Probability of 2 players being dealt the same 'hand' (2 cards), texas holdem, head to head. This happened to me in poker the other week. 2 players going head to head getting the same 2 non-pair cards (of different suits). E.g. a 4 and a 6 of any suit, where the other player gets a 4 and a 6 of any suit. 
I wish to calculate the probability of this occurring, but I'm failing. Using 'combination' and 'counting method' doesn't agree, hence I know I'm making errors. 
With 52 cards in a pack, and 2 cards dealt to each player, I calculated the probability as follows.
Combination method: p=success/total
p=(52C2 x 3C1 x 3C1)/(52C2 x 50C2)
Counting Method: p=success/total
52x51x3x3 =  successes, but two of these are the same as it doesnt matter which card is first or second. hence divide this by 2
52x51x50x49 = total
hence p = (52x51x3x3)/(2x52x51x50x49)
Can someone help me with the answer, and explain my error in thinking please?
Warm Regards.
 A: Based on @remy-jurriens answer, just corrected.
Player1 has a pair and Player2 has same pair,
p(pair_pair) = (3/51)(2/50)(1/49) = 6/(51*50*49)
Player1 has a non-pair and Player2 has same hand,
p(non_pairs) = (48/51)(6/50)(3/49) = 864/(51*50*49)
Chance to get same Texas Holdem hand Heads-Up:
p = p(pair_pair) + p(non_pairs) = (864+6)/(51*50*49) = 0.006962785114045618
or 0.7% - same result
A: I think what you need to take into account is that probabilities depend on whether player 1 (P1) is dealt a pair (p=3/51) or not (p=48/51). If I'm not mistaken, a two-stage probability tree might do the job here:
P1 gets a pair (p=3/51): the chance of P2 getting the same two cards are simply p=(2/50)(1/49). Combined probability: (3/51)((2/50)*(1/49)) = 3/(51*50*49)
P1 doesn't have a pair (p=48/51): the chance of P2 getting the same two cards are p=(6/50), to match either of P1's cards, times p=(2/49), to match the second, p=(6/50)(2/49). Combined probability: (48/51)((6/50)*(2/49)) = 18/(51*50*49)
Add those, e voila: p = (3/(51*50*49)) + (18/(51*50*49)) = 435/62475 = .70%
