probability question from Sklansky book
I was showing a friend who is new to poker Sklansky's Hold'em Poker beginner book and specifically going over the probability section in the back. One of David's probabilities didn't seem right to me, or maybe I wasn't reading it correctly. He says on page 108:
"In a ten-handed game the chance that someone holds both an ace and another card of a specified suite is about 9 percent..."
I read this to mean that if I pick any suit, there is a 9% chance that someone holds Ax of that suit; the implications are that if there are 3 of the same suit on the board than 9 percent of the time in a ten handed game someone has the nut flush. This seems way too high. I can perhaps see that there is a 9% probability that someone has Axs, but not Ax of a specific suit. BTW, he does say that if you have two of the specific suit for a flush the probability goes down to about 6% that someone else has the A-nut flush.
If David is correct can someone please explain the math on this?
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