Hello,
First i would like to say that i really enjoy the forums,have read them for many years but just recently decided to make an account for few Pm's.
I am really terrible at math so i hope some of the math gurus here would be kind to answer a few questions.Last 6-7 months i play only spin and go,mainly 2-4 tables,rarely up to 6.Something that really made me wonder was the amount of times i get the exact same hand ,same suits on 2 tables at the same time,and one time the exact hand on 3 tables while playing 5.Can someone tell me what are the odds of getting the exact same hand on 2 tables when you are 2,3 or 4 tabling?I imagine the odds to be millions to one if not more yet that have happened to me more than 35 times.I would really appreciate if someone do the math and answers .
Greetings

Welcome to the Probability forum. Glad you are interested enough in your question to post it here. Poker math can be done by anyone; it does not require any special knowledge or secret password.

How about if we make your question even simpler? You will then be able to tackle it yourself and see how the more complex analogs can be tackled.

What if poker hands (hole cards) consisted of exactly one card? Then what is the probability of getting the exact same hand (one card) on 2 tables when 2-tabling, say on the very next deal?

Once you have that answer you can branch off into other related questions. What would be the probability of getting the exact same hand (one card) on 2 tables when 2-tabling at least once in the next 10 hands? Of course, over more deals something like that is more likely to occur.

What is the probability of getting the exact same hand (one card) on 2 tables when 2-tabling at least 3 times in the next 100 deals? These types of questions are fairly straightforward to answer and require no special knowledge.

Or you could pretty easily figure out what would be the probability of getting the exact same hand (one card) on 2 tables when 3-tabling in the very next deal. Again, these types of questions require some care but are pretty straightforward and are important to understand in many non-poker environments in real life.

With the knowledge and confidence of answering the first question above concerning one-card hands, you can ask yourself the probability of getting the exact same hand (two cards) in 2 tables when 2-tabling in the very next deal. This might require some back-and-forth but it is pretty straightforward and eminently understandable. Basically ask yourself how many two-card poker hands are there.

Then you can extend to 3-tabling or 4-tabling pretty easily. And/or extend the question to how likely is the occurrence over multiple deals such as 10 deals or 100 deals or 1000 deals, etc. Very soon you will build up enough confidence and understanding to answer your original questions.

Numeracy vs. innumeracy is a question we all must face. Many people had bad math experiences somewhere in their past and now believe that math is closed off to them. But nothing could be further from the truth. Anybody who plays poker knows and understands a great deal of math even if they don't know it!

If OP is curious enough to post the question in the first place, I hope he is curious enough to try to answer some of the questions I posed above, starting with the following:

What if poker hands (hole cards) consisted of exactly one card? Then what is the probability of getting the exact same hand (one card) on 2 tables when 2-tabling on the very next deal?

Thank you very much for your time.
Once again i have to say that i have very very low understanding of math,probabilities and estimations.I have been playing online poker for little under 10 years and i have never really gotten into the concrete math approach.
The reason i decided to post in that specific sub forum is that i am sure there are enough people here who can answer my questions,and i have asked them numerous time in few forums, also to few poker pro's but still havent got answer.Main reason to wonder is that i really feel that such coincidences should be really rare and still i get them so often.With years of playing you kinda start to recognize same patterns happening over and over again,stuff that should be absolutely random but happen all the time.I would really appreciate if someone takes the time and give answers.I am completely honest when i say that have close to no idea how to answer any of your questions .I consider myself a failed pro wannabe,i am familiar with concepts like pot odds,fold equity,ranges,gto,standard stuff yet i have no clue when it comes to calculating probabilities.
Greetings

There are 1326 starting hands. The chance to see the same one dealt on two simultaneous deals is 1/1326. And if we allow for deal overlap before and after (since tables aren't synchronized) then double that to 1/663. Adding another table gives you 3 ways to match, 6 with the deal overlap, so with three tables we're at about a 1 in 110 chance for a match.

If you interpret the question as” what is the probability that at some specific point in time while playing I get two identical hands in a row?”, then I don’t think the doubling is appropriate.

With two tables that alternate evenly, every hand has a chance to match with two other hands that appear while it is showing. If they were synchronized then it only has one chance to match. In real play with random timing, I believe we end up at least 1.5.

I would think you can approximate very well by Poisson. For example 4 tables would be n=C(4,2)=6 and the probability that a pair (of tables) is dealt a matching hand is 1/1326. Then lambda=n*p=6/1326 and P(1 match)=exp(-lambda)*lambda ~ 0.45% (almost the same prob as getting dealt pocket aces preflop) . Actually it is the same problem as the birthday problem as I understand what OP is asking.

Hello,
here are some of the examples for the questions ( sorry cant make the screens to automatically appear) :

http://imgur.com/wRdAtUG

http://imgur.com/NDC11Ak

http://imgur.com/XxNewDc

http://imgur.com/zwhzxp4

http://imgur.com/TKGUPIC

http://imgur.com/A8l6L68

NewOldGuy ,suggested that odds of getting the above screens are 1/663 ,do i understand that correctly?
Also that one is 3 identical hands on 3 tables while 5 tabling :

http://imgur.com/Cihv43E

This is same hand,two hands in a row :

http://imgur.com/59IhQP1
Thank you for your replies.
Greetings

There's just not anything unusual about these. If you average 100 hands per hour, per table, and you play several hours a day, you should expect to see this every day or two.

To your question about the same hole cards twice in a row on the same table, that happens 1 in 1326 hands dealt to you, on average.

This is something that people like the OP who are for the first time thinking about the math for "rare" events should learn right away. If you play enough (and usually enough is a relatively small amount) the chance of seeing one of these "rare" events at some point is pretty much 100%.

Thanks for the replies.If really odds of getting the same hand on 2 tables at the same time is 1/1326 that i was totally wrong in my estimations.As i said i have played online poker with maybe 10 month brakes since 2006 and i still find these kind of stuff really strange even after million hands.
Greetings

I calculated the exact answer and it confirms that your 0.45% was a good approximation.

P(at least one group of at least two same hands) =

C(4,2)/1326 - 2*4/1326^2 + 3/1326^3 - 1325*C(4,2)/2/1326^3

= 0.45186333 %

Funny, that's about 1 in 221.3, so nearly the same as being dealt AA at one table.

A similar, but harder problem is here.

Next, I'll try the version of the problem NewOldGuy considered, which to me does seem more realistic.

If we approximate reality by saying each table alternates and that there is overlap, then I get the following.

Two tables: call them A and B and suppose A gets the 1st hand.

Consider the overlap during B's first hand (which I'll call B1). Hand B1 will be occurring at the same time as hands A1 & B2.
Hand progression = A1, B1, A2, ...

Let p = 1/1326
Let q = 1-p

Then P(at least one simultaneous match during B1) = p + qp = 1 - q^2 =~ 1 in 663.25

That's only from the perspective of a hand that isn't the beginning or end of a session, but that's most hands so that's fine.

Three tables: Call them A, B and C, and consider the sample of hands (A1, B1, C1, A2, B2)

This is where I part ways with NewOldGuy: whereas he said the overlap allows for 6 ways to match, I count 7.

A1 can match with B1 or C1 = 2 ways
B1 can match with C1 or A2 = 2 ways
C1 can match with A2 or B2 = 2 ways
A2 can match with B2 = 1 way

Therefore, a reasonable (and, it turns out, accurate) estimate of P(at least one simultaneous match during C1) is 7/1326.

To calculate it exactly I did the following. Let r = 1-2p

answer = p + 2pq(1 + r + r^2) = p + q(1 - r^3) =~ 1 in 189.8

For 4 tables I'll just count the possible matches to get the estimate. It helps me visually to list the hand progression.

A1, B1, C1, D1, A2, B2, C2

4(3)+2+1 = 15, so the estimate is 15/1326

And now I see a formula for the #ways to match with n tables: (n P 2) + C(n-1, 2)

Probability for n tables: [1-(n-2)p]*[1-(1-(n-1)p)ⁿ] + Sum_{k=1 to n-2} kp[1-(k-1)p]

So for 4 tables that's p + 2p(1-p) + (1-2p)[1-(1-3p)^4] =~ 1.1267% or about 1 in 88.756
By comparison, 15p = 1 in 88.4

For 5 tables my formula says ~1.948% or 1 in 51.34
Compare to 26p = 1 in 51

There's an easy way to remember my formula from last post (for determining the #'s 7, 15, 26, 40, ...)
It's like n! except after one multiplication, the rest of the * signs become + signs.
15 = 4*3+2+1
26 = 5*4+3+2+1

Edit: it's OEIS A005449, "second pentagonal numbers".

Thank you very much !

Nice work deriving that.