I've been playing hold'em on SWC lately, and have noticed that two players (across different tables with different numbers of players) get identical starting hands quite a bit. Like, more than once a night when only playing maybe 100 hands a night. This strikes me as very strange.
Well on two different tables I think you have to take each table as a individual, independent scenario. It is unlikely to happen but it happens.
So for a single specific hand the odds are:
(2 / 52) * (1 / 51) = (2 / 2652) => (1 / 1326)
Which in percentage gives you a probability of 0.0754% for this event to happen once.
From here you can multiple this by the number of times the event/same hands occurred.
As krikara said this is very similar to the birthday problem. Which means the more hands you get the higher the probability that this occurs. It is easier to think of the probability of NOT having two identical hands.
To simplify, lets suppose you are playing 5 tables simultaneously and each table has 6 people. There are 52 choose 2 ways to make a two card hand from a deck, which is 1326.
Consider first round (hand you play)
The first table gets 6 different hands.
The second table then has a probability of (1320/1326)^6, of NOT getting the previous 6.
The third table then has a probability of (1314/1326)^6, of NOT getting the previous 12.
The fourth, (1308/1326)^6, of NOT getting the previous 18.
The fifth, (1302/1326)^6, of NOT getting the previous 24.
Multiply these together and you get the probability of two players NOT getting an identical hand in a round, which is ~.76. Thus the probability of two players getting an identical hand on each round is ~.24.
Now if you play 100 rounds the probability of two people NOT getting an identical hand is .76^100, which is 1.2^-12, or virtually 0.
So if you play 100 rounds, there WILL be identical hands occurring, now whether or not you see them is a different story, but still probable.
Also, you can adjust this problem for different number of tables played and players at each table.