Hand Probability Expanded - Is My Thinking Correct

[ligastar]

Fact: The chances of being dealt a "premium hand" A-A, K-K, Q-Q, A-K, A-Q or K-Q preflop in Hold 'Em is 19:1.

Scenario: You are playing Hold 'Em at a full table of 10 players (you are seated with 9 other players).

Question: Is it factual to say that, on average, on every other hand (actually slightly more frequently than every other hand since the chances are 19:1) someone at the table is holding one of these "premium hands"?

Edit: I think I made a mistake above. It would not be slightly more frequently than every other hand @ 19:1 . . . it would be exactly every other hand, on average, at a 10 player table - correct?

Greg H.

[olavfo]

Makes sense to me. When you deal random starting hands, about 1 hand in 20 will be a premium hand from your list. So with ten players there should be one such hand per every two deals on average.

olavfo

[LinusKS]

Well, if I'm doing this right (you can check my math), and if the odds of having a premium are 1:19, that's one out of 20, or 5%.

That means the chances that no one at a full table has a premium are .05 to the power of 10, or 60%.

The chances that no one would hold a premium after two full orbits would be .95 to the power of 20, or 36%.

So after two full orbits someone would have been dealt a premium 64% of the time.

Edit: I realize that doesn't really answer the question. To figure the average number of times someone would have been dealt a premium over two orbits, I suppose you'd have to calculate how often zero, one, two, three, etc. premiums would have been dealt over two orbits, and then take the weighted average of that. I don't doubt it's close to one, but I'd be willing to bet it's not exactly that.

[ligastar]

[ QUOTE ]
I suppose you'd have to calculate how often zero, one, two, three, etc. premiums would have been dealt over two orbits, and then take the weighted average of that. I don't doubt it's close to one, but I'd be willing to bet it's not exactly that.

[/ QUOTE ]

LinusKS

Thanks for the reply. I'm not a math whiz but logically I was thinking that the hand probability for an individual could be spread over a 10 table game (10 person to keep the math easy for my question). I don't know if my logic is flawed. I think the weighted average angle you delved into would provide some clarity to the question. I'm trying to think this through but am still stuck and not sure if this logic is correct.

Greg H.