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I like to play TH poker. I usually play online. Usually I play on many tables and I noted that sometime hole cards are the same on different table. So i just wondering (as a math fun) how is the probability of hold the two hands of the same class? To explain better, we define hands as JsTd and JcTh as the same class hand, because thay have the same strong. 9c9h and 9d9s are the same class too, and so on. If at table $A$ i have two hole cards and at table $B$ other two hole cards, which is the probability of holding the same class two hole cards on two tables?

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Don't you also have to consider that JsTd is not the same as JcTc? Same suited hands are stronger. – Tim Seguine Jan 19 '13 at 11:36
of course, indeed in the example are all off suited. – emanuele Jan 19 '13 at 11:47
The probability wil depend on the type of hand you have; if you have a pocket pair, the chance that someone eles has the 'same' hand is much less than if you had non-pairing cards. – Elements in Space Jan 19 '13 at 11:52
I am sorry, you misundertood my questiion. I fixed the formulation. – emanuele Jan 19 '13 at 11:57
ok, then the number of hands dealt at each table is irrelevant, assuming the cards were fairly shuffled. And the comment from @UnkleRhaukus applies if you want to know after seeing your hand at one table, what are the odds that it is the same at the other table. – Tim Seguine Jan 19 '13 at 12:02
up vote 3 down vote accepted

There are $13$ different pairs that come in $6$ different suit combinations and $\binom{13}2=78$ different non-pairs that come in $4$ suited and $12$ offsuit varieties ($6\cdot13+(4+12)\cdot78=1326=\binom{52}2$).

The probability of getting a particular one of these classes twice is the square of the probability of getting it once, so we need to sum the squared probabilities:

$$ 13\left(\frac6{1326}\right)^2+78\left(\frac4{1326}\right)^2+78\left(\frac{12}{1326}\right)^2=\frac{83}{11271}\approx0.007364\;. $$

share|cite|improve this answer, in 100 hands, if i found for example 5 of that same hole cards on average, means that the number generator does not work properly? – emanuele Jan 19 '13 at 14:47
@emanuele: Yes, if that's the long-term average, something's wrong. However, if it only happens once that you get $5$ matches in $100$ hands, it might be OK, the probability for that to happen is about $1$ in $1000$. – joriki Jan 19 '13 at 16:21
Where is the term for the number of players at the table? – Elements in Space Jan 19 '13 at 16:22
@Unkle: There isn't one; as Tim pointed out in a comment under the question, the number of players is irrelevant here. – joriki Jan 19 '13 at 16:23
@joriki, ah. I see now (the question has changes slightly) that the player is at two tables and comparing their own hands. – Elements in Space Jan 19 '13 at 16:30

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