Quote:
Originally Posted by indigobuffalo
Has anyone on the Columbus board figured out the statistical probability that Columbus gets the 1st, 2nd and 3rd Overall picks in the NHL's 2013 Entry Draft?
Assuming there is no season and the league adopts a draft lottery (with the same rules as the 2005 draft).
They have three 1sts, afterall.
So teams with a 3/50 chance would be:
Calgary, Carolina, Columbus, Dallas, Minnesota, Toronto and Winnipeg/Atlanta (3/50 = 6%, 7*6% = 42%)
Teams with a 2/50 chance would be:
Anaheim, Colorado, Florida, New York Islanders, St Louis and Tampa Bay (2/50 = 4%, 6*4% = 24%)
Teams with a 1/50 chance would be:
Boston, Buffalo, Chicago, Detroit, Edmonton, Los Angeles, Montreal, Nashville, New Jersey, New York Rangers, Ottawa, Philadelphia, Phoenix, Pittsburgh, San Jose, Vancouver and Washington (1/50 = 2%, 18*2% = 36%)
So Columbus has a 10% chance of getting the 1st overall. Assuming best possible odds (where LA is selected 1st, NYR selected 2nd and CLB selected 3rd):
0.1 * 0.08 * 0.06 = 0.00048, or 0.048% chance, or 3/6,250 for those preferring fractions.

Not a stat whiz,but I don't think your probabilities are right. In your scenario either the Kings or Rangers have a 2% chance of winning #1; the loser then has a slightly better than 2% chance of 2nd leaving the Jackets with a 3/48 or 6.25% chance of picking third. Multiply them together and the probability that this scenario happens is .0051 per cent.
The highest chance of the Jackets getting all three is .051%. 10% (5/50) x 8% (4/49) x 6% (3/48) (rounding ignored here). That is assuming the Rangers and Kings happen to win the first two. If you say for sure that is going to happen then I think its the scenario above that dictates the probability of that happening.
Corrections welcomed.