[QUOTE]
*Originally posted by sdimbert *
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No, because he has a SF to the king.
This might not help:
Ok, that’s cool. Her taking the four 10’s excludes my undercutting theory. If she takes the 10’s, the best I can possibly wind up with is a SF to the 9. No matter what I choose then she can always beat me. (If I try to undercut her by taking the 9’s, she just draws to get a Royal Flush).
I had a whole long post written, but then when I previewed it, I saw the link to the answer, which got in there while I was writing.
So now I’ve read it, and it makes sense except that it only works as a story, where Ray gets to explain his “two Royal Flushes are tied” strategy. It does not work as a riddle, because there are two critical features needed:
After Ray convinces his wife that it will always end in a tie, the wife switches the order so that she will go first. And not only does she pick first, but it is also she who gets the option of exchanging first.
Once you see that, it doesn’t really work as a riddle any more.
Gardner’s version of the problem also specifies that the discards are set aside from the game. Click and Clack probably left that part out, thinking it unimportant or obvious.
For a slightly different version (real poker drawing rules), click here