2017-2018 Standings Tracker

[Go Wings]

The percentage chance is what determines what happens. Nothing else. What you're describing is called sampling, and it is not at all separate from the original set of probabilities. What actually happens is a function of the probabilities specified by the NHL. Well, unless you think it's rigged.

Well the 2 years they had generational players the lottery balls were behind closed doors, every other year since 2005 it was on tv. So it might be rigged tough to say.

 

[Henkka]

What actually happens is the only thing that matters not the percentage chance of it happening.

1: 8.5%
2: 8.7%
3: 8.9%
4: not possible
5: 8.4%
6: 34.5%
7: 26.7%
8: 4.3%

Here are probabilities for every spot.

 

[ridilon]

Does anyone know exactly how the lottery works? I've read that they draw the balls and the number sequence corresponds to a table? or chart? that matches to a particular team. But what I'm curious about is what would DET chances be after a team wins the 1st pick and after another team wins the 2nd. I get the overall percentages, but let's say Arizona wins the lottery - what chance do the wings have at that time of moving up a spot or 2? I'm assuming our chance of moving up to 2 or 3 is better if a team in front of us wins. I'd just like to see the odds of all the winning scenarios.

 

[jkutswings]

Does anyone know exactly how the lottery works? I've read that they draw the balls and the number sequence corresponds to a table? or chart? that matches to a particular team. But what I'm curious about is what would DET chances be after a team wins the 1st pick and after another team wins the 2nd. I get the overall percentages, but let's say Arizona wins the lottery - what chance do the wings have at that time of moving up a spot or 2? I'm assuming our chance of moving up to 2 or 3 is better if a team in front of us wins. I'd just like to see the odds of all the winning scenarios.

It's from a Chicago perspective, but here are all the details for this year:

Everything you need to know about the 2018 NHL Draft Lottery

"The league uses a system where 14 balls numbered 1-14 are placed into a chamber. Balls are drawn one at a time to create a four-digit sequence that corresponds to a board where 1,001 possible combinations have been divvied up at random among the 15 teams based on the lottery odds. The team with the matching combination wins the pick."

"This process is repeated twice more for the second and third overall picks, with the combinations for the teams that already won picks being ineligible to win again. The fourth through 15th picks are then ordered by worst records."

So they first assign each of the non-playoff teams combinations (one combination is made void, so there are an even 1000 to distribute). For the #1 pick, Buffalo gets 185 of the 1000 possibilities, Ottawa gets 135, Arizona gets 115, Montreal gets 95, Detroit gets 85, and so on. They draw four numbers - let's say it's 7, 12, 3, and 9 - and whichever team has those numbers in that order gets the #1 pick. Then they repeat the process for the #2 pick, then for the #3 pick, although I'm not sure if the odds are the same for those, since (using Buffalo as an example) 18.5% x 3 = 55.5%, not 49.4%. Maybe somebody who understands statistics better can explain that one.

 

[Lil Sebastian Cossa]

Does anyone know exactly how the lottery works? I've read that they draw the balls and the number sequence corresponds to a table? or chart? that matches to a particular team. But what I'm curious about is what would DET chances be after a team wins the 1st pick and after another team wins the 2nd. I get the overall percentages, but let's say Arizona wins the lottery - what chance do the wings have at that time of moving up a spot or 2? I'm assuming our chance of moving up to 2 or 3 is better if a team in front of us wins. I'd just like to see the odds of all the winning scenarios.

I don't think that's at all relevant. I think that the odds are still the same.

Also, the Wings actually can't move up one. They either get 1-3, 5-8.

 

[Lil Sebastian Cossa]

1: 8.5%
2: 8.7%
3: 8.9%
4: not possible
5: 8.4%
6: 34.5%
7: 26.7%
8: 4.3%

Here are probabilities for every spot.

This is one thing that is embarrassing about the lottery. The Red Wings have a less than 10% chance to hold the pick they already have. They are three times as likely to drop 2 spots in the draft than they are to get the pick that they "earned".

And the team with the worst record actually has a greater chance of picking #4 OA than they do of staying in the top 3. I know they wrangled with it after Edmonton did what they did, but it seems like the odds could really use a tweak.

 

[njx9]

I don't think that's at all relevant. I think that the odds are still the same.

Also, the Wings actually can't move up one. They either get 1-3, 5-8.

The odds would change if the team with the most combinations drops out. Buffalo has 185 possible combinations out of the 1001. If they get picked first, that leaves 816 combinations. Ottawa still has 135 possible combinations, which now yields something like a 16% chance, a 3% bump in odds of being selected second (or 'next').

I think, if Buffalo wins, Detroit would move to a roughly 10% chance of getting the 2nd overall pick. If Buffalo wins, then Ottawa picks second, Detroit has roughly 12% chances of picking 3rd. Not substantially better, overall.

 

[jkutswings]

The odds would change if the team with the most combinations drops out. Buffalo has 185 possible combinations out of the 1001. If they get picked first, that leaves 816 combinations. Ottawa still has 135 possible combinations, which now yields something like a 16% chance, a 3% bump in odds of being selected second (or 'next').

I think, if Buffalo wins, Detroit would move to a roughly 10% chance of getting the 2nd overall pick. If Buffalo wins, then Ottawa picks second, Detroit has roughly 12% chances of picking 3rd. Not substantially better, overall.

But they're not taken out. They're just void. There are only 14 lottery balls in the machine for each of the 3 drawings, and the odds shouldn't change.

Say Buffalo gets #1 on lottery night. Then they draw for #2. The Sabres' various combinations haven't gone anywhere, but if the 4-number sequence for pick #2 comes up as one of Buffalo's combinations, they just redraw.

I think people are imagining there are 1000 lottery balls in the machine, and that's not the case. There's just 1000 possible sequences that 4 of 14 lottery balls can result in (14! / (4! x 10!)) = 1001, and they void one combination, to leave it at 1000).

4 Choose 3 ; 4 Choose 2; and 4 Choose 0

 

[njx9]

But they're not taken out. They're just void. There are only 14 lottery balls in the machine for each of the 3 drawings, and the odds don't change.

Say Buffalo gets #1 on lottery night. Then they draw for #2. The Sabres' various combinations haven't gone anywhere, but if the 4-number sequence for pick #2 comes up as one of Buffalo's combinations, they just redraw.

... That means those combinations are irrelevant and void. The odds of a number coming up that is non-void is as above. Yes, Ottawa still has a 13.5% chance of having their number come up on the first try for second, but "on the first try" is utterly irrelevant when 18.5% of the numbers have literally been removed from the valid list.

There is, thus, a 16ish% chance that a valid number that gets drawn belongs to Ottawa.

 

[ridilon]

Thanks jkutswings for the link. This is what I wanted to know.

 

[MintBerryCrunch]

I agree with njx9, if a Buffalo combination is drawn after they "won", then the combination is void.
If we don't get the #1 pick, we would want Buffalo to get it. This would result in a lower denominator (1000-185) of 815 for the second pick, where our chances would be 10.43% to get the #2 pick.
If it goes Buffalo #1 and Ottawa #2, our chances jump to 12.50% to pick #3. (815-135 = denominator of 680)

 

[Lil Sebastian Cossa]

I agree with njx9, if a Buffalo combination is drawn after they "won", then the combination is void.
If we don't get the #1 pick, we would want Buffalo to get it. This would result in a lower denominator (1000-185) of 815 for the second pick, where our chances would be 10.43% to get the #2 pick.
If it goes Buffalo #1 and Ottawa #2, our chances jump to 12.50% to pick #3. (815-135 = denominator of 680)

But I don't think it works that way.

Each pull is an individual instance. It's kind of like how each coin flip has a 50-50 shot when you've got a fair coin. Your chances don't improve, they just draw again if a void selection is made. But the "void" selections can still be chosen. Technically speaking, you have an increased chance because if an invalid number is drawn they go again, but you'll always just have that 8.4% chance at #1 or 8.5% at #2, regardless of what happens in front of you.

At least if I'm understanding it... they don't wholly remove the lotto balls for Buffalo if they're pulled, they just make it down as an invalid draw and go again if they are pulled for pick #2.

 

[Red Stanley]

But I don't think it works that way.

Each pull is an individual instance. It's kind of like how each coin flip has a 50-50 shot when you've got a fair coin. Your chances don't improve, they just draw again if a void selection is made. But the "void" selections can still be chosen. Technically speaking, you have an increased chance because if an invalid number is drawn they go again, but you'll always just have that 8.4% chance at #1 or 8.5% at #2, regardless of what happens in front of you.

At least if I'm understanding it... they don't wholly remove the lotto balls for Buffalo if they're pulled, they just make it down as an invalid draw and go again if they are pulled for pick #2.

Wouldn't the fact that less teams are competing in each successive drawing technically improve the chances for the remaining teams? Teams with more combinations would affect and be affected more by this, no?

 

[MintBerryCrunch]

I think it would be different if the "void" selection was an option, as in "hey, we drew Buffalo again, so no one gets the #2 pick"... but it's not; they simply redraw. So any of Buffalo's combinations are removed from the available combinations, leaving in my example above 815 valid combinations when it comes to the the #2 pick.

But who knows, my stats professor in college just talked about tennis all day so I may have missed out.

 

[jkutswings]

If Buffalo wins the first draw, the remaining teams are now X chances out of 815 for the valid possibilities, yes. But that doesn't change the fact that, for draw #2, Detroit would have 85 assigned combinations out of the 1000 possible combinations.

Call it semantics if you like, but in that hypothetical, it's an 8.5% chance that Detroit gets #2, and an 18.5% chance that they redraw. And those probabilities would be the same for as many redraws as it required, because the lottery balls used are the same 1-14 each time, regardless of what combination(s) that came up before. (For example, although EXTREMELY unlikely, they could draw the same combination several times in a row, since it's not what's on the team by team chart that's being drawn, but the same set of 1-14 balls.)

 

[MintBerryCrunch]

Let's simplify it. There are 4 teams. Each team has one ball with their name on it. At this point, Team D (Detroit) has a 25% chance of getting the #1 pick.
They go ahead and draw and it comes out Team A. All 4 balls are left in the pool, and they draw again for the #2 pick, knowing that if Team A is drawn again it will result in a redraw.
Does Team D still only have a 25% chance to get the #2 pick? I think it's reasonable to say no, as now there are only 3 viable winners... the chances would jump up to 33%.

 

[Red Stanley]

Let's simplify it. There are 4 teams. Each team has one ball with their name on it. At this point, Team D (Detroit) has a 25% chance of getting the #1 pick.
They go ahead and draw and it comes out Team A. All 4 balls are left in the pool, and they draw again for the #2 pick, knowing that if Team A is drawn again it will result in a redraw.
Does Team D still only have a 25% chance to get the #2 pick? I think it's reasonable to say no, as now there are only 3 viable winners... the chances would jump up to 33%.

Yes, as per your example. No, as in the initial chance distribution is a lot different and there a ton of variables involved such as the number of combinations being voided after each draw plus the redraws if they land on a void combination.

 

[MintBerryCrunch]

I understand that. My example was to prove that Detroit's percentage of getting the #2 pick if Buffalo got #1 would be higher than 8.5%, even though the number of winning Detroit combinations (85) does not change. A fun discussion nonetheless.

 

[Red Stanley]

I understand that. My example was to prove that Detroit's percentage of getting the #2 pick if Buffalo got #1 would be higher than 8.5%, even though the number of winning Detroit combinations (85) does not change. A fun discussion nonetheless.

Oh, I'm not disagreeing. Just wondering about the exact numbers myself since my math isn't all that sophisticated.

 

[njx9]

Hm. You made me think about it a bit more... I still think that the chances come out the same, since there are still only X valid combinations, and nothing else matters. I suppose the odds would be different where, if we said "On combination two, Detroit still has 8.5% chances", but given that 18.5% or whatever of the combinations don't matter, the fact that it's combination 2 is also irrelevant. The only next combination that matters is the one that's out of 851 or whatever combinations, and so forth (i.e. they could pull 234,897,289 ball combinations and none of them matter a whit if they don't match a valid, remaining combination).

That said, this feels like one of those math problems where my way feels too easy, so...

Either way, sorry tsweeney, if I seemed overly snarky in my response. Wasn't intended personally.

 

[waltdetroit]

This chart shows the odds for each pick but may not account for whoever is withdrawn... or may

edit add url HV Article

 

[MintBerryCrunch]

Oh, I'm not disagreeing. Just wondering about the exact numbers myself since my math isn't all that sophisticated.

Myself as well. It's especially interesting in year's past where a team with a low percentage at #1 does indeed get #1, and how that affects the rest of the teams that were ahead of them.
Like if Florida gets #1 this year and St Louis #2, how it changes the likelihood for Detroit to pick 3. In that example, we'd be 8.59% to pick #2 after Florida wins and 8.74% to pick #3 after Florida AND St Louis win.
Whereas, if it stays chalk with Buffalo #1 and Ottawa #2, we'd be 10.42% to pick #2 after Buffalo wins and 12.41% to pick #3 after Buffalo AND Ottawa win.

The true % of all scenarios lies somewhere in the middle, which is why we see the 8.7% chance for #2 and 8.9% chance for #3.

 

[Zetterberg4Captain]

Vegas' pick, what's the highest it can be assuming they go out first round?

 

[Ezekial]

Vegas' pick, what's the highest it can be assuming they go out first round?

24-27 depending on who makes the cfs

 

[Zetterberg4Captain]

24-27 depending on who makes the cfs

Okay well let' hope for 24