# Probability of 2 players being dealt the same 'hand' (2 cards), texas holdem, head to head.

by OhStatistics   Last Updated June 28, 2017 09:19 AM

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.

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