Multiple Balls: Disadvantage?

[bcrt2000]

your claim is useless though since a forward lottery is done entirely different than how you explained. This isn't about mathmatical proofs, this is abouts odds and percentages. Shows what you know about lotteries. Take it from somebody who gambles, odds are everything. You apparently know nothing about odds.

You still didn't disprove my proof.

A forward lottery is done by picking 48 balls. A reverse lottery is done by picking 48 balls. The probability for any outcome is 1/48!. So the probability of getting outcome any 48 ball outcome A is 1/48!. Let A' be the reverse of A. The probability of getting the outcome A' in the reverse lottery is 1/48!. Okay, This shows that for ANY 48 ball outcome, its reverse is as equally likely to happen.

The second part of my proof showed that if you have ANY outcome A for a forward lottery, and you count only the FIRST ball for each team it will give you the same Draft Order as taking A' and counting only the LAST ball for each team. The <== direction of this is trivial.

Any argument that the a team has more odds given a certain situation of balls WILL NOT disprove this. The only way you can disprove me is by contradicting:
- the fact that in any outcome of successive picking of X balls, where you remove each ball after it has been picked is as equally as likely to happen as the reverse of this outcome
- counting the FIRST ball in outcome A is equivalent to counting only the LAST ball in its reverse outcome, A'

I'm obviously the fool here for continuing to argue pointlessly

 

[helicecopter]

Wow. I don’t know if this thread is more irritating (most people apparently open the mouth without reading what already written above) or hilarious.

Here we have another amazing example:

Easiest way to put it, in a forward lottery they have a 50+ chance at getting picked first. In a reverse lottery, their odds of winning are equal. It is 1/3......at best. In a reverse lottery the absolute best case scenario is equal odds to the other teams. There is no way you can say this is wrong.
…Take it from somebody who gambles, odds are everything. You apparently know nothing about odds.

Both systems give the teams exactly the same odds of winning the #1 seed.

that justy simply can't be true. …crap…crap…crap….crap….crap…..Until the multiple ball teams are down to a single ball, they have a higher chance at being drawn, simple as that.

For the first overall pick:
-if you go straight forward the Rangers have 3/48 chances to get it. It seems everyone agree on this!!!
-going by reverse order: since there are 48 balls, each one of the Rangers balls has 1/48 chances to be the last one left. The rangers have three balls, so their chances overall to possess the last ball left will be 1/48 + 1/48 + 1/48 = 3/48.

don't know whether to laugh or cry

I feel and share your pain/hilarity brother!!

 

[rocketpop]

The systems are statistically the same.

I'm not even good at math. Some of the people who aren't grasping this are hilarious. One of the reasons I hated math was because of situations like this where, even after the mathmatical proof, they'll still say "but that just can't be right..."

I remember how there was a problem like "There are 50 people in a room, what are the odds that 2 of them share the same birthday". The chances are, of course, over 90% likely that at least 2 of those 50 people share a birthday. So many people just couldn't grasp it.

 

[the_gman83]

In case rigorous proofs just don't do it for you, I hope this helps:

I wrote a program to simulate the draft lottery(picking 1 to 30). It starts with a pool of 48 balls, and, every time a ball is picked, that ball and all others belonging to the same team are removed from the pool, until all 48 balls are removed. After 10,000 simulations, these are the results:

Team 'A', with 3 balls, won the first overall pick about 6% of the time.(600/10000)
Team 'B', with 2 balls won the first overall pick about 4% of the time.(400/10000)
Team 'C', with 1 ball, won the first overall pick about 2% of the time.(200/10000)

Team 'A', with 3 balls, on average finished with the 10th pick.
Team 'B', with 2 balls, on average finished with the 13th pick.
Team 'C', with 1 ball, on average finished with the 18th pick.

Seeing this thread, I decided to do some more programming to test the various theories presented. After rewriting the program to simulate a reverse drawing, where only each team's last ball is used to determine the draft order, these were the results:

Team 'A', with 3 balls, won the lottery 6% of the time and finished at an average spot of 10th.
Team 'B', with 2 balls, won the lottery 4% of the time and finished at an average spot of 13th.
Team 'C', with 1 ball, won the lottery 2% of the time and finished at an average spot of 18th.

Remarkably, the two sets of results are almost exactly the same.

Of course, without an infinite amount of trials, this doesn't really prove anything, but it does strongly suggest that the method of picking balls doesn't matter. (This is also assuming that the programs accurately simulate the two versions of the lottery, but I feel confident that they do).

And, I think we've all learned a valuable lesson in this thread... mathematics have absolutely no place in hockey.

 

[Cole Caulifield]

Simple mathematical exemple for Dummies :

2 teams !
Team A : 2 balls
Team B : 1 ball

Probability of Team A picking 1st with the original system : 2/3
Probability of Team A picking 1st with the reverse system : 1 - (2/3 * 1/2) = 2/3

Probability of Team A picking 2nd with the original system : 1/3
Probability of Team A picking 2nd with the reverse system : 2/3 * 1/2 = 2/6 = 1/3

GASP ! THE ODDS ARE THE SAME !!!!!

 

[mas0764]

Well, it appears I was wrong. Apologies for all the pain and hardship I caused in this thread.

I still think there's something to the chances over the long term, but I'll have to look into it.

 

[phacochere]

This may be stupid, or someone may have suggested this before, but what if they simply went with 30 balls, and whenever a multiple ball team ball was picked, it just went back into the bag until their ball came out one or two more times? That way all teams have an equal chance of getting picked first (1/30), and it seems to me that way the reverse draw has the same odds as the conventional draw, no?

 

[NJDevils#4]

My head hurts from reading this thread.

 

[the_gman83]

This may be stupid, or someone may have suggested this before, but what if they simply went with 30 balls, and whenever a multiple ball team ball was picked, it just went back into the bag until their ball came out one or two more times? That way all teams have an equal chance of getting picked first (1/30), and it seems to me that way the reverse draw has the same odds as the conventional draw, no?

If you were to do that for the reverse draw, it would be almost the same as having the extra balls in there(because you're just putting the extra ones back in). If you did it for the forward draw, it would disadvantage the teams that are supposed to have 2 or 3 balls, because the first ball drawn is the one used to determine the order, so putting the ball back in would be pointless.

 

[Patman]

I already did contradict it. The Rangers have a 60% chance at number 1 forward, how is it even remotely possible for them to have a 60% chance in the reverse? I put the numbers down. In a draw where the goal is to not be drawn, the Rangers in every single drawing have a more chance at being drawn, or at best even. I showed you the numbers, and the numbers don't lie. As for your last part, a team with 3 balls is more likely to lose 2 balls and be down to one before a team with 1 ball is eliminated. This is about stats and odds, and the odds do not favor the teams with multiple balls. The only way it could work is if they put in 30 balls, and teams with multiple balls had their's simply not count until there's came up. At least that way the teams with fewer balls would be just as likely to be drawn.

ok, rocket scientist... what about my proof?
If the pickset of one method is always equivalent to the pickset of another method it must follow that the probability of obtaining a pick under the first set must be equivalent to the 2nd set.

If A=B then P(A)=P(B)

 

[Patman]

The systems are statistically the same.

I'm not even good at math. Some of the people who aren't grasping this are hilarious. One of the reasons I hated math was because of situations like this where, even after the mathmatical proof, they'll still say "but that just can't be right..."

I remember how there was a problem like "There are 50 people in a room, what are the odds that 2 of them share the same birthday". The chances are, of course, over 90% likely that at least 2 of those 50 people share a birthday. So many people just couldn't grasp it.

edit: The systems are the same... statistical similiarity is a notion for hypothesis testing of parameters and such... just nitpicking.

Oddly enough I'm one of those that thought reverse order would make a difference... funny thing happens when you put the pen to paper.

As for the birthday problem... again, funny things happen when pen meets paper. I won't go through the proof, but conditional probabilties can be really helpful at times and this is why in prob/stat as much as physics, chem, or engineering that it is wise to write down all your assumptions and go to work. The most powerful tool in math is the eraser.

 

[phacochere]

If you were to do that for the reverse draw, it would be almost the same as having the extra balls in there(because you're just putting the extra ones back in). If you did it for the forward draw, it would disadvantage the teams that are supposed to have 2 or 3 balls, because the first ball drawn is the one used to determine the order, so putting the ball back in would be pointless.

Ok, but look at it this way: If picking in reverse, with 48 balls the 3 ball teams have a 3/48 (6.625%) chance of losing a ball on the first pick, whereas with 30 balls in there they have 1/30 (3.33%). This cancels out the disadvantage that 3-ball teams have of doing it in reverse.

This is all intuitive for me though, I haven't even tried to figure it out statistically (probably because I can't....)

Can someone help with the math?

 

[Tuggy]

So is the NHL actually using a reverse draft selection or is this just a big pissing contest?

Personally when I think of "winning" the lottey, I think of the first name drawn out of a hat. But I guess I'm just old fashion that way.

 

[Seph]

Isn't that what matters, what ball comes next?

Cause we're not just trying to figure out who goes first.

It looks like you get it now, but just to answer your question:

Of course who comes next matters. But the odds for each individual slot have stayed the same.

Think about if they had 48 guys, each one wearing a different shirt numbered 1-48. Each guy reaches into a bag and grabs a ball at the same time. Now, would the odds have changed at all if they revealed which ball they had from first to last, last to first, all at the same time or even at random? Of course not.

This is how the lottery essentially works. All we're doing is assigning at random a number to each team.

If we revealed which of these people had which ball in order, it would change the odds on what ball the next guy had. But only because we are working from a different set of information, that is to say we know which balls have already been assigned. But it doesn't mean that in the end there was anything other than a 1/48 chance that each person had any one ball. And if there are three of one ball, there's a 3/38, etc.

 

[Mess]

FYI

Rogers Sportsnet has had a mock draft draw each day and based it on the weighting of the real one ..

Here are the Results

MOCK DRAFT(S)
Possible destinations for Sidney:
July 15: Toronto
-----------------------------
July 16: Chicago
-----------------------------
July 17: Anaheim
-----------------------------
July 18: Carolina
-----------------------------
July 19: Minnesota
-----------------------------
July 20: Calgary
-----------------------------

http://www.sportsnet.ca/hockey/article.jsp?content=20050715_152444_5152


The thing of interest here is that over the 6 days of drafts the 4 teams that are the most heavily weighted did not win on any day.. NYR, Pitts, Columbus, Buffalo

 

[f1nn]

I'm sorta out of the loop on all the CBA stuff as I've been on vacation... but if they're drawing the balls behind closed doors isn't it very possible that the drawing could be rigged or something?

 

[Emerald City Bruin]

Simple mathematical exemple for Dummies :

2 teams !
Team A : 2 balls
Team B : 1 ball

Probability of Team A picking 1st with the original system : 2/3
Probability of Team A picking 1st with the reverse system : 1 - (2/3 * 1/2) = 2/3

Probability of Team A picking 2nd with the original system : 1/3
Probability of Team A picking 2nd with the reverse system : 2/3 * 1/2 = 2/6 = 1/3

GASP ! THE ODDS ARE THE SAME !!!!!

I like how you mis-spelled example in a post for "Dummies"

 

[CRAZY_FAN]

I like how you mis-spelled example in a post for "Dummies"

Mis-spelled or wrote in french....Who knows !

 

[PecaFan]

Well, it appears I was wrong. Apologies for all the pain and hardship I caused in this thread.

Don't worry about it. At least you were genuine in your misunderstanding, and was a very easy mistake to make.

 

[DJA]

Yes, yes, yes! Finally!!!! Bettman says at the press conference that "the first ball drawn will be the #1 pick."

 

[Crosbyfan]

Quote:
Originally Posted by Seachd
No, they'd count the last ball, not the first one. So the more balls a team has, the more likely it is that their last one comes out later.

That has to be one of the dumbest "theories" ever read on this board.

Seriously.

In which Vlad impales himself!

 

[incawg]

The thing of interest here is that over the 6 days of drafts the 4 teams that are the most heavily weighted did not win on any day.. NYR, Pitts, Columbus, Buffalo

The thing of interest is that people actually care about these "mock" lotteries. Odds are odds. These are about as useful as running a "mock" lottery for your local 649 or Powerball.

 

[Crosbyfan]

Yes, yes, yes! Finally!!!! Bettman says at the press conference that "the first ball drawn will be the #1 pick."

Then they can show us a replay in reverse.

This would show us what could have happened in a reverse lottery!

 

[Cole Caulifield]

I like how you mis-spelled example in a post for "Dummies"

My bad !

My first language is french...

Anyway, it's maths for dummies, not an english lesson

 

[Louiss]

let's assume there is 65 balls... ( I dont remember the exact number)

so every ball has 1/65 of being the last one pulled...

so the team with 3 balls have 3 x 1/65 of having the last balls...

so it's claerly an advantage for them compare to those with one ball....