So, regardless of the coin flip, our most likely outcome is:
1. Ending up with a top 3 pick (33 percent)
2. Ending up with the 5th pick (30 percent)
3. Ending up with the 6th pick (26 percent)
Anything else is highly unlikely and the coin flip doesn't change that.
The flip is for more than just a ping pong ball--it ups the C's chances of landing in the top four by about a third. Here's an article explaining it: http://www.nba.com/celtics/news/sidebar/inside-numbers-draft-lottery-qa
That's a little misleading - as Harry said, the odds of top 3 stay about the same, the increased odds of "top 4" are almost entirely just increased odds of #4. Which is still good, but it doesn't move the needle much on the odds of 1-3.
The biggest difference to me is that if we don't move up and get leapfrogged by one team (the most probable outcome in either spot, I believe) we pick 5th instead of 6th. And there's nearly no chance of falling out of the top 6, whereas that's more likely in the 5 spot.
If neither team gets leapfrogged and the top 3 goes as the odds dictate (Bucks, Philly, Orlando), winner of the coin flip will pick 4th and the loser will pick 5th. So if you see this as a 4 player draft... the coin flip is kind of important.
Winner of the coin flip can't fall below 7. Loser can't fall below 8.
Yes, though the odds dictate that the top 3 WON'T go that way. Collectively, the field has greater than 50% odds of cracking the top-3, forcing one of those three teams down to the 4th, 5th or 6th spot.
Remember, Orlando has a 53.1% chance of falling out of the top 3.
Plus Philadelphia hasa 44.2% chance of falling out of the top 3.
Plus Milwaukee has a 35.8% chance of falling out of the top 3.
'Just need one of those to occur so even without thinking about the exact probability addition method to use, I can see that the odds strongly favor a team from the field cracking into the top 3.