[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.