Unpaired hand vs probability that the opponent has a pocket pair

Question

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 :

Knowing that :

Direct calculation of the probability of event A :

This calculation can be verified by determining the complementary case :

Attempt to calculate with two opposing players

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 :

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.

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.