then 333 then 444? KKK then JJJ then KKK again and so forth? There is no significance to the fact that the cards in this case were 9s.

Yeah, I realise the statistic is very meaningless.

And I also realise it depends on the sample size as to how many players you would expect to experience this situation over X hands.

I was just wondering, more specifically if this helps people out:

Hold 'Em,
9 Handed,
Odds of hitting a pair, then that pair again, the very next two hands (any combination)
Each of these times flopping a set.

Something along the lines of

Same pocket pair 3 times in a row and flopping a set every time: (1/17 * 1/7) * (1/17 * 1/7) * (1/17 * 1/7) = 1685159 to 1

Or roughly 0.6 times the odds of hitting a royal....I might be wrong but thats why I asked you guys!

Well, the probability of getting a particular pocket pair on any given hand is 1 in 221.

Then the probability to flop a set with it is 1 - (48/50)(47/49)(46/48) = 0.117551

So the probability of this exact event happening on a single given hand is:

(0.117551)/221 = 0.0005319 = 1 in 1880.

For it to be back to back to back in only 3 hands of play: (1/1880)^3 = 1 in 6.64 billion.

But, before we start saying, "zomg, poker is rigged!", we need to consider a couple of things. One, you play more than 3 hands of poker in your lifetime.

So in this case the probability nearly scales with the number of hands you have played since the event is a rare occurrence. Exact calculations may be done using a streak calculator like this one:

http://www.pulcinientertainment.com/...tor-enter.html

So if you have a sample of 1,000,000 hands, he chances of seeing exactly 99 3 times in a row and flopping a set 3 times would become 1 in 6649.

Now the other thing to consider is that you would have made this post with any pocket pair where you would have flopped a set 3 consecutive times. So considering this would increase the overall probability of this occurring by a factor of 13 [(1/17)*(1/221)*(1/221)*(0.117551)]^3. So for a sample of 3 hands, it would become 1 in 511 million. For a sample of a million hands, 1 in 511.5.

So this is still a rare occurrence and virtually all live poker players won't see it in their lifetime, but some of the online grinders will definitely be seeing it.

Awesome explaination. Im guessing you get a lot of people on hear who lack the fundamental understanding of probability, every reply seemed to include explanations as to not assume over a small sample. I realise it's the wrong wh to approach things but I wanted to just see the hard numbers

Sometimes. When dealing with rare events calculations there are often multiple ways to look at the problem and depending on which way you look at it, the answer can drastically change. When people ask questions like this, I try to fill in the missing details as I would think the OP would want it, or if I can't I'll ask the OP to clarify his or herself.

Not trying to be mean or offensive (just factual), but it is impossible for someone to "understand probability" and not know how to solve this problem. It is possible, but still relatively unlikely, for someone to understand probability and realize that there are many solutions to this problem and many (if not all) of them are meaningless without context.

You've assumed that a particular player is getting the hand 99, but with 9 players, someone being dealt 99 is much higher.

You've also assumed that the 99 hand is always getting to see a flop. Unless the player with 99 is in the BB, it is by no means assured that he will see a flop even if he does not bet. And then there is the possibility that, the action in front of him reaches the 4bet level, when you are not seeing a flop, because you will fold (or least should).

It's not so easy.

Yes, those facts are also true. It's never easy to include everything. The preflop fold issue makes each of these calculations table dependent.