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.

  • $\begingroup$ is non-pair important ? $\endgroup$
    – el Josso
    Jun 28, 2017 at 8:59
  • $\begingroup$ for the way I have asked the question, yes. But i would be interested to see answers incorporating the possibility of the pair as well. Thanks for the reply. $\endgroup$ Jun 28, 2017 at 13:45

2 Answers 2


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

  • $\begingroup$ Thanks for the replies. I see that I was using the wrong approach, and that probability tree and the multiplication rule were better suited to answer the question. Was there a way of using the combinations function? Or was I overcomplicating the situation? $\endgroup$ Jun 29, 2017 at 9:32
  • $\begingroup$ I believe trying to search 'combinations' function is over-complicated there. Probability tree is very simple and obvious: 3 to 48 to have a pair on 51 cards. Same for each tree node. $\endgroup$ Jun 29, 2017 at 10:05

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%

  • $\begingroup$ (6/50)(3/49) not 2/49 $\endgroup$ Jun 28, 2017 at 15:07
  • $\begingroup$ and (3/51)((2/50)*(1/49)) != 3/(51*50*49) :) $\endgroup$ Jun 28, 2017 at 15:13
  • $\begingroup$ Thank you for your corrections! And sorry for the sloppy typos. $\endgroup$ Jun 29, 2017 at 19:25

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