Dynasty, you are cheating and I think you know it
I'll admit, I didn't read the whole debate on the beginners forum. It started to make my head hurt. If my point has already been covered there, I appologize.
You are playing games with the numbers, and I'm sure you are smart enough to know it. Did you and bernie actually agree that there will be exactly one A and exactly one K out in the other players hands?
I'm assuming bernie's point was that there are probably some A's and K's already dead. Assuming that there's exactly one of each is just silly. The math is totally meaningless.
If you are really interested in this question, you shouldn't just randomly pick two of your cards as dead. If all 9 players are playing random cards (which I think is a good assumption since limping doesn't make someone more or less likely to have an A or K) then on average 2.16 A's and K's will be in their hands, I think. That's 6/50 (the chance that any random card is an A or K, given that you have AK) * 18 (the number of random cards). Using 2.16 dead cards instead of 2 in your equations below, and altering the number of unknows accordingly makes the chances of floping a pair 28.89%, or slightly lower than if all cards are unknowns.
To answer your question though, ifyou really are interested in that specific problem your math is correct.
Boy, did that sound like a Dynasty post or what? [img]/forums/images/icons/smile.gif[/img]
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