Odd Coincidence in East

[amin723]

The East actually has a perfect bracket setup by conference standing in spite of the divisional setup.

Wonder how often that will happen. I didn't check any other post re-alignment years so flame away if it happened just last year.


By points:
1. NYR
2. MTL
3. TBL
4. WSH
5. NYI
6. DET
7. OTT
8. PIT

Click to read more...

 

[jniklast]

The East actually has a perfect bracket setup by conference standing in spite of the divisional setup.

Wonder how often that will happen. I didn't check any other post re-alignment years so flame away if it happened just last year.


By points:
1. NYR
2. MTL
3. TBL
4. WSH
5. NYI
6. DET
7. OTT
8. PIT

It happened just last year and the odds are 25% if we use simplified probabilities. So on average it is expected to happen every other year in one of the conferences. BTW this is just the 2nd post-alignment year.

 

[SladeWilson23]

That is pretty neat.

 

[CaliforniaBlues310]

Under the old alignment, the West would go something like this

1 Ducks v Flames 8
2 Blues v Jets 7
3 Predators v Wild 6
4 Blackhawks v Canucks 5

Those would be some interesting matchups. Especially that NSH/MIN series.

 

[IvanMalison]

It happened just last year and the odds are 25% if we use simplified probabilities. So on average it is expected to happen every other year in one of the conferences. BTW this is just the 2nd post-alignment year.

How do you figure that the odds are 25%? I'm assuming that you are putting aside the issue of point distribution when you come up with this number.

The configurations that give
1v8, 2v7, 3v6, 4v5

are of the form

1(X=one of A/M)
2(Y=the other division)
3(K=one of A/M)
4(J=the other division)
5(J)
6(K)
7.(A=one of A/M)
8.(B=one of A/M)

of the six variables used (X, Y, K, J, A, B), four are free to be one of two values, so such a configuration can be obtained in 2^4 = 16 ways.

There are 2^8 = 256 total possible seeding arrangements. 16/256 = 1/16 != 1/4, so unless you are making some assumption about the likelyhood of certain distributions vs others, the probability of this happening is not 1/4.

ALSO, even the odds of it happening in each conference were 25% the chance of it happening would not be 50% -- Thats just not how probability works. You can't just add the odds together to get the total probability that either will happen.

The actual probability that either of two 25% chance events will occur is

1-(the probability that neither will occur) = 1-(3/4*3/4) = 1-(9/16) = 7/16

No one seems to understand even very basic probability around here.

 

[amin723]

*snip*

I had mind to make a similar post, but decided against it. I still appreciate yours nonetheless because my main deterrant was in writing out how it's NOT 25%...also it didn't happen last year.

That being said...I'm not so sure your picture of the combinations is accurate either. It doesn't feel right, but I have no real argument against your description.

I went back to 2010-2011 in the Western Conference to find the last time it happened, but it was a lot more likely in the previous alignment with the 3 division winners and then everybody else. You just had to get #3 in the right spot.

In the current setup you would think the 1 v 8 matchup is always correct, but it's not because the 2 wild cards could have better points numbers than the other division's 1-3. See: Central Division. So it's pretty convoluted how it would need to come together.

It's a decent puzzle. I'm not sure the map you laid out tells the full story.

I think this year is the cleanest possible way to make it work. I think the most convoluted way for it to be fulfilled is below:

#1 from ATL (1) plays #2 WC
#2 from ATL (2) plays #7 from ATL (3) div
#3 from MET (1) plays # 6 who is #1 WC and must be from MET
#4 from MET (2) plays #5 from MET (3)

I also want to add that it's more than just playing the proper seed, they're also in proper order so that 2/7 plays 3/6 in Round 2 and 1/8 plays 4/5.

 

[jniklast]

I was assuming that it should simply be the same bracket as if the old format was used (with 2 divisions obviously) which means, that the first two seeds are always correct. That increases the probability.

I made the following assumption to simplify it:
- the probability for the team in WC1 to have more points than the last divisional playoff team (D6) is 50% if they are from different divisions (otherwise not possible obviously)

That assumption is necessary as there is no solution without knowing the probability that WC1 finishes above D6. 50% seems to be a reasonable number to me, but you can tweak it if you know a better approximation.


So our first two spots are set and always correct. Next are the four divisional playoff spots. The odds for the right distribution is 1/3 (two correct layouts out of six possible).

Now there's the two wildcard spots and the only thing to spoil it is if WC1 has more points than the last placed divisional playoff team (D6). There's four possible permutations of the wildcard spot. Two of those permutations make it possible that the WC1 team has more points than D6 (only those permutations with teams from different divisions in WC1 and last D6). If we use the 50% possibility for each of those from above, that leads to a probability of 1/3*3/4=3/12=1/4

You are of course right, that the probability for that in both conference isn't quite 50%.

 

[Makar Goes Fast]

Under the old alignment, the West would go something like this

1 Ducks v Flames 8
2 Blues v Jets 7
3 Predators v Wild 6
4 Blackhawks v Canucks 5

Those would be some interesting matchups. Especially that NSH/MIN series.

bolded the series i would rather see

maybe calgary and Van i would prefer most, but overall with canucks playing the BHAWKS and the rest divisional... thats some good series

 

[IvanMalison]

I was assuming that it should simply be the same bracket as if the old format was used (with 2 divisions obviously) which means, that the first two seeds are always correct. That increases the probability.

You still can't make that assumption even if you are only comparing it to the old point format. Consider the following regular season (raw point total) finish (A represents a team from one division and M represents a team from another)

1A
2A
3A
4A
5A
6M
7M
8M

The seedings in the current format would end up being

1 v 5, 6 v 4, 2 v 3, 7 v 8

The seedings in the old format would end up being

1 v 8, 6 v 7 (6 would get bumped up as the division leader), 2 v 5 3v4

As you can see, it is not guaranteed that the first two matchups will always be correct.

I made the following assumption to simplify it:
- the probability for the team in WC1 to have more points than the last divisional playoff team (D6) is 50% if they are from different divisions (otherwise not possible obviously)

That assumption is necessary as there is no solution without knowing the probability that WC1 finishes above D6. 50% seems to be a reasonable number to me, but you can tweak it if you know a better approximation.

Instead of just making up numbers, I just counted permutations. This probably overestimates the probability of some of the stranger seeding arrangements, but I would still guess that it is closer to being correct.

So our first two spots are set and always correct. Next are the four divisional playoff spots. The odds for the right distribution is 1/3 (two correct layouts out of six possible).

Now there's the two wildcard spots and the only thing to spoil it is if WC1 has more points than the last placed divisional playoff team (D6). There's four possible permutations of the wildcard spot. Two of those permutations make it possible that the WC1 team has more points than D6 (only those permutations with teams from different divisions in WC1 and last D6). If we use the 50% possibility for each of those from above, that leads to a probability of 1/3*3/4=3/12=1/4

You are of course right, that the probability for that in both conference isn't quite 50%.