Anyone interested in explaining the math behind this?

I have a decent understanding of math and it is really starting to grind my gears that I cant come to the same percentages that are listed online. How would you calculate the percentage of pairing up either card by the river?

Using AK as the hole cards, my take is 6/50 + 6/49 + 6/48 + 6/47 + 6/46. This however would give me roughly 65% chance. I know that can't be right. I have seen other people explain it by taking the chances of each card NOT pairing and multiplying those odds but how come the calculation cannot be done this way? Can someone explain why my method is wrong? Its killing me!!

You can't calculate the chance of it happening directly. For your question, the correct method of calculating the answer is to calculate the chance of it not happening then subtracting that from 1.

So, it would look like this: 1 - (44/50)*(43/49)*(42/48)*(41/47)*(40/46)= 1 - 130320960/254251200 = 1 - 0.5125 = 0.4875 = 48.75% to pair one of your hole cards by the river.

48.75% is an excellent number but is not the percentage for pairing one of the hole cards. That percentage is less than the 48.75%, but probably not much. I'm sure I have the exact percentage at home, or can figure it out here at work, but time doesn't allow it right now.

The difference between that percentage and 48.75% is the additional chances of extra Aces and Kings resulting in two pair, trips, full houses, or quads.

The technique that Jack Daniels used is great for calculating, or at least estimating, the answer to a lot of poker math questions.

Well the OP was about pairing one and only one of your hole cards by the river. So I'd alter my previous equation as follows which can be calculated directly:

6/50*44/49*43/48*42/47*41/46 = 19548144/254251200 = 0.0769 = 7.69%

So chance of pairing exactly one of your hole cards (no more no less and no care about straights or flushes) by the river is 7.69% unless I botched typing on the calculator.

I guess I couldn't let this go until I figured it out. It's been almost 40 years since my advanced statistics course at xPI so this probably isn't as pretty as it could be.

The odds of pairing exactly one card with the first card of the flop is (6/50) x (44/49) x (43/48) x (42/47) x (41/46) or approximately 7.69%.

The odds of pairing exactly one card with the second card of the flop is (44/50) x (6/49) x (43/48) x (42/47) x (41/46) or the same 7.69%.

Because the card can be paired in any of the five positions, the percentage is 5 x 7.69% or approximately 38.45%.

That means that the chance of two pair (AAKK), trips (AAA, KKK), full house (AAAKK, KKKAA), and quads (AAAA, KKKK) is 48.75-38.45 or 10.30%.

Looking at my Excel spreadsheet, it looks like the odds of pairing exactly one of these cards on the flop is almost 29%.

If anyone finds fault with MY math, please attribute it to getting ooooooooolderrrrrr.

I guess it's not that easy. After all, it took two of us to figure it out!

Basically they already did all the math but just wanted to comment on the logic of the OP a lot of people seem to think they can add the probabilities of 2 independent events and get the probability that one of them happens. Note that this means if there's a 70% chance of rain on Saturday and a 70% chance of rain on Sunday there must be a 140% chance of rain for the weekend. You can only add when the events are mutually exclusive, thus it cannot be both of them. To solve this you can either add them together then subtract when both are (in the rain example 140% - 70%*70% = 91%) or do like JD did which is usually easier and find the odds that they don't both not happen (1 - 30%*30% = 91%).

Joeslick much appreciated so this would be the equation of calculating it directly. And as Zachvac stated it would be easier to calculate the percentages of it not happening, I was absolutely frustrated that I could not come up with the proper formula of the "harder" way. Thank you so much guys. I can play poker in peace now.