Who can do the math?

[exodus]

[Diesel]

Yeah. this is about the time where you try to insult me and then you give up... is that what's going on?

INSECURE..

You can't answer because you don't know.

[Diesel]

You're right, I mean 1000 combinations.

But after each team has moved up, the number of combinations decrease.

So Utah having what 2 combinations move up and get first...

The pot now has 998 combinations. Of those combinations, PTL has 250.

There probability of getting the 2nd pick will be closer to 25%.

However, if a team like Chicago moves up... There goes 200 combinations..

PTL still has 250 out of the 800 combinations left...

Their probability of getting the 2nd pick has increased to more than 25%...

Am I right so far?

[Teeareess]

Quote:

---

If these numbers provided by Lascar are true:

THE ODDS THAT PORTLAND WOULD FINISH AT #1 IN THE 2006 NBA DRAFT WERE EXACTLY 25%.

-THE ODDS THAT PORTLAND WOULD GET THE #2 PICK GIVEN THAT TORONTO GOT THE #1 WERE EXACTLY 27.4%.

-THE ODDS THAT PORTLAND WOULD GET THE #3 PICK GIVEN THAT TORONTO-CHICAGO GOT THE #1-#2 WERE EXACTLY 35%.

AND

The probability that PTL picks 1-4 must be 100%...

Then shouldn't the odds for PTL getting the fourth pick be simple math:

100-25-27.4-35 = 12.6?

WHO CAN DO THE MATH?

---

No. The number you came up with means nothing. The three percentages you're trying to use were each derived making different assumptions, and mean nothing in relation to each other.

Quote:

---

-THE ODDS THAT PORTLAND WOULD FINISH AT #1 IN THE 2006 NBA DRAFT WERE EXACTLY 25%

-THE ODDS THAT PORTLAND WOULD FINISH AT #2 IN THE 2006 NBA DRAFT WERE EXACTLY 21.5%

-THE ODDS THAT PORTLAND WOULD FINISH AT #3 IN THE 2006 NBA DRAFT WERE EXACTLY 17.7%

---

Adding up those three numbers and taking the difference from 100 would get you the odds for Portland getting the 4th pick.

[exodus]

[Diesel]

[Diesel]

What was wrong about what I said?

[Diesel]

me why??

Come on Fanatic.. Since you jumped in.. explain where I'm wrong on this?

[exodus]

[exodus]

[Diesel]

There are 1000 combinations!

[Diesel]

You're a lighthouse of misinformation Fanatic..

From Wikipedia..

As of 2005, with 30 NBA teams, 16 qualify for the playoffs and the remaining 14 teams are entered in the draft lottery. These 14 teams are ranked in reverse order of their regular season record and are assigned the following number of chances:

1. 250 combinations, 25% chance of receiving the #1 pick

2. 199 combinations, 19.9% chance

3. 156 combinations, 15.6% chance

4. 119 combinations, 11.9% chance

5. 88 combinations, 8.8% chance

6. 63 combinations, 6.3% chance

7. 43 combinations, 4.3% chance

8. 28 combinations, 2.8% chance

9. 17 combinations, 1.7% chance

10. 11 combinations, 1.1% chance

11. 8 combinations, 0.8% chance

12. 7 combinations, 0.7% chance

13. 6 combinations, 0.6% chance

14. 5 combinations, 0.5% chance

[Diesel]

[Diesel]

"But after each team has moved up, the number of combinations decrease."

You said that this statement is wrong, but that's not true.

The combinations can come up, but if the team who won the lottery, whoose number it was original pops up, it is disregarded...

So in practice, each time a team moves up, the number of combinations decrease.

[Diesel]

I think that's the definition of an [censored].

[exodus]

[exodus]

[Diesel]

So Utah having what 4 combinations move up and get first...

The pot now has 998 usable combinations. Of those combinations, PTL has 250 combinations.

There probability of getting the 2nd pick will be closer to 25%.

However, if a team like Chicago moves up... There goes 200 combinations..

PTL still has 250 out of the 800 usuable combinations left...

Their probability of getting the 2nd pick has increased to more than 25%...

Am I right so far?

[Lascar78]

Oh man you're so clueless it's hilarious. You think you're exposing me but you're only digging yourself deeper and deeper. I guess I'll bite and educate you and waste 1/2 an hour that should be going to writing my dissertation. Oh well.

LESSON 1: LASCAR EXPLAINS THE LOTTERY ODDS TO DIESEL:

I explained it to you in the previous post. Those 4 numbers are the 4 exact odds of portland getting either of the top 4 picks. Because the odds of portland getting #1 or #2 or #3 or #4 are 100%. Therefore you can add those 4 odds up and obtain 100%. They are not averages.

Every team's odds on the first drawing are straightforward. 250 combinations out of 1000 means a 25% chance.

To get the odds that portland gets the #2 pick, it's (the odds that 1-2 will be chicago-portland) + (the odds that 1-2 will be charlotte-portland)+...+(the odds that 1-2 will be utah-portland). That comes out to .215 for Portland.

For instance, the odds that the 1-2 would be CHI-POR is :

(odds that chicago gets #1)*(odds that portland gets #2 given the fact that chicago got #1)

=(199/1000)*(odds that portland gets #2 given the fact that chicago got #1)

=(.199)*(250/(1000-199))

=(.199)*(.312)

=.0621

=6.21% overall chance of chi-por being 1-2 in order

The odds of portland going #3 the same way. odds that portland gets the #3 pick, are = (the odds that 1-2-3 will be chicago-charlotte-portland) + (the odds that 1-2-3 will be chicag-atlanta-portland)+...+(the odds that 1-2-3 will be philly-utah-portland). That comes out to .177 for Portland.

Basically you can calculate the exact odds of every 1-2-3 ordered combination, and that tells you the exact odds of each team ending up in an exact spot. No averages, no magic.

Because portland must end up 1st, 2nd, 3rd, or 4th, and each situation is mutually exclusive, the sum of the odds of each outcome must come out to 100%.

To repeat it again, the odds add up to 100% is because exactly one of the following four outcomes must occur:

-Portland draws the #1 spot

-Portland draws the #2 spot

-Portland draws the #3 spot

-Portland draws the #4 spot

LESSON 2: LASCAR EXPLAINS CONDITIONAL PROBABILITY

Conditional probablility is the probability that event B will happen, given the fact that event A has previously occurred. For instance the conditional odds that came across portland's path last night:

-(the odds that portland got the #1): 25%

-(the odds that portland will go #2) GIVEN THE FACT THAT (Toronto went #1) were: 250 combinations left out of the 911 combinations still in play after toronto's combinations are dead = 250/911= 27.4%

-(the odds that portland will go #3) GIVEN THE FACT THAT (Toronto went #1 & Chicago went #2) were: 250 combinations left out of the 911 combinations still in play after toronto's combinations are dead = 250/712= 35.1%

-(the odds that portland will go #4) GIVEN THE FACT THAT (Toronto went #1 & Chicago went #2 & Charlotte went #3) were: 100% because of the lottery's set up.

25%+27.4%+35.1%+100% is not equal to 100% because there is no reason for it to. They would only add up to 100 if we knew that one and exactly one of these scenarios would occur

If you want to look at the odds that portland

LESSON 3: LASCAR SHOWS DIESEL HOW CLUELESS HE IS

To say that the "true odds" of Portland getting the #4 pick (call that P4) is equal to 12.4% because 25+27.4+35+P4=100% is to say that you believe that before the draft happenned, you were sure that one and exactly one of these events was sure to happen last night over the course of the 3 drawings:

-portland would get #1

-(portland will go #2) IF (Toronto gets #1)

-(portland will go #3) IF (Toronto gets #1 & Chicago gets #2)

-portland will get #4

Unfortunately for you, that makes no sense. There is no reason to add up some exact stats and some conditional probabilities because no one knew that toronto would get #1 or that chicago were going to get #2.

For instance if you want to include the odds that portland will go #2 if toronto gets #2, you have to multiply by the odds that toronto actually would get #2. Then you'd have to do the same for all other teams. And you'd get ... the same numbers obtained in Lesson 2.

I notice you stop using conditional probablities for the last line. If you carry that out, why not just add:

-portland would get #1

-(portland will go #2) IF (Toronto gets #1)

-(portland will go #3) IF (Toronto gets #1 & Chicago gets #2)

-portland will get #4 IF (Toronto gets #2 & Chicago gets #2 & Charlotte gets #3.

Add them up, you'll get 187.5%

LESSON 4: LASCAR EXPLAINS FURTHER:

If you want to consider the conditional probabilities again, and factor out how the actual drawings happenned, and look not at the pre-lottery odds of Portland going #4, but look at the odds of Portland losing the 3 drawings that actually took place, you can look at the 3 drawings:

-odds that portland would lose drawing #1: 75%

-odds that portland would lose drawing #2 if they're in it: 661/911=72.6%

-odds that portland would lose drawing #3 if they're in it: 462/712=64.9%

Now it's true that IF you assume Portland to participate in all three lotteries and IF you assume that they have the odds that they ended up getting in each lottery, then the odds of them losing all 3 is:

75%*72.6%*64.9%=35.3%

However this number has limited significance because you have to assume that they were in all three lotteries and that they get the odds that they got. In reality, there was no way of knowing this ahead of time. Still, you can see that it's very close to the 35.8% predicted by the pre-lottery numbers

Enjoy the lesson dumbass!

[exodus]

Just simplify it for him.

Man it sucks to be on his filter right now lol.