SABRES WIN LOTTERY!!!! Will pick #1 overall in the 2018 Draft

[Bryanbryoil]

Congrats guys! You have a solid core going forward. Time for your GM to insulate your youngsters with some quality veterans and things could change for the better in a hurry.

 

[tripleX]

Why do the permutations matter? If you're trying to determine who the number one pick is, and there are only three possible teams that could pick one, and pick one is drawn before picks 2 and 3, and the drawings for pick 2 and 3 in fact depend on pick 1, what relevance are the drawings for picks 2 and 3 to the drawings for pick 1, which is the one that we care about.

You need to take into account the information known before first 3 picks were revealed, ie. the teams in top 3 are Buffalo, Carolina, and Montreal. This is very important in calculating the odds and needs to be emphasized.

Let's see how wrong it is by using (18.5% / (18.5% + 9.5% + 3%)) = 59.7% to calculate the odds Buffalo wins first overall pick (before top 3 were announced).

Among the 18.5% percentage, it also includes the probability that Carolina and Montreal aren't in top 3 and these possibilities needs to be excluded because we already know these two teams are in top 3.

Let's use an example, before today where none of the lottery result is revealed, what is the odd of Buffalo getting 1st pick, Ottawa getting 2nd pick, and Arizona getting 3rd pick? It's 18.5% * (13.5% / (100% - 18.5%)) * (11.5% / (100% - 18.5% - 13.5%)) = ~0.52%. In other words, among the 18.5% chance that Buffalo gets 1st pick, 0.52% is when Buffalo wins 1st pick, Ottawa wins 2nd, and Arizona wins 3rd. Once we know the top 3 teams are Buffalo, Carolina, and Montreal, should we still count this 0.52% chance? No, we don't count it anymore because we know this scenario (Buffalo:1, Ottawa:2, and Arizona:3 ) won't happen.We should subtract 0.52% from 18.5% due to this particular possibility is eliminated and continue to subtract the percentages of other possibilities when top 3 teams aren't Buffalo, Carolina, and Montreal. We also need to do the same to Caroline's and Montreal's percentages based on the information that top 3 teams are known.

Basically we should only take the possibilities of Buffalo, Carolina, and Montreal in top 3 into account and exclude the possibilities otherwise, which is the post misterchainsaw posted in main board.

 

[HaNotsri]

Best swedish threesome since Kristian Huselius, Henrik Tallinder and Andreas Lilja.
yesiknowimoffby1

If that happened in 2018 they’d be locked up for a while.

 

[valet]

Yes, I am. And yes, it does. Buffalo's chance to get the 2nd pick if they did not get the 1st pick, for example, was around 19.7%. (It varies depending by the identity of the team that got the 1st pick, that's the percentage if the identity of the team was not known)

You're ignoring the information we have that Carolina did, in fact, win one of the three draws, which they were much less likely to do than Buffalo. That adjusts significantly their odds vs. a plain look at the original odds. I literally showed mathematically in the post above the one you responded to how your method does not work.

EDIT: Drop the base rate fallacy crap. You're using it wrong. The fact that Carolina made the top three at long odds compared to Buffalo making the top 3 at much shorter odds absolutely does increase the chance that Carolina won the whole thing, because of the fact that all of those draws were made with all 15 teams balls still in the pot. Before we knew anything, was it 6X more likely that Buffalo won the #1? Yes. Once we knew that both Carolina and Buffalo's number hit somewhere in the top 3, it was no longer 6X likely for Buffalo to win over Carolina, because Buffalo was 5X more likely to be in the final 3 in the first place. Carolina has surmounted much of those 6X odds by the fact that we know that one of the 3 numbers drawn was theirs.

DOUBLE EDIT: And yes, Buffalo has 6X the balls in the machine as Carolina, but there are 13 other teams with balls in there. Buffalo only has 18.5% of the balls in the machine. You cannot just pretend the other balls from teams who finished 4-15 weren't there. Carolina only had a 10% chance to get into the top 3. Once we KNOW that they got into the top three, that increases their chance to win the #1 by quite a bit. Their chance to win the #1 given that they're in the top 3 is 3%/10% = 30%. That number fluctuates a little depending on the teams in with them, but it's not a very big fluctuation.

the top 3 being revealed does not change the original lottery probability. there is no 'given that they're in the top 3', the top 3 is decided by the odds, the rest of the picks are determined by points. carolina never increased it's odds of picking #1 by being lucky to be in the top 3 . the original odds hold up with buffalo having a significantly better chance than both montreal and carolina. your argument would also imply that carolina's 69.6% chance of picking 11th got smaller in the lottery because they were in the top 3... while buffalo's 0% chance of picking 11th got larger.... which it clearly didn't, because buffalo was already guaranteed a top 4 pick.

buffalo had about twice the chance at #1 than montreal, and about 5 times the chance than carolina... that is literally how the lottery works, the top 3 teams are determined by the balls and the rest of the teams fall in line after points. you are operating on incorrect assumptions and applying bayes theorem incorrectly (and also improperly to boot)

 

[misterchainsaw]

the top 3 being revealed does not change the original lottery probability. there is no 'given that they're in the top 3', the top 3 is decided by the odds, the rest of the picks are determined by points. carolina never increased it's odds of picking #1 by being lucky to be in the top 3 . the original odds hold up with buffalo having a significantly better chance than both montreal and carolina. your argument would also imply that carolina's 69.6% chance of picking 11th got smaller in the lottery because they were in the top 3... while buffalo's 0% chance of picking 11th got larger.... which it clearly didn't, because buffalo was already guaranteed a top 4 pick.

Um. Carolina's 69.6% chance to pick 11th got MUCH smaller once it was revealed they were in the top 3. It was, in fact, 0.

That 69.6%, along with their other non top-3 odds that we now know did not happen gets spread proportionally to their #1, #2, and #3 chances. Which results in something like a 30-33-37 split, which are marginally affected by the other two teams who are in the top 3 with them.

Buffalo's chance of picking 11th did not get larger because that remained 0. Their chance of picking 4th also became 0.

buffalo had about twice the chance at #1 than montreal, and about 5 times the chance than carolina... that is literally how the lottery works, the top 3 teams are determined by the balls and the rest of the teams fall in line after points. you are operating on incorrect assumptions and applying bayes theorem incorrectly (and also improperly to boot)

At the start of the process, Buffalo had twice the chance as Montreal and 6 times the chance as Carolina. This changes once we know which teams made the top 3. I already showed where your method breaks down. Show me where I'm wrong. I am flat out telling you where your assumption is wrong, and have already shown you that your method results in absurd numbers in some cases.

PS: By brute forcing it, I didn't use Bayes Theroem at all. Stop throwing around terms like you understand them.

 

[tripleX]

the top 3 being revealed does not change the original lottery probability. there is no 'given that they're in the top 3', the top 3 is decided by the odds, the rest of the picks are determined by points. carolina never increased it's odds of picking #1 by being lucky to be in the top 3 . the original odds hold up with buffalo having a significantly better chance than both montreal and carolina. your argument would also imply that carolina's 69.6% chance of picking 11th got smaller in the lottery because they were in the top 3... while buffalo's 0% chance of picking 11th got larger.... which it clearly didn't, because buffalo was already guaranteed a top 4 pick.

buffalo had about twice the chance at #1 than montreal, and about 5 times the chance than carolina... that is literally how the lottery works, the top 3 teams are determined by the balls and the rest of the teams fall in line after points.

The information of top 3 teams does change the odds. Let's use an even simpler example, before top 2 picks were revealed, what is the chance of Buffalo winning 1st overall? Is it 18.5% vs. 3%, which is (18.5% / (18.5% + 3%)) = 86%?

Of the 18.5% chance Buffalo winning the lottery, we can break it down to two possibilities, (1) "Buffalo 1st, Carolina 2nd", and (2) "Buffalo 1st, Carolina not 2nd". What are the chances of these two possibilities, they are:

(1) Buffalo 1st, Carolina 2nd: 18.5% * (3% / (100% - 18.5%)) = 0.68%
(2) Buffalo 1st, Carolina not 2nd: 18.5% * (100% - (3% / (100% - 18.5%))) = 17.82%

Note that 17.82% + 0.68% = 18.5%.

So when we know the top 2 teams are Buffalo and Carolina, should we still use 18.5% to calculate the odds? No, we shouldn't because we already know (2) won't happen once we know Carolina is in top 2. Thus, we should only use 0.68% in (1).

Let's continue to calculate Carolina's percentage:

(3) Carolina 1st, Buffalo 2nd: 3% * (18.5% / (100% - 3%)) = 0.57%
(4) Caroline 1st, Buffalo not 2nd: 3% * (100% - (18.5% / (100% - 3%))) = 2.43%

Should we still use 3% for Carolina? No, we shouldn't because we know (4) won't happen. Thus, we should only use 0.57% in (3).

Therefore, before top 2 picks were revealed, Buffalo winning chance is (0.68% / (0.68% + 0.57%)) = 54.4%, which is much lower than most people thought.

 

[misterchainsaw]

The information of top 3 teams does change the odds. Let's use an even simpler example, before top 2 picks were revealed, what is the chance of Buffalo winning 1st overall? Is it 18.5% vs. 5%, which is (18.5% / (18.5% + 5%)) = 78.7%?

Of the 18.5% chance Buffalo winning the lottery, we can break it down to two possibilities, (1) "Buffalo 1st, Carolina 2nd", and (2) "Buffalo 1st, Carolina not 2nd". What are the chances of these two possibilities, they are:

(1) Buffalo 1st, Carolina 2nd: 18.5% * (5% / (100% - 18.5%)) = 1.13%
(2) Buffalo 1st, Carolina not 2nd: 18.5% * (100% - (5% / (100% - 18.5%))) = 17.37%

Note that 17.37% + 1.13% = 18.5%.

So when we know the top 2 teams are Buffalo and Carolina, should we still use 18.5% to calculate the odds? No, we shouldn't because we already know (2) won't happen. Thus, we should use 1.13% in (1).

Let's continue to calculate Carolina's percentage:

(3) Carolina 1st, Buffalo 2nd: 5% * (18.5% / (100% - 5%)) = 0.97%
(4) Caroline 1st, Buffalo not 2nd: 5% * (100% - (18.5% / (100% - 5%))) = 4.03%

Should we still use 5% for Carolina? No, we shouldn't because we know (4) won't happen. Thus, we should use 0.97% in (3).

Therefore, before top 2 picks were revealed, Buffalo winning chance is (1.13% / (1.13% + 0.97%)) = 53.8%, which is much lower than most people thought.

That's all great, but the Canes original odds were 3% ;-).

Method is correct, the real numbers end up crunching to 54.34% (I made an excel file, lol).

 

[tripleX]

That's all great, but the Canes original odds were 3% ;-).

Method is correct, the real numbers end up crunching to 54.34% (I made an excel file, lol).

Ha, let me edit the post to use correct 3% in calculation. The point stays the same that the odds are affected by information revealed on what top 2 or 3 teams are.

Original post is now edited.

 

[Icicle]

I have a goober at my work that still says NHL rigged the McDavid draft. Even after i logically explianed that if the NHL did rig it, it wouldn't be for a small market near the Arctic circle.

You’re thinking about it wrong. NHL is about money. The biggest money out there are the TV contracts: USA and Canada have separate ones. Prior to the Canada one being renewed is when you saw Edmonton and Toronto win the lotteries and get franchise players. Last summer the deal was renewed and now it doesn’t matter for a while which market gets the franchise player.

 

[gallagt01]

It wasn't a dream, guys. They really won it.

 

[26CornerBlitz]

 

[sabresfan129103]

I feel like I'm living in some alternate reality where the Sabres won the lottery. This stuff doesn't happen to Buffalo. It still doesn't feel real.

 

[old kummelweck]

I went to the Montreal board and they are arguing over who to take... Zadina or Tkachuk.

There by the grace of god go I.

 

[OcAirlines]

Whoooooaaaa I didn't realize that this happened last night, thought it would be today! Well that's a nice change of pace and some luck for once

 

[OkimLom]

Kind of feel I took the sacrifice last night to win the lottery. Dealing with a tooth infection all night i missed the lottery....

 

[Jim Bob]

Hockey Twitter reacts to NHL Draft Lottery results

 

[BananaSquad]

What a weekend for Buffalo!

 

[brian_griffin]

So do they tank again next year too? That's like four years running, right?

they have not tanked and certainly not for 4 years. They intentionally iced a marginally competitive roster in 2014-15. Improved in ‘15-16 and 16-17, and unexpectedly took a step back this season. Many expected them to challenge for a wild card or at least not be in the basement of the east. Bad coaching, limited d-depth, injuries, and a bottom-6 of AHL-NHL tweeners contributed to their underachieving season.

I could make a snide remark that as an Oilers fan you know full well the experience of drafting first year over year yet not improving the following season despite trying. I could also make a snide remark that unlike the Oil, the Sabres finished last 3 seasons in 5, yet only drafted 1st once, unlike the Oil who drafted first overall 4 times in 6 years despite only once being last in the standings one time. But as an Oilers fan you know this.

 

[varano]

Just wanted to say congrats to the sabres fan base. This will turn the franchise around big time.

 

[26CornerBlitz]

TSN: Sabres win top pick in NHL Draft; Habs move up to No. 3
TORONTO — The Buffalo Sabres will select first overall at the NHL Draft for the first time in 31 years after winning Saturday's lottery. Buffalo entered the lottery with the best odds at 18.5 per cent and become the 16th team to win the sweepstakes.

"It's awesome, it's a great feeling. I wasn't too nervous about the whole process," said Sabres general manager Jason Botterill. "I hope (the fans) are ecstatic. They've been extremely supportive. You felt in the building, a lot of quietness coming into the games, but whenever we did something they wanted to cheer, they wanted to be supportive."

The Sabres have picked first overall twice previously taking Pierre Turgeon in 1987 and selecting Gilbert Perreault in 1970.

 

[Gabrielor]

I'm waking up still in shock that this is real.

 

[mmalady]

congrats guys !!!!

 

[La Cosa Nostra]

Passed out at 645 last night since I didn't sleep the night before and woke up to pandemonium on my phone...had to fully wake up and pinch myself to make sure this wasn't part of a dream.

We actually won a draft lotto? And for arguably the best D prospect ....ever?

 

[boots electric]

I heard that the Canucks asked if their mascot could represent the team at the lottery and were turned down, but now I can't stop thinking about the Sabres sending Rob Ray

 

[sincerity0]

Watching people say the probabilities didn't change from the original after gaining information on the top 3 teams is so hilarious. It may be more entertaining than watching Dahlin/Eichel/Mittelstadt in every OT game for the next 15 years