Post-Game Talk: Mission Accomplished

[Leafsman]

No.

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

Not if you factor in the 20% odds of us not even picking at all in that round having won the round previous. That is why after picking 4th, picking 1st has the best odds because there is a possibility in rounds 2 and 3 we would not be involved because we've already won.

Your talking about the redistributed weighting but that can not be determined at this point and is very small compared to teh carried over weight of us not picking at all.

 

[Gary Nylund]

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.

2nd = 17.49% ... that factors in that 20% of the time we win the first pick and our odds of getting the 2nd pick in that case are 0. If we don't win the 1st pick, then the odds of us getting the 2nd pick at that moment go up to over 20%.

 

[Leafsman]

2nd = 17.49% ... that factors in that 20% of the time we win the first pick and our odds of getting the 2nd pick in that case are 0. If we don't win the 1st pick, then the odds of us getting the 2nd pick at that moment go up to over 20%.

The odds of 2nd can be shown more like

17.49% winning
2.5% not involved having won 1st overall
80% not winning


*not including redistributed weighting system

 

[BigBlu]

~~~~~

 

[Pyromaniac3]

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.

Nope, those are weighted averages of all scenarios before the lottery. We have 200 balls out of a 1000. If EDM wins, their 135 balls get removed. So now the odds are 200/865. All the scenarios are simulated (teams from 2-14 winning) and their probabilities are averaged with weights. Then we come up with the odds of winning 2nd overall = 17.49%.
A good explanation is here: http://canucksarmy.com/2015/12/9/let-s-talk-about-the-2016-draft-lottery

Leafsman, when people say we have 52% chance of being top 3 it means, the odds of winning 1st or 2nd or 3rd is 52%. Simple example would be how likely am I going to get one tails if we flip the coin 4 times. What is the probability that all are heads? 0.5^4 = 6.25% which means at least one tail is 93.75%.

 

[go_leafs_go02]

Forgot that last night's game is the last time we see that logo worn on the ice, minus any future retro nights.

 

[BigBlu]

If we keep losing and (for example), our odds go 20%, 24%, 26%... this =/= a 70% chance of picking top 3. In the same way that flipping a coin 2 times in a row =/= 100% chance of hitting a heads.

 

[Ciao]

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.

I think these numbers are correct, but there might be another way of interpreting them.

What if you asked yourself this question: "What are the odds that the Leafs will get its pick after this round?"

According to these tables, the answer is as follows:

Round 1: 80%.
Round 2: 62.51%
Round 3: 47.5
Round 4: 0%

If that's true, then the odds of winning round 1 are 20%.

If they don't win round 1, then the odds of winning round 2 are 37.49%.

If they don't win rounds 1 or 2, then the odds of winning round 3 are 52.5%.

And, of course, those are all a priori odds that will go up or down after each round.

One way or another, they are very likely to have a top-three pick which is important in this particular draft.

 

[Leafsman]

Nope, those are weighted averages of all scenarios before the lottery. We have 200 balls out of a 1000. If EDM wins, their 135 balls get removed. So now the odds are 200/865. All the scenarios are simulated (teams from 2-14 winning) and their probabilities are averaged with weights. Then we come up with the odds of winning 2nd overall = 17.49%.
A good explanation is here: http://canucksarmy.com/2015/12/9/let-s-talk-about-the-2016-draft-lottery

Leafsman, when people say we have 52% chance of being top 3 it means, the odds of winning 1st or 2nd or 3rd is 52%. Simple example would be how likely am I going to get one tails if we flip the coin 4 times. What is the probability that all are heads? 0.5^4 = 6.25% which means at least one tail is 93.75%.

Your not flipping a coin 4x's.

Yoru flipping a coin once 3x's in a row. The prior coin flip has no consequence on the one following.

People are falsely accumulating the odds.

We have a 20% chance of winning or 1:5.

People are saying there are 3 draws which means we have a chance of winning 3:5 (subtract times you won and you'd get your 52%).

This is false. The odds do not increase to 3:5, they remain 1:5 3 times.

 

[Ciao]

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).

I think the actual math behind the table is very complicated (because it takes into account all possible outcomes), but the outcome is simple: at this point in time, the Leafs really do have 52.5% chance of a top-three pick, just as you say

 

[Budsfan]

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.

I'm a little confused with those percentages, we are guaranteed 4th, so that would be a 100% sure thing, unless they are working backwards calculating we should be picking in the first 3 and so our odds are 52.5% to pick in that group of 3.

I admit I'm not a great mathematician but I still don't understand that calculation, unless that's the way it's calculated and if you add up 20%+17.49%+15.02%=52.51% and because with each position 1-3 being picked by another team our chances are reduced to pick in the top 3, but it is confusing.

 

[Leafsman]

I'm a little confused with those percentages, we are guaranteed 4th, so that would be a 100% sure thing, unless they are working backwards calculating we should be picking in the first 3 and so our odds are 52.5% to pick in that group of 3.

I admit I'm not a great mathematician but I still don't understand that calculation, unless that's the way it's calculated and if you add up 20%+17.49%+15.02%=52.51% and because with each position 1-3 being picked by another team our chances are reduced to pick in the top 4, but it is confusing.

Why are you adding the odds from the first round and incorporating them in the second???

If you lose the first round, the odds do not carry over. They are erased. We do not gain any advantage from our first draw loss what soever so adding them into the overall odds is wrong.

 

[Semantics]

Nope, those are weighted averages of all scenarios before the lottery. We have 200 balls out of a 1000. If EDM wins, their 135 balls get removed. So now the odds are 200/865. All the scenarios are simulated (teams from 2-14 winning) and their probabilities are averaged with weights. Then we come up with the odds of winning 2nd overall = 17.49%.
A good explanation is here: http://canucksarmy.com/2015/12/9/let-s-talk-about-the-2016-draft-lottery

Nothing you said is wrong, but I'm not sure you're interpreting Mess's point correctly. The odds of getting top 3 are 52.5% right now, but if we don't win the lottery for #1, then the probability of top three (before the lotteries for #2 and #3) will be lower than that because there are only two chances left to win.

If you flip a coin four times, the odds of at least one tail are 93.75% as you said, but if the first flip comes up heads then the odds of at least one tail in the remaining three flips is 1 - (0.5)^3 = 87.5%. I believe that is the point Mess was making.

That said, I don't think there's much value in thinking about these conditional probabilities. As far as I know, the results of all three lotteries will be revealed at once in the form of the draft order. So we'll never be presented with the situation where the Leafs lost the #1 and there are two lotteries remaining. It makes more sense to think of all three drawings as one simultaneous event. 20%, 17.5%, 15%, 47.5%.

 

[Pyromaniac3]

Your not flipping a coin 4x's.

Yoru flipping a coin once 3x's in a row.

What's the difference between these statements?

The prior coin flip has no consequence on the one following.

You're correct. But in draft lottery case, it does because you're eliminating possibilities of the team winning the 1st overall lottery.

People are falsely accumulating the odds.

We have a 20% chance of winning or 1:5.

People are saying there are 3 draws which means we have a chance of winning 3:5 (subtract times you won and you'd get your 52%).

This is false. The odds do not increase to 3:5, they remain 1:5 3 times.

The odds are not increasing as we draw the lottery. The odds before lottery starts of picking top 3 is 52%. If we lose first overall, the odds are less of picking in the top 3 because the possibility of winning first is gone.

The other thing is the table shows odds that are independent of each other. So it's basically like a series of coin flips where the odds are independent of each other.


I don't know if you know how the lottery works but here it is:
There are 1000 balls where each team is allocated a percentage of balls according to the lottery odds. So TOR = 200, EDM = 135, etc. So if EDM wins the first lottery, EDM balls are removed meaning total number of balls left is 1000-135 = 865. But TOR still has 200 balls remaining. So new odds are 200/865 = 23.1%.

 

[Leafsman]

I'm not sure you're interpreting Mess's point correctly. The odds of getting top 3 are 52.5% right now, but if we don't win the lottery for #1, then the probability of top three (before the lotteries for #2 and #3) will be lower than that because there are only two chances left to win.

If you flip a coin four times, the odds of at least one tail are 93.75%, but if the first coin comes up heads then the odds of at least one tail in the remaining three flips drops to 1 - (0.5)^3 = 87.5%. I believe that is the point Mess was making.

That said, I don't think there's much value in thinking about these conditional probabilities. As far as I know, the results of all three lotteries will be revealed at once in the form of the draft order. So we'll never be presented with the situation where the Leafs lost the #1 and there are two lotteries remaining.

I believe what Mess is saying and this is my last thought on the matter is the odds are as follows:

1 st - 20%
2nd - 17.5%
3 rd - 15%
4th - 47.5%

You can not sum teh odds of the three rounds and say you have 52% chance of winning because at no point do the three rounds interact with each other. Each round is an event independent of itself with no bearing on the following round (except redistributed weightings),

At the time of the 1st draw - we will have a 20% chance of winning.

Going into the second draw There will be a 17.5% chance we will be winning, a 2.5% chance we will be sitting with the 1st overall, and a 80% chance of not winning.

Going into the third draw. there will be a 15% chance we win, a 5% chance we will have either th 1st or the 2nd and an 80% chance we will not win.

Based on this the most likely outcome is picking 4th which is the remainder of 20,17.5 and the 15% or 47.5%.

At no time does our odds ever increase above low 20%'s and because of this it is over 2x's more likely we will be picking 4th than picking at any other percentage.

 

[Pyromaniac3]

Nothing you said is wrong, but I'm not sure you're interpreting Mess's point correctly. The odds of getting top 3 are 52.5% right now, but if we don't win the lottery for #1, then the probability of top three (before the lotteries for #2 and #3) will be lower than that because there are only two chances left to win.

If you flip a coin four times, the odds of at least one tail are 93.75% as you said, but if the first flip comes up heads then the odds of at least one tail in the remaining three flips is 1 - (0.5)^3 = 87.5%. I believe that is the point Mess was making.

Yes overall odds of being in the top 3 drops. But the odds of winning 2nd is now around 23% rather than the 17% he said it was.

 

[Leafsman]

What's the difference between these statements?



You're correct. But in draft lottery case, it does because you're eliminating possibilities of the team winning the 1st overall lottery.



The odds are not increasing as we draw the lottery. The odds before lottery starts of picking top 3 is 52%. If we lose first overall, the odds are less of picking in the top 3 because the possibility of winning first is gone.

The other thing is the table shows odds that are independent of each other. So it's basically like a series of coin flips where the odds are independent of each other.


I don't know if you know how the lottery works but here it is:
There are 1000 balls where each team is allocated a percentage of balls according to the lottery odds. So TOR = 200, EDM = 135, etc. So if EDM wins the first lottery, EDM balls are removed meaning total number of balls left is 1000-135 = 865. But TOR still has 200 balls remaining. So new odds are 200/865 = 23.1%.

No one's arguing the increase in weighting?!?! You can't account for something you don't know. At best it accounts for a few percent added each subsequent round.

What I am saying is Toronto does not carry over the 200 balls from the first round and add it to the 200 balls from the second round.

So when you say we have a 52% chance of winning, where are you getting the 520 balls from????

 

[Ciao]

Why are you adding the odds from the first round and incorporating them in the second???

If you lose the first round, the odds do not carry over. They are erased. We do not gain any advantage from our first draw loss what soever so adding them into the overall odds is wrong.

You are right that the outcome of one round doesn't affect the probable outcome of the next.

Keeping in mind that I am probably a moron at this, here is one way I've come to think of it:

Imagine that you try something three times in a row, with a 20% chance of winning on each try, and you repeat that process 100 times.

You would expect that in 20 times out of 100, you would have succeeded on the first try and would not go on to the second or third tries.

Of the 80 times you failed on the first try, you would also fail 80% of the time on the second try. Therefore, you would expect to succeed 16 times and fail 64 times in those 80 tries.

Of the 64 times that you failed on the first and second tries (combined), you would also expect to fail 80% of the time on the third try, resulting in success about 13 times (rounded) and failure about 51 times out of those 64 tries.

After doing this 100 times, you would expect to succeed 20 times on the first try; 16 times on the second try; about 13 times on the third try; and there would be about 51 times left over that you did not succeed on the first, second or third tries.

I think this logic resembles the tables showing a likelihood of success of 52.5% in the first three rounds, when you consider that the odds are not static and will change from round to round as the field is thinned.

I can't do the actual math, but that table sure looks right.

 

[Budsfan]

Why are you adding the odds from the first round and incorporating them in the second???

If you lose the first round, the odds do not carry over. They are erased. We do not gain any advantage from our first draw loss what soever so adding them into the overall odds is wrong.

Why would our percentage at 4 be 47.5% for that pick, I'm saying that calculating that way, that they are saying, the odds of getting the 1 through 3 picks, the odds are collectively 52.50% and even though we still have 20% chance or more depending on which team wins on each round, with each draw the reason the odds go down are because of number of chances left until we pick 4th.

Don't foget, we have an 80% chance of not picking first but we would have a 52.50% chance to pick in the top 3, or I think that's what they are saying?

 

[Leafsman]

You are right that the outcome of one round doesn't affect the probable outcome of the next.

Keeping in mind that I am probably a moron at this, here is one way I've come to think of it:

Imagine that you try something three times in a row, with a 20% Chance of winning on each try, and you repeat that process 100 times.

You would expect that in 20 times out of 100, you would have succeeded on the first try and would not go on to the second or third tries.

Of the 80 times you failed on the first try, you would also fail 80% of the time. Therefore, you'd would would succeed 16 times and fail 64 times in those 80 tries.

Of the 64 times that you failed on. The first and second tries (combined), you would also expect to fail 80% of the time on the third try, resulting in success about 13 times (rounded) and failure about 51 times out of those 64 tries.

After doing this 100 times, you would expect to succeed 20 times on the first try; 16 times on the second try; about 13 times on the third try; and there would be about 51 times left over that you did not succeed on the first, second or third tries.

I think this logic resembles the tables showing a likelihood of success of 52.5% in the first three rounds, when you consider that the odds are not static and will change from round to round as the field is thinned.

I can't do the actual math, but it sure looks right.

Imagine you try something three time in a row with an 80% chance of not winning each time and you try it 100 times. Tell me how that somehow becomes a lower likelihood of happening.

Somehow the 20% odds of winning are more valuable than the 80% odds of not winning and become a higher likelihood of happening.

So if our odds were 80% of winning the 1st overall, we'd somehow be worse off???

 

[Ciao]

Is this more complicated than CORSI?

I never understood that either.

 

[Leafsman]

Why would our percentage at 4 be 47.5% for that pick, I'm saying that calculating that way, that they are saying, the odds of getting the 1 through 3 picks, the odds are collectively 52.50% and even though we still have 20% chance or more depending on which team wins on each round, with each draw the reason the odds go down are because of number of chances left until we pick 4th.

Don't foget, we have an 80% chance of not picking first but we would have a 52.50% chance to pick in the top 3, or I think that's what they are saying?

It's the percentage that is throwing it off.

The odds are approximately as follows picking 1/2/3/4
2:2:1.5:4.5.

How many times do you see odds displayed as percentages.

The highest likelihood is picking 4th overall by almost 2.25x

Edit: Look at the logic in your sentence. We have an 80% chance of not picking first but 52.5% chance of picking 1-3???!?!?!?

 

[Ciao]

Imagine you try something three time in a row with an 80% chance of not winning each time and you try it 100 times. Tell me how that somehow becomes a lower likelihood of happening.

Somehow the 20% odds of winning are more valuable than the 80% odds of not winning and become a higher likelihood of happening.

So if our odds were 80% of winning the 1st overall, we'd somehow be worse off???

No, 80% odds of success would be fabulous

That way you would expect to succeed 80 times on the first try, and to be unsuccessful only 20 times.

Of the 20 times you had to go on to a second try you would also expect to succeed 80% of the time. You would expect 16 wins and four losses in those 20 tries.

Ditto for the four times you had to go on to a third try: you would expect about three wins (rounded down from 3.2, or 80% of 4) and one loss (rounded up from 0.8, or 20% of 4) in those four tries.

In the end, you would have succeeded 80 times on the very first try, 16 times on the second try, and three times on the third try for a total of 99 times; and after 100 tries you would have expected to fail only once after all three tries. That would be a failure rate of 1% and a 99% of picking in one of the first three rounds.

 

[Pyromaniac3]

No one's arguing the increase in weighting?!?! You can't account for something you don't know. At best it accounts for a few percent added each subsequent round.

As I said before, every single scenario was simulated and combined together in a weighted fashion to produce that table. Because it makes zero sense to have lower absolute odds to win 2nd overall than 1st overall. That 2nd overall odds includes the odds that TOR loses the first lottery.

What I am saying is Toronto does not carry over the 200 balls from the first round and add it to the 200 balls from the second round.

So when you say we have a 52% chance of winning, where are you getting the 520 balls from????

You don't add balls at all. What I am saying is the chances of TOR winning *any* one of the 3 lotteries is 52%. This is because the odds of picking 4th is 48%. The other possibility is picking top 3. They both have to add up to 100%. So it is 52%. There is no other possibility.

The odds to pick top 3 continue to drop to as you lose lotteries. So the odds of being second in this table = the odds of losing 1st lottery and odds of winning second lottery with eliminating the balls of winner in the subsequent lottery.

These are all independent absolute odds not dependent on winning or losing. As soon as 1st lottery is done, all the odds change in the table.

I know it's a bit counter intuitive but the example of 4 coin flips is easier to understand. The probability of seeing a tails in 4 coin flips is 93%. But the odds of seeing one tails in the remaining flips decrease if I see heads in the first coin flip.

 

[burpsalot]