Author Topic: Question about the lottery odds  (Read 5640 times)

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Re: Question about the lottery odds
« Reply #30 on: May 14, 2016, 01:29:44 AM »

Offline PaulP34

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1st - 5 times
2nd - 0 times
3rd - 3 times
4th - 4 times
5th - 9 times 😨
6th - 1 time (4% chance Boston lands here)

This shows you how many times each % happened. I'm not a numbers guy and I know every lottery is different but you cannot hide the fact that each % has landed in a certain position more times then another for a reason. This is why we record statistics so we can look back and see what is most probable. Now I did not say we are going to end up with the fifth pick. It was just to show what the most probables will most likely be. If we are playing a game of % then its in the hands of the % chances we got. 5 times the 3rd worse fell into the #1 pick. That may be what happens Tuesday. I like our chances but if it falls to the 5th pick then the statistics shown us that, that was the most likely senerio and we cannot look away from those % statistics...


« Last Edit: May 14, 2016, 11:18:04 AM by PaulP34 »

Re: Question about the lottery odds
« Reply #31 on: May 14, 2016, 02:09:43 AM »

Offline Csfan1984

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So odds= What are the chances C's get #1 pick?
Probability=What is the most likely spot C's draft?
Um no, not really. Odds and probability are two different numerical representation of an uncertain event.

Roughly speaking, 10 to 1 odds of X over Y means that for every 10 occurences of event X there is one occurence of event Y. This is another way of saying that the probability of X is 90.9% (10/11) and the probability of event Y is 9.1% (1/11).

In NBA lottery terms, the worst record in the NBA has a 3 to 1 odds to not get the top pick. In other words, for every ball that gives the pick to the worst record, there are three balls that give the pick to someone else (which is the same as saying they have a 25% chance of getting the top pick).
The whole post explains the same thing.

EDIT ADD= Just read over my post. The ratio is the same as example you gave of 10 to 1. Occurrence is chance. The percentage you stated as did I is always how probability is given. And there is the way you stated it as high and low which I also wrote. It's the same explanation. My point which you singled out isn't even the problem in fact it's the most sane part of the whole thing. The difference in what to use odds vs probability is based on the question being asked.
« Last Edit: May 14, 2016, 03:08:41 AM by Csfan1984 »

Re: Question about the lottery odds
« Reply #32 on: May 14, 2016, 04:10:09 AM »

Offline guava_wrench

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OK so here's an easy way to look at it and not get a headache from all the percentages and combinations. History tends to repeat itself. Since 1994, here's how many times the team with the 3rd worse record has landed in each spot.

1st - 5 times
2nd - 0 times
3rd - 3 times
4th - 4 times
5th - 9 times 😨
6th - 1 time (4% chance Boston lands here)
This is not helpful. Each lottery drawing is an independent result. Past results are irrelevant. The odds are all the matter.

Re: Question about the lottery odds
« Reply #33 on: May 14, 2016, 10:44:03 AM »

Offline hwangjini_1

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OK so here's an easy way to look at it and not get a headache from all the percentages and combinations. History tends to repeat itself. Since 1994, here's how many times the team with the 3rd worse record has landed in each spot.

1st - 5 times
2nd - 0 times
3rd - 3 times
4th - 4 times
5th - 9 times 😨
6th - 1 time (4% chance Boston lands here)
not really. this lottery, as with every nba lottery, is discrete and not in any way connected to the odds of past lotteries. they are independent of one another.

as an example of this, if you flip a coin 9 times and it comes up heads, what are the chances of it coming up heads on the 10th flip?

see? not dependent upon history at all, is it?
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Re: Question about the lottery odds
« Reply #34 on: May 14, 2016, 10:52:16 AM »

Offline SHAQATTACK

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OK so here's an easy way to look at it and not get a headache from all the percentages and combinations. History tends to repeat itself. Since 1994, here's how many times the team with the 3rd worse record has landed in each spot.

1st - 5 times
2nd - 0 times
3rd - 3 times
4th - 4 times
5th - 9 times 😨
6th - 1 time (4% chance Boston lands here)


Oh crap.......we re doomed .........here comes the fifth pick

Lakers automatically pick first .......it's the NBA unwritten law.

Re: Question about the lottery odds
« Reply #35 on: May 14, 2016, 03:48:45 PM »

Offline Greyman

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Pick 1 - minute chance
Pick 2 - minimal chance
Pick 3 - some chance
Pick 4 - good chance
Pick 5 - good chance
Pick 6 - little chance but the most likely as Boston has no luck with these things.

Who needs numbers? Words count.
I thing the odds of 1, 2, and 3 are actual equal for us at around 15.6%. So minute = minimal = some.

Also, the odds of getting #4 is also not good. It is only around 23%. The odds of getting #5 are also low, at around 26%.

If I understand the post right, people are saying (which makes sense to me) that the 'real' odds for picks 2, 3 ... to 6 change in your favour as other teams balls are eliminated. So while in theory the odds of 1, 2 and 3 are of equal value (15.6%), in actuality, as each pick is taken, the odds of you getting the next pick must grow as you retain the same number of balls (or combinations if that reads better) for your team whereas the overall number of combinations has decreased by a number dependent on which teams have had already taken picks.

So while the odds of your pick being say pick 5 are low-ish (close to 1 in 4), by the time that pick 6 comes around there is a nearly 95% chance the nets pick has been taken. Lets assume you don't end up with the 6th pick - the most likely pick you would have is 5. While the odds for any particular pick in the first 5 picks may be be 26% or less, the odds that the pick ends up as one of them are quite high.
You are making this more complicated than it is. It isn't like a GM can make a trade after pick #1 is determined and before #2 is determined. Just stick with the 15.6 odds each for 1, 2, and 3.

If we wanted to look at the odds after the first pick happens, our outcomes are not as good as previously because previously, they included the scenario where we would be #1.

In fact, our expected outcomes could get way worse. If Chicago's pick won't the lottery, we would have a significantly increased chance of pick #6. In fact, any picks with worse odds than the Nets pick getting #1 is bad news for use since we lost the chance at #1 but we still only have the 3 most combinations.

If Philly gets #1, we can no longer drop to #6 because dropping to #6 requires that LA, PHI and NJ picks all don't end up top 3.

But none of those details matter because nothing of significance happens until all the picks are made and announced. There is no point in concerning ourselves with anything but the probabilities as they stand now.

Yeah, I looked into it a bit more and now I understand the process better I don't even want to consider the odds. TP to you.

Re: Question about the lottery odds
« Reply #36 on: May 14, 2016, 04:01:41 PM »

Offline fairweatherfan

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OK so here's an easy way to look at it and not get a headache from all the percentages and combinations. History tends to repeat itself. Since 1994, here's how many times the team with the 3rd worse record has landed in each spot.

1st - 5 times
2nd - 0 times
3rd - 3 times
4th - 4 times
5th - 9 times 😨
6th - 1 time (4% chance Boston lands here)


Oh crap.......we re doomed .........here comes the fifth pick

Lakers automatically pick first .......it's the NBA unwritten law.

Just like last year...though it was the Minneapolis Lakers...technicalities...

Re: Question about the lottery odds
« Reply #37 on: May 14, 2016, 04:27:26 PM »

Offline alldaboston

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Do lottery results ever leak out before its shown on tv? I know Woj reports the draft picks before they happen on tv, was wondering if that also has happened before with the lottery?
I could very well see the Hawks... starting Taurean Prince at the 3, who is already better than Crowder, imo.

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