I have a buddy who insists that anytime one player at a table has a pocket pair, the odds of others at the table also having pocket pairs are increased. Is he full of **** or is there validity to this claim?

This is probably a good question for the Probability forum. I'd say that if your friend IS right, it's not a mathematically significant difference to influence any decision you're making at the table, so it doesn't really matter.

Right. Mods feel free to move if appropriate.

Looks like your friend is right.

The odds of having a pocket pair dealt to you with all other cards being unknown are 52/52 * 3/51 = 5.88% (any card as first card, and one of the 3 of the same rank for the second card).

Let's say you're heads-up and you have a pocket pair. What are the odds of your opponent having a pocket pair? We know 2 cards so there are 50 left. 48 of those cards have 3 mates, 2 are the mates to your pair. (48/50 * 3/49) + (2/50 * 1/49) = 5.96%

Not a big difference. Don't really feel like working out the math for 3 players and up though I would expect that it gets slightly higher for each added player.

Your friend is definitely correct.

If you start with a full deck, it's possible to deal a pair of any rank in 6 ways.

If someone is dealt a pair of, say, 3's, there's one pair of 3's left in the deck, so the number of pairs possible has gone down by 5.

However, if someone is dealt, say, 75, then the number of pairs of 7s or 5s - which is 12 in a full deck - has gone down to just 6 (3 pairs of 7's and 3 pairs of 5's) which is a reduction of 6.

Long story short, dealing a non-pair hand, reduces the number of pairs still possible by more than if you deal a pair.

Yes, but the difference is so small that it's not significant. It's like when you get KK at a full ring table. There's about a 4% chance that someone will have aces, but you can't be scared of that. They are just as likely to have QQ or worse.

Your friend is correct.

There are 78 unique ways to make a pair (since AdAc And AcAd are the same).

If the two cards we remove are both the same, there are 73 ways to make pairs with the remaining cards (6 for each of the cards not in our pair, and the one remaining pair of whatever we removed).

If the two cards are different, there are only 72 (6 for each of the 11 we haven't removed, plus 3 for each of the two that are missing one card).

It's about .08% more likely any given hand will be a pair if you have a pair.

At a six max table, it's .32% more likely there's at least one other pair out there when we have a pair. Really doesn't seem worth bothering about.