Why are the odds 16 to 6 that any given player does not hold a pocket pair?

In the book Ken Warren teaches Texas Holdem, its explained that there are 1,326 possible combinations of hole cards. Then its explained that if you disregard suits, there are 169 possible two-hole cards. Then its explained that 1 of 17 hole cards is a pair. Then its explained that the odds of being dealt any specific pocket pair is 220 to 1. Finally, he says that "if you know nothing of your opponents hand, you can know that odds are 16 to 6 or 8 to 3 that he does not have a pocket pair. The unpaired hand is two and two-thirds more likely than the pair (6 × 2 2/3=16)."

Can someone explain to me this jump in logic? How is it that 16 to 6 odds are that any given opponent does not hold a pocket pair?

I don't see the jump in logic, either.

Your opponent's odds of having a pocket pair should be the same 1/17 that you have.

I have no idea where he gets the 16 to 6 figure. There are 16 ways to have any given non-pair and 6 ways to have any given pair, but there are 78 non-pair rank combinations and only 13 pair ranks.

If that's how he's calculating it, it's wrong.

Most of that section seems good but the part where he mentions 13x13 = 169 card combinations doesn't really say much. No applicability at all to calculating odds of anything.

One possible angle, though, is that he's talking about the odds of any one of five opponents having a pocket pair. Taking (16/17)^5 as about 0.7385 is pretty close to the 8 to 3 probability of 0.7273. I'm oversimplifying the calculation a bit so that may be what he was going for.

• I'm thinking that there are 16 ways to make a non-pair and 6 ways to make a pair. So 16 nonpaired hands/6 paired hands = 8/3. I guess he's saying it's more likely to not have a pair and not be up against a pair. But in light of what you said about 78 non-pair rank combinations and only 13 pair ranks, it doesn't seem like good math, or even that relevant to actually playing the cards, because you have to assume people are more likely to actually play their hand when they have a pocket pair. Dec 22, 2020 at 0:49
• He makes no mention of 5 players so I doubt that the probability of any one in five opponents having a pocket pair is not his point, but that's an interesting angle and props for finding some scenario to fit the number in! Dec 22, 2020 at 0:53
• @EvanWelch regarding your first comment, if that is the math he's using, it's just plain wrong. As for your second comment, if he's a winning player obviously he understands poker; it could be that he was trying to fill pages for his book. It's straightforward to figure out if you're playing heads-up but more complicated with more than that, because it depends on whether you have a pocket pair or not yourself, whether two opponents have common ranks among their cards, etc.
– John
Dec 22, 2020 at 1:37