NBA draft lottery is one week away! We have best odds of landing the #1 pick!

[trickybilly]

Maybe a maths geek can help with this, but do the odds change for the our likelihood of getting #2 immediately after #1 has gone to someone else? And do those odds change if it goes to a team with more balls or fewer balls or that doesn't matter?

[Ogaju]

Maybe a maths geek can help with this, but do the odds change for the our likelihood of getting #2 immediately after #1 has gone to someone else? And do those odds change if it goes to a team with more balls or fewer balls or that doesn't matter?

the Cs odds will depend on the number of balls left at any given time prior to the Cs ball coming up.

[Smokeeye123]

If we drop to 2,3, or 4 we will pick Tatum and that's A-OKAY by me!

Second best player in the draft.

[RockinRyA]

Maybe a maths geek can help with this, but do the odds change for the our likelihood of getting #2 immediately after #1 has gone to someone else? And do those odds change if it goes to a team with more balls or fewer balls or that doesn't matter?

Isnt the number 2 called out before number 1?

Also the odds don't change even if partial results have been shown. Your odds of picking rock over paper and scissors are 1 of three even if you know someone already got scissors.

[Donoghus]

Maybe a maths geek can help with this, but do the odds change for the our likelihood of getting #2 immediately after #1 has gone to someone else? And do those odds change if it goes to a team with more balls or fewer balls or that doesn't matter?

Isnt the number 2 called out before number 1?

Also the odds don't change even if partial results have been shown. Your odds of picking rock over paper and scissors are 1 of three even if you know someone already got scissors.

In the actual lottery done behind the scenes, #1 is drawn first. 

On the tv broadcast, they come back from break and pull the cards #3, #2, and finally #1.

[bdm860]

Maybe a maths geek can help with this, but do the odds change for the our likelihood of getting #2 immediately after #1 has gone to someone else? And do those odds change if it goes to a team with more balls or fewer balls or that doesn't matter?

Yes the odds will change no matter what.  They will always improve, but by how much depends.

Here's the math.

There's 1,000 eligible ping pong ball combinations (1,001 total but the last potential combo is ignored).

C's have 250 of them for a 25% chance of #1. 250 / 1000 = 25% chance.
Phoenix has 199 combinations (19.9% chance)
Lakers have 156 combinations (15.6% chance)
All the way down to Miami with 5 combinations (0.5% chance).

Once one of your combinations is selected, all remaining combinations for that spot are eliminated from contention.

So if Phoenix gets #1, then the C's still have 250 combinations, but the total eligible pool of combinations is decreased by 199 down to 801.  So now the C's odds of #2 increase to 250 / 801 = 31.2%.

But if Miami gets #1, the potential pool only shrinks by 5, so the C's odds for # 2 are now 250 / 995 = 25.1%.

Starting off, C's have a 25% chance of #1.

If they don't get #1, their odds for #2 increase to somewhere between 25.1% (least likely) and 31.2% (most likely). 

If they don't get #1 or #2, their odds for #3 increase to somewhere between 25.3% (least likely) and 38.8% (most likely).

[hwangjini_1]

Maybe a maths geek can help with this, but do the odds change for the our likelihood of getting #2 immediately after #1 has gone to someone else? And do those odds change if it goes to a team with more balls or fewer balls or that doesn't matter?

the Cs odds will depend on the number of balls left at any given time prior to the Cs ball coming up.

it also depends upon which team gets the number one pick since that teams' "ping pong balls" are removed from the next selection. i.e., the 14th team has fewer ppb than the 4th team. hence, fewer balls are removed if the 14th team gets the first pick.

here is something from SoSH i read earlier.

Q: How likely are the Celtics to get a top-2 or top-3 pick?

MZ: Brooklyn finished with the worst record this year. Figuring out the overall probability of getting any particular pick is a bit tricky, because the odds in the 2nd and 3rd pick drawings depend on who wins the first drawing (for example, if the 2nd worst team, which has 199 combinations assigned to it, wins the 1st pick, we have a 250/801=31.2% chance to win the second pick, but if the 14th-worst team, which has only 5 combinations assigned to it, wins, we have only a 250/995=25.1% chance at the second pick -- and the math gets even more complicated when you start working on the third pick).

However, David’s done the math, and it turns out that entering the lottery, with 250 combinations, the Celtics have a 25.0% chance of getting the top pick, a 21.5% chance of getting the 2nd pick, and a 17.8% chance of getting the 3rd pick. This means that entering the lottery, we have a 46.5% chance of getting a top 2 pick, and a 64.3% chance of getting a top 3 pick.

http://www.nba.com/celtics/news/sidebar/2017-draft-lottery-qa?sf77193347=1

[trickybilly]

Maybe a maths geek can help with this, but do the odds change for the our likelihood of getting #2 immediately after #1 has gone to someone else? And do those odds change if it goes to a team with more balls or fewer balls or that doesn't matter?

Yes the odds will change no matter what.  They will always improve, but by how much depends.

Here's the math.

There's 1,000 eligible ping pong ball combinations (1,001 total but the last potential combo is ignored).

C's have 250 of them for a 25% chance of #1. 250 / 1000 = 25% chance.
Phoenix has 199 combinations (19.9% chance)
Lakers have 156 combinations (15.6% chance)
All the way down to Miami with 5 combinations (0.5% chance).

Once one of your combinations is selected, all remaining combinations for that spot are eliminated from contention.

So if Phoenix gets #1, then the C's still have 250 combinations, but the total eligible pool of combinations is decreased by 199 down to 801.  So now the C's odds of #2 increase to 250 / 801 = 31.2%.

But if Miami gets #1, the potential pool only shrinks by 5, so the C's odds for # 2 are now 250 / 995 = 25.1%.

Starting off, C's have a 25% chance of #1.

If they don't get #1, their odds for #2 increase to somewhere between 25.1% (least likely) and 31.2% (most likely). 

If they don't get #1 or #2, their odds for #3 increase to somewhere between 25.3% (least likely) and 38.8% (most likely).

2 TP's!!!

[Rosco917]

We're due!

[csfansince60s]

We're due!

I agree....and our "trend" is in the right direction.

Last year, we could have had as bad as 5 (I think) but got 3 as the third worst record. First time in a long time that in any drafts where we had a significant chance at a top pick (Jeff Green and Billups/Mercer come to mind) that we didn't do as bad as possible.

#1 this time, baby. The basketball gods owe us.

[CelticsElite]

imagine next year. hayward, Fultz or Tatum, zizic, yabu. And were about to reach the ecf without these guys. Imagine what we can accomplish with them

[droopdog7]

I would be happy with any pick in the top three because that guarantees either fultz, Jackson, or Tatum.

[Quetzalcoatl]

I would be happy with any pick in the top three because that guarantees either fultz, Jackson, or Tatum.

We also need Danny to pick in that order

[Ogaju]

I would be happy with any pick in the top three because that guarantees either fultz, Jackson, or Tatum.

We also need Danny to pick in that order

I like that order... those three are Ball players.   See what I did there?

[Somebody]

Fingers crossed for the first pick