Post-Game Talk: Mission Accomplished

[Leafsman]

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.

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.



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.

If you guys can answer this then I'll concede.

If we do the same calculations but instead calculate the odds of not picking 1-3, it comes out to approximately 248%?????

If the odds are 20,17.5,15% for winning.

The odds then should be 80%,82.5%,85% for not winning

 

[Ciao]

If you guys can answer this then I'll concede.

If we do the same calculations but instead calculate the odds of not picking 1-3, it comes out to approximately 248%?????

If the odds are 20,17.5,15% for winning.

The odds then should be 80%,82.5%,85% for not winning

I can't answer that.

I'm just not smart enough to even try.

 

[Pyromaniac3]

If you guys can answer this then I'll concede.

If we do the same calculations but instead calculate the odds of not picking 1-3, it comes out to approximately 248%?????

If the odds are 20,17.5,15% for winning.

The odds then should be 80%,82.5%,85% for not winning

The odds of picking 4th is right in the table. So to pick 1-3, the probability is 100%-47.5% = 52.5% as there is no other possibility.

20%, 17.5% and 15% odds are absolute odds. Meaning that if you lose 1st overall, the odds are recalculated. 17.5% represents the absolute odds that you lose the first lottery and win the second lottery. But if you calculate the odds of winning 2nd lottery after the first one, it is >20%. It's impossible to drop below 20% since you always have 200 balls and any subsequent lotteries will have less than 1000 balls.

 

[Budsfan]

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???!?!?!?

That's by their calculation not mine but evidently their calculations are supposedly based on scientific evaluation, frankly I don't know.

#1 pick odds would be 5 to 1, they would go down slightly for the next pick depending on who wins the pick.

I will leave you with this thought and as a lot of my misspent youth
was at the racetracks around Toronto and the old saying was "You can usually pick the best horse but it doesn't mean it will win" and believe me that's true, all I really know, is that we will for sure have the 4th OA pick but even that's not sure, because, Hunter may trade it away.

G'night.

 

[Pyromaniac3]

The odds then should be 80%,82.5%,85% for not winning

Those odds can be explained this way:

80% - chance of not picking 1st overall meaning picking either 2nd, 3rd, or 4th
82.5% - chance of not picking 2nd overall meaning picking either 1st, 3rd or 4th
85% - chance of not picking 3rd overall meaning picking either 1st, 2nd or 4th.

 

[Leafsman]

The odds of picking 4th is right in the table. So to pick 1-3, the probability is 100%-47.5% = 52.5% as there is no other possibility.

20%, 17.5% and 15% odds are absolute odds. Meaning that if you lose 1st overall, the odds are recalculated. 17.5% represents the absolute odds that you lose the first lottery and win the second lottery. But if you calculate the odds of winning 2nd lottery after the first one, it is >20%. It's impossible to drop below 20% since you always have 200 balls and any subsequent lotteries will have less than 1000 balls.

You didn't even come close to answering the question, you just recalculated the initial question.

If your formula and method is correct, I should be able to calculate the odds for winning and the odds for losing and they should coincide.

The 20% is no more absolute than the 80%.

The math should work out either way.

 

[Leafsman]

Those odds can be explained this way:

80% - chance of not picking 1st overall meaning picking either 2nd, 3rd, or 4th
82.5% - chance of not picking 2nd overall meaning picking either 1st, 3rd or 4th
85% - chance of not picking 3rd overall meaning picking either 1st, 2nd or 4th.

You accumulate when it comes to winning but not accumulating when it comes to losing. It's not either/or.

If you accumulate the additional chance of winning then you have to account for the accumulating chance of losing. Each round carries approx. an additional 80% chance of losing, The 2.5% and the 5% account for the winning in the previous rounds.

So really my math was a little wrong and the chance for losing is 240%

 

[Pyromaniac3]

You accumulate when it comes to winning but not accumulating when it comes to losing. It's not either/or.

You don't accumulate when you're losing because the odds are structured that way. Losing first lottery = 80%, losing second lottery = 63% which includes the odds of both losses. You don't lose second lottery unless you have already lost the first one. So total odds of losing 2 lotteries = 63% and losing all 3 lotteries = 47.5%.

 

[Leafsman]

You don't accumulate when you're losing because the odds are structured that way. Losing first lottery = 80%, losing second lottery = 63% which includes the odds of both losses. You don't lose second lottery unless you have already lost the first one. So total odds of losing 2 lotteries = 63% and losing all 3 lotteries = 47.5%.

I think we'll have to agree to disagree.

 

[TheOtherSide]

Know what would be awesome? The Leafs winning the lottery. That would be awesome

 

[Ciao]

I think we'll have to agree to disagree.

Okay.

Sounds good

(I think the mods must be ready to shut us down anyway. If they aren't, they should be.)

 

[Ciao]

Know what would be awesome? The Leafs winning the lottery. That would be awesome

What are the odds of that?

 

[Leafsman]

What are the odds of that?

Hahahahaha!!!! NICE!!

 

[Macallan18]

The table predicts what our odds are BEFORE a ball is dropped. That is why the second and third columns still list all the teams and yet still totalsl 100%. What changes is our odds on getting the second or third pick if we don't win the first pick, dependent on who actually wins the first pick. So if (once?) Edmonton wins first overall, our odds of getting the second pick will be higher that the 17.5% quoted. Similarly for the third pick.
That was my only point, sorry for any confusion.

 

[Ari91]

I can't even be a bit excited that they actually finished with the best odds at the lottery. The lottery could be exactly as it was years prior and with this team's luck, they would lose it. I'll just expect that we pick fourth but I have no honest expectation that our chances for the first overall is any good. It would be really nice though. The franchise is due another #1 overall I think.

 

[hockeyes]

You accumulate when it comes to winning but not accumulating when it comes to losing. It's not either/or.

If you accumulate the additional chance of winning then you have to account for the accumulating chance of losing. Each round carries approx. an additional 80% chance of losing, The 2.5% and the 5% account for the winning in the previous rounds.

So really my math was a little wrong and the chance for losing is 240%

To calculate the odds of losing each round you simply multiply the percent of losing, so an 80% chance to lose each of the 3 rounds is just .8x.8.x.8 = .512 or 51.2% of losing 3 draws where you have a 20% chance of winning. Like you said though, the odds increase each round depending on who won the previous which is why the odds of picking top 3 end up over 50%., taking your example, .8x.775x.75 = 0.465 or 46.5% chance to lose all 3 rounds with those increasing odds.

 

[Leafsman]

To calculate the odds of losing each round you simply multiply the percent of losing, so an 80% chance to lose each of the 3 rounds is just .8x.8.x.8 = .512 or 51.2% of losing 3 draws where you have a 20% chance of winning. Like you said though, the odds increase each round depending on who won the previous which is why the odds of picking top 3 end up over 50%., taking your example, .8x.775x.75 = 0.465 or 46.5% chance to lose all 3 rounds with those increasing odds.

https://www.youtube.com/watch?v=XllKz2cX73Q&nohtml5=False
https://www.youtube.com/watch?v=XllKz2cX73Q&nohtml5=False
https://www.youtube.com/watch?v=5VM1LZx_fpE&nohtml5=False
http://www.mathplanet.com/education/pre-algebra/probability-and-statistic/finding-the-odds
http://www.wikihow.com/Calculate-Odds

No where can I find not only teh basis for this formula but anywhere it says to even express the odds as a percentage?????

Everywhere you look it stats to express the odds as a ratio.

Each round the odds of winning are 1 in 5, there are 3 rounds, therefor our odds of winning are 3 in 15 or more realistically teh odds of winning are 3:12 or 1:4.

 

[Macallan18]

View attachment 89383

from
http://theleafsnation.com/2016/4/11/tln-power-rankings-final-edition

 

[crump]

Every day when I run the damn thing for the first time we get 4th. That's what I'm expecting.

The way I see it is we are picking 4th with a chance to go higher, but not even half as much chance as staying put at 4th.

 

[Pyromaniac3]

No where can I find not only teh basis for this formula but anywhere it says to even express the odds as a percentage?????

Everywhere you look it stats to express the odds as a ratio.

Each round the odds of winning are 1 in 5, there are 3 rounds, therefor our odds of winning are 3 in 15 or more realistically teh odds of winning are 3:12 or 1:4.

Percentages are just odds expressed in decimal form multiplied by 100. So 1:5 -> 0.20 -> 20%.

Also if you wish to understand draft lottery further, take a look here: http://nhllotterysimulator.com/

It shows you the updated odds of each team to win the next lottery after removing the winner's balls. If you manage to get EDM win the first lottery, take a look at TOR's odds to win 2nd lottery. It will be > 20% or > 1:5

 

[Mugzy97]

Just let it happen and then we'll discuss how mad we are.

 

[sxvnert]

Leader of the inmates