Post-Game Talk: Mission Accomplished

[Ciao]

So... who do we pick with our shiny 4th overall selection?

That will be the best kept secret on Bay St.

At least the Bruins will not have to decide how to use the Leafs pick this time around.

 

[schenneuf]

Where's Hunter/Lou/Shanahan vid?

 

[TorMapleJays]

I'm happy that they accomplished exactly what i hoped they would, 30th place and certain players traded away.

The lotto is pure dumb luck , it would be childish to get bothered by it's out come

This is nothing more then a stick at 17 on a blackjack table.

I hit on soft 17.. I know...

 

[Kelly]

Where's Hunter/Lou/Shanahan vid?

Shanahan
Lamoriello

 

[Macallan18]

I hit on soft 17.. I know...

depending on what the dealer is showing, that is perfectly acceptable.
BUT, if he is showing a 6................... not so much

 

[Macallan18]

That's coming from a realist who has said numerous times there's almost a 75% chance we pick 4th.

Most numbers published about the lottery percentages are not correct.
Ok, first pick is a 20% for us, lesser for everyone else.
BUT after that, our odds are dependent on WHO wins the first lottery.
As an example, if Edmonton wins, they are not elegible obviously to win any further slots.
So their percentage at the second pick goes into the pool and gets divided up, and we get 20% of that.
HOWEVER, if someone at the very bottom with a 1% chance wins the first pick, our odds are pretty much what has been printed in the various tables.
Haven't worked out the actual numbers (hey, Sunday afternoon, having some scotch) but they are AT LEAST better then 50% that we will pick in the top three, and can move up a few points from the numbers that are published by math challenged folks.

lol, edit, your friend, can he count?

 

[Halla]

All 14 teams have a chance and even the Bruins with 93 points and missed the playoffs on a tie breaker do have a chance to select Matthews.

The odds of us picking 4th is almost a coin flip 50/50 (52.5% & verses 47.5%)

Here is a great table the shows the lottery odds : http://www.pensionplanpuppets.com/2...-nhl-draft-lottery-final-rules-and-odds-table.

Leafs odds for Toronto .. 1st = 20% ... 2nd =17.49% ... 3rd = 15.02% ... 4th = 47.5%

So picking 4th has the highest odds of happening.

Last year the Leafs finished 4th last and drafted 4th and this year we finished 30th and last and could still pick 4th overall.

4th is the most likely individually....but we are more likely to pick top 3 than 4th....

our worst case scenario is picking a guy that has 38g/85a/123pts in 64 games, whose father put up 500+g/1000+pts in the NHL

 

[pucky]

They chose the wrong year to finish last overall?

 

[Macallan18]

The odds of us picking 4th is almost a coin flip 50/50 (52.5% & verses 47.5%)

Here is a great table the shows the lottery odds : http://www.pensionplanpuppets.com/2...-nhl-draft-lottery-final-rules-and-odds-table.

Leafs odds for Toronto .. 1st = 20% ... 2nd =17.49% ... 3rd = 15.02% ... 4th = 47.5%

So picking 4th has the highest odds of happening.

No, the table is being wrongly interpreted.
It doesn't factor in that after the first pick, a team drops out of the running and it's percentages on the second pick and third pick go back into the pool. Ditto for the team that wins the second pick, their percentages on the third pick go back into the pool.
Now, the odds are still the greatest at picking fourth AS A SINGLE RESULT, but the odds are BETTER than 52.5% that we will pick in the top three.

 

[Leafsman]

Most numbers published about the lottery percentages are not correct.
Ok, first pick is a 20% for us, lesser for everyone else.
BUT after that, our odds are dependent on WHO wins the first lottery.
As an example, if Edmonton wins, they are not elegible obviously to win any further slots.
So their percentage at the second pick goes into the pool and gets divided up, and we get 20% of that.
HOWEVER, if someone at the very bottom with a 1% chance wins the first pick, our odds are pretty much what has been printed in the various tables.
Haven't worked out the actual numbers (hey, Sunday afternoon, having some scotch) but they are AT LEAST better then 50% that we will pick in the top three, and can move up a few points from the numbers that are published by math challenged folks.

lol, edit, your friend, can he count?

Everyone knows how the odd system works as it has been posted all over and we've already gone through this. So coming in far after it's done while saying the same thing then rudely insinuating someone can't count is just insulting.

The 50% people are discussing incorporate a formula in which predicts the outcome of the entire situation as a whole from this point in time. Ie: it sums the possibility of the odds of the three rounds while eliminating the outcome of winning the round before it (incorporating the multiplier .8).

In real life, the odds of winning the first round have nothing to do with the odds of winning the second round. Meaning you can not carry the 80% chance from teh first round.

The only affect the first round has on the second round is the redistribution of the weight allocated to whomever won. Ie: If Edmonton wins their heavier weight is redistributed more than if Montreal wins.

Come time of the draft lottery in no round will Toronto have odds better than 20-25%. So while predicting outcomes today you can say there is 50% chance of us winning, during the actual lottery round there is only 20-25% chance of winning.

If you buy 20 lottery tickets to one lottery, 17 to another and 15 to a third which all have 1 in 100 odds. You don't have a 50% chance of winning the lottery. You have a 20% chance of winning the first lottery, 17% chance of winning the second and 15% chance of winning the third.

If you want to see this demonstrated - go to the NHL lottery Simulator and run it ONCE, do not run it multiple times. Run it once and that is all. If you run it multiple times the numbers will start to align to the percentages discussed.

People are calculating that if you run the scenario and calculate the results you will see that around 50% of the time Toronto wins a top 3. However, this event will not occur multiple times, it will occur only once.

If in any given round, Toronto never has a better weighted advantage of 20% of the balls. How can you possibly state they are 50% likely to win????

 

[Ovate]

No, the table is being wrongly interpreted.
It doesn't factor in that after the first pick, a team drops out of the running and it's percentages on the second pick and third pick go back into the pool. Ditto for the team that wins the second pick, their percentages on the third pick go back into the pool.
Now, the odds are still the greatest at picking fourth AS A SINGLE RESULT, but the odds are BETTER than 52.5% that we will pick in the top three.

Actually, that's exactly what the table factors in. That's why the table is a little more complicated. We have exactly a 52.5% chance of picking in the top 3 (and a 47.5% chance of picking 4th).

 

[Pyromaniac3]

No, the table is being wrongly interpreted.
It doesn't factor in that after the first pick, a team drops out of the running and it's percentages on the second pick and third pick go back into the pool. Ditto for the team that wins the second pick, their percentages on the third pick go back into the pool.
Now, the odds are still the greatest at picking fourth AS A SINGLE RESULT, but the odds are BETTER than 52.5% that we will pick in the top three.

I am pretty sure all of the factors you mentioned are in the the calculation. So 2nd and 3rd overall odds are the weighted average of all scenarios.

 

[Ciao]

No, the table is being wrongly interpreted.
It doesn't factor in that after the first pick, a team drops out of the running and it's percentages on the second pick and third pick go back into the pool. Ditto for the team that wins the second pick, their percentages on the third pick go back into the pool.
Now, the odds are still the greatest at picking fourth AS A SINGLE RESULT, but the odds are BETTER than 52.5% that we will pick in the top three.

I've already demonstrated my complete and utter ignorance on this topic, but the published 52.5% chance of a top-three pick sounds about right to me.

The engineers in the room will know that probability of a safety device failure is the product of the probability of failure of each independent safety mechanism. For example, if an automobile has independent front and rear brakes each with a probability of failure of 1%, then the likelihood of both front and rear brakes failing is 1% of 1%, or 1/100 of 1%.

I wonder if if the probability of the last place team NOT winning each round might work along those lines.

If there were three draws, and the last place team had exactly an 80% chance of not winning each draw, the the chance of not winning the first draw would be 80%; and of not winning the first or second draws would be 80% of 80% or 64%; and of not winning the first, second or third draws would be 80% of that, which would be about 51.2%.

If you add into the equation that the probability of loss will change after each of the first and second draws depending on which teams are eliminated from the pool, then a 47.5% chance of not winning the first, second or third draws sounds about right to me.

Mind you, I wouldn't know the actual answer because I'm not capable of performing the math or knowing if it's right. It just sounds right to me -- that's all.

 

[Leafsman]

I've already demonstrated my complete and utter ignorance on this topic, but the published 52.5% chance of a top-three pick sounds about right to me.

The engineers in the room will know that probability of a safety device failure is the product of the probability of failure of each independent safety mechanism. For example, if an automobile has independent front and rear brakes each with a probability of failure of 1%, then the likelihood of both front and rear brakes failing is 1% of 1%, or 1/100 of 1%.

I wonder if if the probability of the last place team NOT winning each round might work along those lines.

If there were three draws, and the last place team had exactly an 80% chance of not winning each draw, the the chance of not winning the first draw would be 80%; and of not winning the first or second draws would be 80% of 80% or 64%; and of not winning the first, second or third draws would be 80% of that, which would be about 51.2%.

If you add into the equation that the probability of loss will change after each of the first and second draws depending on which teams are eliminated from the pool, then a 47.5% chance of not winning the first, second or third draws sounds about right to me.

Mind you, I wouldn't know the actual answer because I'm not capable of performing the math or knowing if it's right. It just sounds right to me -- that's all.

the outcome of not winning the first round has no bearing on whether or not you'd win the second round except for the redistribution of the weighting.

You would still be 80% likely to not win the following round.

Determing our odds based on today is irrelevant. The only odds that matter are the ones each time the lottery is run, at that point the results from the previous round have zero impact.

 

[Gary Nylund]

Most numbers published about the lottery percentages are not correct.
Ok, first pick is a 20% for us, lesser for everyone else.
BUT after that, our odds are dependent on WHO wins the first lottery.
As an example, if Edmonton wins, they are not elegible obviously to win any further slots.
So their percentage at the second pick goes into the pool and gets divided up, and we get 20% of that.
HOWEVER, if someone at the very bottom with a 1% chance wins the first pick, our odds are pretty much what has been printed in the various tables.
Haven't worked out the actual numbers (hey, Sunday afternoon, having some scotch) but they are AT LEAST better then 50% that we will pick in the top three, and can move up a few points from the numbers that are published by math challenged folks.

lol, edit, your friend, can he count?

Makes sense to me. Figuring out the odds of us picking second is thus quite complicated and calculating the odds of us picking third even more so because there are just so many possibilities.

 

[crump]

I've already demonstrated my complete and utter ignorance on this topic, but the published 52.5% chance of a top-three pick sounds about right to me.

The engineers in the room will know that probability of a safety device failure is the product of the probability of failure of each independent safety mechanism. For example, if an automobile has independent front and rear brakes each with a probability of failure of 1%, then the likelihood of both front and rear brakes failing is 1% of 1%, or 1/100 of 1%.

I wonder if if the probability of the last place team NOT winning each round might work along those lines.

If there were three draws, and the last place team had exactly an 80% chance of not winning each draw, the the chance of not winning the first draw would be 80%; and of not winning the first or second draws would be 80% of 80% or 64%; and of not winning the first, second or third draws would be 80% of that, which would be about 51.2%.

If you add into the equation that the probability of loss will change after each of the first and second draws depending on which teams are eliminated from the pool, then a 47.5% chance of not winning the first, second or third draws sounds about right to me.

Mind you, I wouldn't know the actual answer because I'm not capable of performing the math or knowing if it's right. It just sounds right to me -- that's all.

Mind numbing. I think your chances go up with each ball...so I would imagine it's more like 53.5 to 47.5 between the last two teams seeing as the leafs as 6.5 % chance more than the next team. Heheh.

 

[Ciao]

I am pretty sure all of the factors you mentioned are in the the calculation. So 2nd and 3rd overall odds are the weighted average of all scenarios.

This also sound right -- that the tables would be calculated based on all possible outcomes before a single ball is drawn. If so, they are a priori odds.

After the first draw, the a priori odds of the second and third draws will go up or down based on the actual outcome of the first draw. If you knew in advance which team would win each round, you could recalculate the odds of each subsequent round using the same formula with different variables. If the the second-last team gets the first-overall pick, then the Leafs chance of getting a top-three pick would go up; but if by chance the team with the least chance of getting the first-overall wins it then the Leafs chance of getting a top-three pick would go down.

I read the tables as taking every possible outcome into account before a single ball is drawn.

 

[teeder333*]

Shanahan, what a job he has done. Have we seen ONE Leaf jersey thrown onto the ice? And we finished lower that where we did last year.

Babcock's best answer could have come quicker, his biggest thrill this year should have been the way MLSE new braintrust got the media and the fans to buy into the new mantra, there will be pain, but after, there will be gain.

But having said all that,man Babcock is just outstanding, just a great hokey dokey orator, sounds like he came off a farm a year ago, from anywhere from Manitoba to Iowa. Guy just has the skill to get the viewer to buy in.

Lets face it I am sure he is good enough to get a room full of players to buy in as well and we saw that.

I think another crappy year would be good to really stock up, but I really don't see this guy accepting easily, another year of being out of the playoffs, but money talks, and maybe he does buy in

If we finish near last again next year, our farm system will be sick our ECHL team might make the AHL playoffs.

 

[Ciao]

One tank is enough for me.

 

[Gary Nylund]

the outcome of not winning the first round has no bearing on whether or not you'd win the second round except for the redistribution of the weighting.

You would still be 80% likely to not win the following round.

Determing our odds based on today is irrelevant. The only odds that matter are the ones each time the lottery is run, at that point the results from the previous round have zero impact.

This seems clearly wrong. No matter who wins the 1st pick, assuming it's not us, our chances of winning the 2nd pick will be slightly more than 20%, how much more depends on who wins the first pick.

There are 14 teams in the lottery for the 1st pick, there will only be 13 in the running for the 2nd pick. That alone makes it obvious that all those 13 teams will have a slightly better chance at winning the 2nd pick than they did the 1st pick. If a team with say a 1% chance wins the 1st pick than the effect is quite small but if a team with a 10% chance wins it, essentially that 10% gets divided up between the other 13 teams for the next pick.

Or you can think about it another way, if Boston wins the 1st pick, that doesn't change things much at all. But if Edmonton wins, our odds of winning the 2nd pick go up as our main competition is eliminated. We would still have say 20 ping pong balls left, but now the total is only 90 or whatever as opposed to 99 or whatever if Boston had won.

What's more likely, winning with 20 chances out of 90 or 20 changes out of 99?

 

[Leafsman]

This seems clearly wrong. No matter who wins the 1st pick, assuming it's not us, our chances of winning the 2nd pick will be slightly more than 20%, how much more depends on who wins the first pick.

There are 14 teams in the lottery for the 1st pick, there will only be 13 in the running for the 2nd pick. That alone makes it obvious that all those 13 teams will have a slightly better chance at winning the 2nd pick than they did the 1st pick. If a team with say a 1% chance wins the 1st pick than the effect is quite small but if a team with a 10% chance wins it, essentially that 10% gets divided up between the other 13 teams for the next pick.

Or you can think about it another way, if Boston wins the 1st pick, that doesn't change things much at all. But if Edmonton wins, our odds of winning the 2nd pick go up as our main competition is eliminated. We would still have say 20 ping pong balls left, but now the total is only 90 or whatever as opposed to 99 or whatever if Boston had won.

What's more likely, winning with 20 chances out of 90 or 20 changes out of 99?

Your right that it likely won't be 80% but without knowing the results you can not say for sure. It changes anywhere from 75ish% to 78ish%. I just generalized and used the 80% the poster was using in his example.

I was also referring to carrying over the odds in each round which is how people are deriving at the 52%

 

[Mess]

This seems clearly wrong. No matter who wins the 1st pick, assuming it's not us, our chances of winning the 2nd pick will be slightly more than 20%, how much more depends on who wins the first pick.

There are 14 teams in the lottery for the 1st pick, there will only be 13 in the running for the 2nd pick. That alone makes it obvious that all those 13 teams will have a slightly better chance at winning the 2nd pick than they did the 1st pick. If a team with say a 1% chance wins the 1st pick than the effect is quite small but if a team with a 10% chance wins it, essentially that 10% gets divided up between the other 13 teams for the next pick.

Or you can think about it another way, if Boston wins the 1st pick, that doesn't change things much at all. But if Edmonton wins, our odds of winning the 2nd pick go up as our main competition is eliminated. We would still have say 20 ping pong balls left, but now the total is only 90 or whatever as opposed to 99 or whatever if Boston had won.

What's more likely, winning with 20 chances out of 90 or 20 changes out of 99?

Leafs odds for Toronto .. 1st = 20% ... 2nd =17.49% ... 3rd = 15.02% ... 4th = 47.5%

The odds drop per round if Leafs don't win.

 

[BigBlu]

Leafs odds for Toronto .. 1st = 20% ... 2nd =17.49% ... 3rd = 15.02% ... 4th = 47.5%

The odds drop per round if Leafs don't win.

No.

If Leafs don't win 1st, odds go up for each subsequent pick. Your numbers factor in a potential 1st overall win.

 

[Leafsman]

Leafs odds for Toronto .. 1st = 20% ... 2nd =17.49% ... 3rd = 15.02% ... 4th = 47.5%

The odds drop per round if Leafs don't win.

Yesssss!!! This is it!!! Nice! You can't though turn around and say there is a 52% chance of winning because each round is independent of teh one prior. It is not cumulative.

The odds are more accurately reflected as 2:2:1.5:4.5 (rounding to the nearest .5). We are over 2x more likely to pick 4th than 1st.

Awesome!!! That was the best way to display it. The drop in odds reflects the notion of not picking because we had won the previous round.

 

[BigBlu]

Odds must go up in each subsequent round with less teams remaining.

We're only given 17.5% chance of picking 2nd at this point, because we may indeed be eliminated from picking 2nd... because we already won 1st.