My question is the following :
What is the probability of facing a pocket pair (preflop phase) with an off-suit hand?

I made some calculations considering a single adversary. I display them here.

Initial problem data :

Only two players are considered.

  • Player 1 receive two offsuit cards (this is important).
  • Therefore the remaining deck of cards contains 50 cards as they remain:
    • 11 card values ​​with 4 suits. So 44 cards.
    • 2 card values ​​with 3 suits. So 6 cards.

Given these data, the following events can be considered :
enter image description here

Knowing that :
enter image description here

Direct calculation of the probability of event A :

enter image description here

This calculation can be verified by determining the complementary case :

enter image description here

Attempt to calculate with two opposing players

enter image description here

Main question :

The simple addition of an opposing player seems to make this calculation much more complicated (or so I'm doing it wrong). Moreover, the calculated probability decreases whereas intuitively it should increase.

Would there exist a calculation formula making it possible to obtain "simply" the result for a number n of players ?

Temporary Answer :

Thanks to BowlOfRed's formula, we can have an approximate but nevertheless precise value of the answer to my question. Moreover, we can perform a relatively reliable linear regression on these values to obtain the following formula :

Probability = 4 x "number of players" + 3

We get the following table :

enter image description here

If we can see that the relative error is not negligible on a certain line, it remains very acceptable for a mental calculation (which is the goal) and the rapid obtaining of a useful value.

I am still working on a numerical calculation method. I will post the result once the work is finished.
Do not hesitate if you notice any calculation errors or for any other remarks or questions.

  • You say that P(pair) + P(offsuit hand) = 1? What about non-pair suited hands? Where does a hand with 2 spades fit into that probability?
    – BowlOfRed
    Commented Jan 16, 2022 at 20:01
  • Thank you for your comment which allowed me to put my finger on a translation error I made. I will soon correct the post with the various advances that I have been able to make on the subject.
    – Doedalos
    Commented Jan 17, 2022 at 19:14
  • Just to be specific, do you intend that player one's offsuit cards must not be paired as well?
    – BowlOfRed
    Commented Feb 1, 2022 at 3:19
  • @BowlOfRed. Absolutely.
    – Doedalos
    Commented Feb 4, 2022 at 10:39

1 Answer 1


I'm assuming here that the offsuit hand of player one also cannot be a pair.

For a full deck, you have 3/51 chances of pairing your hand with the second card. This is the same as the the 78 pair combinations out of the 1326 combinations of 2 cards (~0.05882).

If player one removes a non-pair (suited or offsuit doesn't matter), then there are now only 72 pair combinations out of 1225 combinations for the remaining 50 cards. Player two in this situation has a ~0.05877 chance of a pair.

For all remaining players there is some dependence on the previous draws, but for the first few players, the dependence is small. You won't be far wrong if you assume them to be independent. In which case the chance of facing no player with a pair is nearly (1 - (72/1225))^n.

Holding offsuit, facing no pair probability for n players:

  • 1 other: 0.94
  • 2 other: 0.88
  • 3 other: 0.83
  • 4 other: 0.78
  • 5 other: 0.74

This agrees very well with a simulation where 5000000 deals of an offsuit non-pair were evaluated. The following were the probability that a pair was held by an opponent at position n or lower.

Position - probability of pair - probability of no pair
1 - 0.0585 - 0.9415
2 - 0.1139 - 0.8861
3 - 0.1659 - 0.8341
4 - 0.2149 - 0.7851
5 - 0.2610 - 0.7390
6 - 0.3042 - 0.6958
7 - 0.3448 - 0.6552
8 - 0.3831 - 0.6169
9 - 0.4192 - 0.5808
10 - 0.4531 - 0.5469

If the initial hand is unconstrained (suited cards and pairs allowed), then the probabilities change by very little. A 50 and 52 card deck are very similar.

  • First of all thank you for your reply. This gives an answer that is both simple and accessible. This makes me think about my previous calculations. Two questions about this: - Your result for two players is very different from mine, there is no doubt that your answer is correct, but I cannot find my error in my calculation. Do you have an idea ? - You mention a simulation of 500,000 deals. How did you accomplish it? With a specific software?
    – Doedalos
    Commented Feb 4, 2022 at 10:45
  • 1
    I don't know where the difference is with the first analysis. The program was very similar to gist.github.com/BowlOfRed/7a85fe085582456a2497c914e721aaa8.
    – BowlOfRed
    Commented Feb 4, 2022 at 16:48

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