Did some statistical programming

[MojoJojo]

It would be very hard to figure out with mathematical certainty what the chances are. It's easier this way, running a mock draft a couple hundred thousand times and just counting up the results.

OK, I see how it works now. Thats probably the easiest way, even if its not an explicit solution.

 

[ceber]

http://hfboards.com/showpost.php?p=3069527&postcount=196

I think those were the results of a billion-run simulation. Should be pretty close to theoretical, from what I understand.

 

[MojoJojo]

One thing thats interesting is that all teams have roughly the same odds at getting the 17th pick.

 

[MaV]

You also need to multiply that total by the odds that the team did not get the first pick, which is 1 minus the odds they got it.

No no, no need for that. I mean, 9/48 is the chance of some other three ball holder to get the 1st pick, 20/48 two ball holder and 16/48 one ball holder. You know, then the team in question could not have got the pick.

But anyway, from math to the origianl question. Rangers have only ~54% chance to get top-10 pick. So it's not guaranteed.

 

[Patman]

http://hfboards.com/showpost.php?p=3069527&postcount=196

I think those were the results of a billion-run simulation. Should be pretty close to theoretical, from what I understand.

Yeah, and if news sources want to use those numbers (or want to see the full simulation) I have no problem with it. I am fully confident that those numbers are accurate to the digits listed but after that it's sketchy. Standard error calculations maximized the variance at 0.00158%... 3 standard deviations in either direction is about 0.009% for a ball park in the deviation estimates for each team when done team by team... realize that my calculations are further averaged since we have teams with the same weights binned together so they are even more accurate than the 0.009% figure... but not too much more accurate.

If the media does want to use this I'd just ask them to contact me at [email protected] . I admit this board gets rather close to public domain but I'd love to see my name in print or on TV :p

 

[Skk82]

3-ball teams: there is a 50% chance you will get the 9th pick or higher.
2-ball teams: there is a 50% chance you will get the 13th pick or higher
1-ball teams: there is a 50% chance you'll get the 20th pick or higher

Enjoy

i think your program is off....consider it this way...

follow the draft from a '3 ball team', say columbus. they have 3 chances out of 48 total balls, 6.25%. we'll even assume a fellow 3 ball team (say buffalo) wins the #1 pick. now there are only 45 balls left, columbus still has 3 shots to win. 3/45 is 6.6%.

assume, for the sake of argument, new york rangers win the #2 pick. minus their balls out, and there's 42 left, leaving columbus with a 3/42 shot (7.14%) of getting the #3. let's say again that the last '3 ball team', pittsburgh, wins the #3 slot. now there's 38 balls left to pick from, giving columbus a 3/39 (7.69) shot at 4th overall.

at this point, no team other than columbus has more than two balls. but say anaheim (a '2 ball' team) wins #4. that leaves cbs a 3/37 (8.1%) of winning #5 overall. but we'll assume atlanta (with two balls) wins the 5th pick. minus them out of the equation and columbus still has 3 balls out of the total 35 (8.57%) to get #6. but #6 is won by calgary, new cbs odds at #7 would be 3/33 (9.01%). carolina wins #7 and cbs chances at #8 are 3/31 (9.667%). chicago wins #8, new odds at 9th are 3/29 (10.3%).

say edmonton wins 9th, there still would be 27 balls left, columbus still only has 3 of them, 3/27 odds are 11.1% of getting 10.

and that's the best case scenario. if many of the 15 'one ball' teams won a top 10 pick (and it's sheer luck), there could be as many as 37 lottery balls left to choose from to decide the #10 pick, leaving 3 ball teams with possibly as little as 8.11% chance of getting a top ten pick.

doesn't this make sense once one breaks it down and analyzes it like this?

look at pecafan's simulator when a '3 ball' team (buff, cbs, nyr or pit) gets the 10th or 11th pick, the odds beside it (3/number of balls remaining) isn't anything near 50%, it's usually 10, 11 or 12%.

the one time i ran it, columbus got the 20th selection, and they only had a 18.8% chance of getting that, because the last ten teams in the simulation (sjs, stl, cgy, fla, nyi, dall, phi, chi, mtl and cbs) had a combined 16 balls left in the lottery. columbus had 3 of those balls giving them a significant advantage over two ball teams like chicago (who only had a 2/16 or 12.5% chance) or a one ball team like montreal (1/16 chance or just 6.25%). even then, the field still held a large edge (81.2%) over columbus to win the 20th overall pick.

 

[PecaFan]

look at pecafan's simulator when a '3 ball' team (buff, cbs, nyr or pit) gets the 10th or 11th pick, the odds beside it (3/number of balls remaining) isn't anything near 50%, it's usually 10, 11 or 12%.

My program is showing the odds only at that specific time for that particular ball coming out of the machine, which is why they're so low.

The folks above are talking about cumulative odds, which takes into account all your chances of getting the #1 pick, or the #2 pick, or the #3 pick, etc. So they're much higher than the one specific snapshot.

 

[Skk82]

oh i see. i guess i was just thinking a little differently but i get what you're talking about now.