Probability with draft lottery

Commissioner_rumY · 2012-02-29 07:57:10

Hi,

I'm looking for some help figuring out draft pick probability for my fantasy hockey league draft (math of this level is not my thing).

It's a ping-pong ball draft (4 of 14 lotto style), like in real life, where each team is assigned a certain number of chances (tickets) to weight their odds of getting a better pick. What I need to know is each of the nine team's odds of getting each of the nine draft picks (1st to 9th). There are no duplicates, as in once a team's ticket is pulled they cannot be pulled again. The weighting is as such:

Out of 1001 tickets (all of the 4/14 possible combos):

Team 1: 400 tickets
Team 2: 300 tickets
Team 3: 150 tickets
Team 4: 75 tickets
Team 5: 40 tickets
Team 6: 20 tickets
Team 7: 10 tickets
Team 8: 5 tickets
Team 9: 1 tickets

I'd love to know how to figure this out myself (or how to get Excel to do it) but if that's too much of an ordeal just knowing the percentages for this scenario would be a great help.

Thank you for your time,
MK

bobbym · 2012-02-29 09:42:29

Hi Commissioner_rumY;

The NBA style 14 - 4 has different rules than some of the others as least as far as I can understand. I will need to know, which draw gets dropped, 11,12,13,14? After that all the particular rules of picking.

Commissioner_rumY · 2012-02-29 09:48:56

We don't drop one, we use all 1001 combinations. We have no "can only fall 3 spots" type rules like, say, the NBA has. It's a simplified system that is modeled after the NBA/NHL lottery.

Each team lottery team gets ONE of the nine picks (i.e. no duplicates), and it can be any pick, weighted the way posted above.

Does that tell you what you need to know? And thanks so much for looking at my problem.

bobbym · 2012-02-29 13:10:39

Do not thank me, I have not solved it. So we have nine teams and each is given one of the picks above. Are the picks assigned randomly?

Commissioner_rumY · 2012-02-29 17:29:50

Yes the picks are assigned randomly, with a weighted chance of certain teams receiving the better picks. That is what I'm trying to figure out. Prolly some miscommunication there, I'm not quite sure what you are asking, sorry.

bobbym · 2012-02-29 17:50:13

Just need all the correct information before attempting to solve any problem.

I am assuming the first team that has 400 tickets goes first followed by the second team...?

Commissioner_rumY · 2012-02-29 18:07:11

Yes your assumption would be correct, yes. Sorry I missed your meaning there.

bobbym · 2012-02-29 18:20:33

Okay, we will start at team 1 and progress down to team 9. Now do you require to know just what the pecentage of getting the top draft pick for each team or each teams chance of getting all the picks?

Commissioner_rumY · 2012-02-29 18:23:46

Each team, all of the picks, sadly.

bobbym · 2012-02-29 18:38:04

I still have some questions to ask. Team 1 has a 400 tickets out of 1001. If that means they have 400 / 1001 or roughly .4 chance of getting the first pick? If they do not get the first pick, whom do they get?

Commissioner_rumY · 2012-02-29 18:55:55

If they don't get the first pick, then whoever's "ticket" was pulled first gets the first pick, and the draw continues on until everyone has a draft pick. For example, if the team with 300 tickets (team 2) gets pulled first (the first pick) then the draw continues on for the second pick with the remaining 8 teams (with 701 balls remaining, I think, as team 2 cannot be selected again). That make sense?

bobbym · 2012-02-29 19:00:43

Hi;

Thanks for bearing with me but I never understood the NBA draft so I am finally getting it now. So you are saying they pick one ticket out of the 1001 and whatever team name is on that ticket gets the first round pick, etc?

Commissioner_rumY · 2012-02-29 19:11:28

Yes that is it exactly.

bobbym · 2012-02-29 23:31:57

Hi Commissioner_rumY;

The chance that a team comes up with the first pick.

The chance that a team comes up with the second pick.

The chance that a team comes up with the third pick.

The chance that a team comes up with the fourth pick.

The chance that a team comes up with the fifth pick.

The chance that a team comes up with the sixth pick.

The chance that a team comes up with the seventh pick.

The chance that a team comes up with the eighth pick.

The chance that a team comes up with the ninth pick.

These percentages were done by computer simulation and are only close approximations.

Commissioner_rumY · 2012-03-01 04:23:04

Thank you so much, that's exactly what I've been looking for!

bobbym · 2012-03-01 05:41:31

Hi Commissioner_rumY;

Your welcome!

bobbym · 2013-08-04 02:24:44

Hi Commissioner_rumY;

The update to the problem: A recent poster has an automated to tool to do this but it is not working yet. That inspired me to right my own.

A trick allowed me to get exact answers so I will give to you as they come in. This will replace post #14 which are just approximations. Also, the format is slightly different.

First Pick:

Second pick:

Third pick:

Fourth pick:

Fifth pick:

Sixth pick:

Seventh pick:

Eighth pick:

Ninth pick:

anonimnystefy · 2013-08-04 07:38:31

A recent poster has an automated to tool to do this but it is not working yet.

Who? And which tool? I would like to see it.

bobbym · 2013-08-04 07:48:10

http://www.mathisfunforum.com/viewtopic.php?id=19834

But his tool is erratic.

anonimnystefy · 2013-08-04 07:50:02

So, how did you do it?

bobbym · 2013-08-04 07:52:08

There must be a better solution, one that is entirely combinatoric in nature. I do not have that yet. What I do have is a working tree structure, that allows me to compute exact probabilities.

This a continuation from the discussion that started here.

http://www.mathisfunforum.com/viewtopic … =19268&p=2

Both you and I have forgotten that I have already solved this problem ( post #53 ) for exact arithmetic.

I just came up with a different form.

anonimnystefy · 2013-08-05 03:08:41

Hi;

Here is my M code:

P[team_, place_, l_] := ((Product[#[[i]], {i, 1, place}]/Product[Total[l] - Sum[#[[i]], {i, 1, j}], {j, 0, place - 1}] & /@ Select[Permutations[l,{place}], Last[#] == l[[team]] &]) // Total) // N

It's fairly quick.

Last edited by anonimnystefy (2013-08-05 03:23:44)

bobbym · 2013-08-05 03:09:39

What does n,k,l stand for?

anonimnystefy · 2013-08-05 03:12:05

I have edited the code to put in more significant variable names.

bobbym · 2013-08-05 03:13:46

Does it work?

Math Is Fun Forum

#1 2012-02-29 07:57:10

Probability with draft lottery

#2 2012-02-29 09:42:29

Re: Probability with draft lottery

#3 2012-02-29 09:48:56

Re: Probability with draft lottery

#4 2012-02-29 13:10:39

Re: Probability with draft lottery

#5 2012-02-29 17:29:50

Re: Probability with draft lottery

#6 2012-02-29 17:50:13

Re: Probability with draft lottery

#7 2012-02-29 18:07:11

Re: Probability with draft lottery

#8 2012-02-29 18:20:33

Re: Probability with draft lottery

#9 2012-02-29 18:23:46

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#10 2012-02-29 18:38:04

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#11 2012-02-29 18:55:55

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#12 2012-02-29 19:00:43

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#13 2012-02-29 19:11:28

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#14 2012-02-29 23:31:57

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#15 2012-03-01 04:23:04

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#16 2012-03-01 05:41:31

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#17 2013-08-04 02:24:44

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#18 2013-08-04 07:38:31

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#24 2013-08-05 03:12:05

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