Drawing with a pocket pair

Thanks!

As a side note, these probabilities are exactly the same for a flop hit in NLHE.

Is that actually true? I’ll defer to you; from your posting history I know you are knowledgeable, but it seems like different situations to me. For example the probability of hitting a set:

For five card draw we already know the three cards we discarded do not contain the same rank as our pair. Therefore the probability that the first card we look at after the draw completes our set is 2/47.

For a hold ‘em flop, we only know our two hole cards so the probability that the first card of the flop completes the set is 2/50.

I haven’t done the full calculation, but it seems just from that consideration that it should not come out exactly the same right? Or am I missing something?

Sorry for the confusion.

The results I posted assumed two cards are dealt closed to the player and then three open cards. This is not the same as a draw game.

We know two things about the discards: they don't match the pair, and they're unpaired. Our denominator will be combos from 47 cards, while our numerator must distinguish the 3 discarded ranks from the 9 ranks with all 4 cards remaining.

P(quads) = 3/(47C2) = 3/1081 ≈ .002775

P(boat) = 2[9(4C2) + 3^2]/(47C3) ≈ .00777

P(set) = 2[(9C2)*4^2 + 9*3*4*3 + 3^3]/(47C3) ≈ .11434

Correction: P(boat) = [2(9(4C2) + 3^2) + 9*4 + 3]/(47C3) ≈ .010176

Previously I only counted boats where our pair improved to a set. The other way to make a boat is for all 3 drawn cards to be the same rank; there are 39 combos of such boats.