Dan Tilque:
> > > Mark Brader 1 0 0 0 1 1 0 1 1 1 0 0 6
Mark Brader:
> > Nope. I missed Istanbul.
Dan Blum:
> As did I, now that I look.
Looking again, I find that I somehow looked back at Dan B.'s answer
when I meant to look back at my own. Dan *did* miss Istanbul, but
I *didn't*.
> That puts me in a tie with Peter and
> Erland, assuming their scores are correct.
Agreed. I now get:
1 2 3 4 5 6 7 8 9 10 11 12 TOTALS
Erland Sommarskog 0 1 1 0 1 0 0 1 1 1 1 0 7
Peter Smyth 1 0 0 0 0 1 1 1 0 1 1 1 7
Dan Blum 1 1 1 0 1 1 0 0 0 1 0 1 7
Mark Brader 1 0 0 0 1 1 0 1 1 1 0 0 6
3 2 2 0 3 3 1 3 2 4 2 2
Sorry about the error.
Dan T. didn't specify a tiebreaker criterion, just "usual rules".
If the first tiebreaker is hardest questions and it's computed
numerically the same way "Calvin" does it on his contests, then
I make it that Dan B. has a 19, and since lower is better, he's
beaten by both Peter and Erland, tied at 18.
Peter was the first to post, so if that's the second tiebreaker,
then it's Peter who wins.
Dan T., can you please confirm or correct this, as applicable?
--
Mark Brader, Toronto Carpe pecuniam!
m...@vex.net --Roger L. Smith
My text in this article is in the public domain.