So... not trying to stir anything up just going on record that I now believe I was wrong in my calculation of the sabres odds at the 1st pick after the top 3 were announced, and apologize. The sabres odds were indeed 38% and not 60%. I came to the realization after simplifying it down to hypothetically assuming the top 2 picks were the only picks left unannounced:
If 3 through 15 were known, and it was also known that Buffalo and Carolina were the top 2 picks (but not the order), then 1 of the following 2 scenarios happened:
Buffalo won the 1st lottery, and Carolina the 2nd.
or
Carolina won the 1st lottery, and Buffalo the 2nd.
The odds of either specific event happening are small. But, they're nearly equally small. The first result would have happened just 00.68% of the time, and the 2nd result would have happened just 00.57% of the time. But given one of those 2 outcomes DID happen, Buffalo would have had the top pick 54% of the time. Not much better than a 50/50 split. That's actually what the odds became once Montreal was announced at 3rd, I believe. Certainly not the 185/215=86% odds I thought at the time was the case once Montreal was announced at 3rd.
Given that realization, those who worked out the 6 possible combination of top 3 picks, the odds of each of the 6 combinations occurring, and then the conditional odds of a specific result given that 1 of the 6 results DID occur, were absolutely correct. I apologize for stubbornly refusing to listen, to those of you who were trying to explain. Makes perfect and obvious sense now that I've had some time to think about it.
What I am failing at is understanding where the method that calculated the sabres odds at 60% once the top 3 were revealed went wrong. Or said another way, I'm trying to understand what the correct way to calculate the effect on the odds based on the information that a team did *not* win a lottery. Similar to the 2 ways you can calculate the odds of rolling a 5 or a 6 on a die. Method 1: 5 was rolled *or* 6 was rolled (1/6 + 1/6). Method 2: not 1 *and* not 2 *and* not 3 *and* not 4 (5/6 * 4/5 * 3/4 * 2/3 = 2/6). Those who got it right were using 1 method (Buffalo *and* Montreal *and* Carolina won a lottery), and the rest of us who got 60% were trying to use the other method, but using it incorrectly (Fla didn't win a lottery *and* Philly didn't win *and* Dallas didn't win *and* etc etc etc). The 2 methods calculated properly will give the same answer, and that answer will be that when the top 3 teams were known Buffalo had just a 38% chance at the 1st pick given the known information.
In any event, when I get a spare minute or 2 I'm putting together an excel VBA lottery simulator to verify. I actually started the simulator first, when I thought I'd be proving the 38%ers wrong, but then worked it out for myself before I finished. But I'll probably still finish the simulator anyways just for fun. If I run the simulator 100,000 times, Buf, Car, and Mtl are expected to be the top 3 (in some order) 451 times. Of the 451 times, 172 should have Buf 1st (38%), 150 should have Buf 2nd (33%), and 129 should have Buf 3rd (29%). That was the correct answer all along. Mea Culpa. I see it now.
But Larsson disagrees and says its a good place once you get to know it better and socially it's fine. All the players live close to the rink and there are always things to do although he doesn't specify what. And if there are any problems there will always be someone to help out.
