My buddies and I played hold'em and noticed an unusual number of hands in which the board paired. I decided to do a simple calculation to find out the probability of this happening. I have an answer, but I'm not 100% certain I did it right. Would someone else mind running the calculation quickly? If I get a response, I'll post my answer and method. I think it's very simple, but I'd like to double-check.

Thank you.

Cheers.

I don't know if this is the easiest way, but I'd calculate the probability of NOT getting a pair as follows:

The first card can be any one of 52.
The second card can then be any one of 48 (it just has to have a different rank to the first one).
The third card can then be any one of 44.

The total number of ways of dealing three cards = 52x51x50.

So, probability of three different ranks = (52x48x44)/(52x51x50) = 0.828

So, probability of a pair (or three of a kind) = 1 - 0.828 = 0.172

Very roughly, 1 in 6

(This ignores any knowledge you have of your hole cards. I guess if you're holding a pair in your hand, the probability of a pair on the flop goes down. Wouldn't make much difference, though, is my guess.)

I approached it this way and got a slightly different answer. I think it's because I strictly took into account pairing rather than pairing or tripling.

Total number of flop possibilities = 52*51*50

The first card can be any of the full 52. The second one then must be any of the remaining cards of different rank than the first; i.e., you have 48 choices for the second card. Then the last card--in order to pair the board--must be the same rank as either of the first two cards; i.e., you have 6 choices for the third card:

Number of board pairings = 52*48*6

Probability of strictly pairing = (52*48*6)/(52*51*50) = 11.3%

Thanks for the response.

You only counted ABA and ABB type flops you are missing AAB which is further

(52*3*48)/(52*51*50) = 5.6%

Total 16.9% which is consistent with the first response (trips have 52*3*2/52*51*50=0.2%)

That's a neat way of doing it, but I think you're missing some of the possibilities.

You're only counting those flops where the first and second cards are not paired, but the third card pairs with either the first or second.

But there's another way of flopping precisely one pair: the first two cards have the same rank, and the third card is of a different rank.

That adds another 52 x 3 x 48 = 7488 possibilities, giving a probability of precisely one pair equal to

((52x48x6) + (52x3x48)) / (52 x 51 x50)

= 0.169

We can reconcile that to the total number of flops as follows

Ways of getting no pair: 52 x 48 x 44 = 109 824

Ways of getting precisely one pair: (52x48x6) + (52x3x48) = 22 464

Ways of getting three of a kind: 52 x 3 x 2 = 312

Total ways = 109 824 + 22 464 + 312 = 132 600

Total flops possible = 52 x 51 x 50 = 132 600

(As before, that all ignores anything we know about our hole cards)

The total number of flop possibilities with disregard to order is 52*51*50/(3*2*1) = 22,100 or 52C3

i.e., 234 is the same as 4,3,2

The total number of flops where the order in which the cards are dealt is important is 52*51*50=132,600 or simply 52P3.

i.e., 234 is different that 4,3,2.

Thank you, everyone, for taking the time to respond. Very good replies.

Cheers.