Question about the lottery odds

[jpotter33]

Okay, so here are the lottery odds by pick for us as the third pick:

1st - 15.6%
2nd - 15.7%
3rd - 15.6%
4th - 22.6%
5th - 26.5%
6th - 4%

Okay, so here's my question. I think I understand why the 4th and 5th picks are the most likely scenarios for us, i.e. because the combination of percentages of all of those behind us greatly outweigh our own 15.6% chance. So with that combined greater percentage, it's more likely that one or two of those teams will jump us in the lottery, but it's due to the combination of the overall percentages rather than those particular teams' percentages that jumped us.

However, why are the first three picks essentially the same percentage then? No matter who ends up getting the first pick, everyone is, at least slightly, more likely to get the second pick by virtue of one less team being in the mix. So then shouldn't the second pick be a higher percentage for us, and, consequently, the third pick should be an even higher percentage for us than the second pick?

Perhaps they just label it that way due to not knowing the exact percentages each pick will ultimately manifest as since the percentages themselves would ultimately depend upon who ends up getting the first pick, i.e. the odds for the second pick would be much different for everyone if the Bulls got the first pick than if the Sixers got the first pick.

So where is my quantitative reasoning flawed?

[PaulP34]

Its basicly like this. We have a 50% chance of either the top 3 picks or a 50% chance at the 4-5 picks. Either way we r going to get a top 5 pick. It just depends on how the balls are lifted.

[nickagneta]

The NBA made those percentages when they determined how many balls each team was going to get in the 1000 ball lottery. The math is the math. The NBA could have decided to put a certain finish at any percentage they wanted simply by adding or subtracting balls at various positions

[saltlover]

Okay, so here are the lottery odds by pick for us as the third pick:

1st - 15.6%
2nd - 15.7%
3rd - 15.6%
4th - 22.6%
5th - 26.5%
6th - 4%

Okay, so here's my question. I think I understand why the 4th and 5th picks are the most likely scenarios for us, i.e. because the combination of percentages of all of those behind us greatly outweigh our own 15.6% chance. So with that combined greater percentage, it's more likely that one or two of those teams will jump us in the lottery, but it's due to the combination of the overall percentages rather than those particular teams' percentages that jumped us.

However, why are the first three picks essentially the same percentage then? No matter who ends up getting the first pick, everyone is, at least slightly, more likely to get the second pick by virtue of one less team being in the mix. So then shouldn't the second pick be a higher percentage for us, and, consequently, the third pick should be an even higher percentage for us than the second pick?

Perhaps they just label it that way due to not knowing the exact percentages each pick will ultimately manifest as since the percentages themselves would ultimately depend upon who ends up getting the first pick, i.e. the odds for the second pick would be much different for everyone if the Bulls got the first pick than if the Sixers got the first pick.

So where is my quantitative reasoning flawed?

I think you might be right and I'm going to do the math for the odds of the second pick.  I'll post shortly.

EDIT:  Nope!  My instinct was wrong, the math is correct.  I'll explain shortly.

EDIT 2 - Longer explanation:

When the lottery is conducted, there are 1,000 winning combinations for the first pick.  The Celtics hold 156 of those.  If they win, their 156 combinations are no longer valid to win the second pick.  The same is true with everyone else.  So if the Sixers win, their 250 combinations are removed.  In this universe where the Sixers win, the Celtics would have a 20.8% chance of picking second (156 chances in 750 winning combinations).  However, the Sixers only win in 25% of the draws.  If the Lakers win the first overall, the Celtics would have a 19.5% chance of picking second (156 chances in 801 winning combinations).  But the Laters only win in 19.9% of draws.  So you have to go through what the Celtics' odds would be to pick second given all the various outcomes of the first draw, and weight them by the likelihood of that occurring.  When you do this, the Celtics have a 0.157389 chance of picking second.

It's not the NBA trying to give better or worse odds.  It's just math.

[jpotter33]

The NBA made those percentages when they determined how many balls each team was going to get in the 1000 ball lottery. The math is the math. The NBA could have decided to put a certain finish at any percentage they wanted simply by adding or subtracting balls at various positions

Okay, so it's essentially just a practical thing then. We have 156 combinations/ping pong balls, so they just put 15.6% pretty much across the board for the top-3 spots, even though we have slightly higher chances for the second and third picks, respectively, which is why we have a little over 45% chance of being top-3 altogether.

[jpotter33]

Okay, so here are the lottery odds by pick for us as the third pick:

1st - 15.6%
2nd - 15.7%
3rd - 15.6%
4th - 22.6%
5th - 26.5%
6th - 4%

Okay, so here's my question. I think I understand why the 4th and 5th picks are the most likely scenarios for us, i.e. because the combination of percentages of all of those behind us greatly outweigh our own 15.6% chance. So with that combined greater percentage, it's more likely that one or two of those teams will jump us in the lottery, but it's due to the combination of the overall percentages rather than those particular teams' percentages that jumped us.

However, why are the first three picks essentially the same percentage then? No matter who ends up getting the first pick, everyone is, at least slightly, more likely to get the second pick by virtue of one less team being in the mix. So then shouldn't the second pick be a higher percentage for us, and, consequently, the third pick should be an even higher percentage for us than the second pick?

Perhaps they just label it that way due to not knowing the exact percentages each pick will ultimately manifest as since the percentages themselves would ultimately depend upon who ends up getting the first pick, i.e. the odds for the second pick would be much different for everyone if the Bulls got the first pick than if the Sixers got the first pick.

So where is my quantitative reasoning flawed?

I think you might be right and I'm going to do the math for the odds of the second pick.  I'll post shortly.

Right. So essentially if the Sixers are drawn first, then that takes 250 combinations from the original 1001 combinations out of play. So then instead of a 15.6% chance of getting the second pick, our 156 combinations will represent approximately 20.8% of a chance of getting the second pick from the remaining 751 combinations. If the Lakers then get the second pick removing another 199 combinations from the remaining 751, we'd have approximately 28.3% chance of landing the third pick from the remaining 552 combinations.

But now I'm confused lol Where's the 46.9% chance of the top three pick come from then?

[Csfan1984]

The old odds vs probability dilemma

[bdm860]

The NBA made those percentages when they determined how many balls each team was going to get in the 1000 ball lottery. The math is the math. The NBA could have decided to put a certain finish at any percentage they wanted simply by adding or subtracting balls at various positions

Okay, so it's essentially just a practical thing then. We have 156 combinations/ping pong balls, so they just put 15.6% pretty much across the board for the top-3 spots, even though we have slightly higher chances for the second and third picks, respectively, which is why we have a little over 45% chance of being top-3 altogether.

I'm actually following your math and understand where you're coming from.

Philly for example has 250 combinations out of 1000 for 25% odds at #1.

If they don't win #1, their odds for #2 should increase.  They still have 250 combinations, but the total pool of eligible combinations has decreased somewhere between 199 (if Lakers win) and 5 (if Chicago wins).  So 250 divided by 801 or 995 would equal either 31.2% or 25.1% odds.  This is not reflected in the chart everyone uses though which shows decreased odds of 21.5% for Philly.

[saltlover]

Okay, so here are the lottery odds by pick for us as the third pick:

1st - 15.6%
2nd - 15.7%
3rd - 15.6%
4th - 22.6%
5th - 26.5%
6th - 4%

Okay, so here's my question. I think I understand why the 4th and 5th picks are the most likely scenarios for us, i.e. because the combination of percentages of all of those behind us greatly outweigh our own 15.6% chance. So with that combined greater percentage, it's more likely that one or two of those teams will jump us in the lottery, but it's due to the combination of the overall percentages rather than those particular teams' percentages that jumped us.

However, why are the first three picks essentially the same percentage then? No matter who ends up getting the first pick, everyone is, at least slightly, more likely to get the second pick by virtue of one less team being in the mix. So then shouldn't the second pick be a higher percentage for us, and, consequently, the third pick should be an even higher percentage for us than the second pick?

Perhaps they just label it that way due to not knowing the exact percentages each pick will ultimately manifest as since the percentages themselves would ultimately depend upon who ends up getting the first pick, i.e. the odds for the second pick would be much different for everyone if the Bulls got the first pick than if the Sixers got the first pick.

So where is my quantitative reasoning flawed?

I think you might be right and I'm going to do the math for the odds of the second pick.  I'll post shortly.

Right. So essentially if the Sixers are drawn first, then that takes 250 combinations from the original 1001 combinations out of play. So then instead of a 15.6% chance of getting the second pick, our 156 combinations will represent approximately 20.8% of a chance of getting the second pick from the remaining 751 combinations. If the Lakers then get the second pick removing another 199 combinations from the remaining 751, we'd have approximately 28.3% chance of landing the third pick from the remaining 552 combinations.

But now I'm confused lol Where's the 46.9% chance of the top three pick come from then?

See my edited post above.

[jpotter33]

The old odds vs probability dilemma

But that's merely a rhetorical issue, isn't it? It's like the difference between saying the C's record was 48 AND 34 versus 48 out of 72. How would that dilemma apply here?

[jpotter33]

Okay, so here are the lottery odds by pick for us as the third pick:

1st - 15.6%
2nd - 15.7%
3rd - 15.6%
4th - 22.6%
5th - 26.5%
6th - 4%

Okay, so here's my question. I think I understand why the 4th and 5th picks are the most likely scenarios for us, i.e. because the combination of percentages of all of those behind us greatly outweigh our own 15.6% chance. So with that combined greater percentage, it's more likely that one or two of those teams will jump us in the lottery, but it's due to the combination of the overall percentages rather than those particular teams' percentages that jumped us.

However, why are the first three picks essentially the same percentage then? No matter who ends up getting the first pick, everyone is, at least slightly, more likely to get the second pick by virtue of one less team being in the mix. So then shouldn't the second pick be a higher percentage for us, and, consequently, the third pick should be an even higher percentage for us than the second pick?

Perhaps they just label it that way due to not knowing the exact percentages each pick will ultimately manifest as since the percentages themselves would ultimately depend upon who ends up getting the first pick, i.e. the odds for the second pick would be much different for everyone if the Bulls got the first pick than if the Sixers got the first pick.

So where is my quantitative reasoning flawed?

I think you might be right and I'm going to do the math for the odds of the second pick.  I'll post shortly.

EDIT:  Nope!  My instinct was wrong, the math is correct.  I'll explain shortly.

EDIT 2 - Longer explanation:

When the lottery is conducted, there are 1,000 winning combinations for the first pick.  The Celtics hold 156 of those.  If they win, their 156 combinations are no longer valid to win the second pick.  The same is true with everyone else.  So if the Sixers win, their 250 combinations are removed.  In this universe where the Sixers win, the Celtics would have a 20.8% chance of picking second (156 chances in 750 winning combinations).  However, the Sixers only win in 25% of the draws.  If the Lakers win the first overall, the Celtics would have a 19.5% chance of picking second (156 chances in 801 winning combinations).  But the Laters only win in 19.9% of draws.  So you have to go through what the Celtics' odds would be to pick second given all the various outcomes of the first draw, and weight them by the likelihood of that occurring.  When you do this, the Celtics have a 0.157389 chance of picking second.

It's not the NBA trying to give better or worse odds.  It's just math.

Ooooooooookay, I get it now. I wasn't factoring in the actual probabilities of each team getting the first pick, which makes sense now. TP!

So it makes sense that they put it down that way PRIOR to the lottery, but once those picks actually do happen, our chances actually still do increase for each pick in real time.

EDIT: I also should've looked at the overall odds for each team, because it's only the third spot in the lottery that has that type of consistency across the board (.156, .157, .156).

[saltlover]

Okay, so here are the lottery odds by pick for us as the third pick:

1st - 15.6%
2nd - 15.7%
3rd - 15.6%
4th - 22.6%
5th - 26.5%
6th - 4%

Okay, so here's my question. I think I understand why the 4th and 5th picks are the most likely scenarios for us, i.e. because the combination of percentages of all of those behind us greatly outweigh our own 15.6% chance. So with that combined greater percentage, it's more likely that one or two of those teams will jump us in the lottery, but it's due to the combination of the overall percentages rather than those particular teams' percentages that jumped us.

However, why are the first three picks essentially the same percentage then? No matter who ends up getting the first pick, everyone is, at least slightly, more likely to get the second pick by virtue of one less team being in the mix. So then shouldn't the second pick be a higher percentage for us, and, consequently, the third pick should be an even higher percentage for us than the second pick?

Perhaps they just label it that way due to not knowing the exact percentages each pick will ultimately manifest as since the percentages themselves would ultimately depend upon who ends up getting the first pick, i.e. the odds for the second pick would be much different for everyone if the Bulls got the first pick than if the Sixers got the first pick.

So where is my quantitative reasoning flawed?

I think you might be right and I'm going to do the math for the odds of the second pick.  I'll post shortly.

EDIT:  Nope!  My instinct was wrong, the math is correct.  I'll explain shortly.

EDIT 2 - Longer explanation:

When the lottery is conducted, there are 1,000 winning combinations for the first pick.  The Celtics hold 156 of those.  If they win, their 156 combinations are no longer valid to win the second pick.  The same is true with everyone else.  So if the Sixers win, their 250 combinations are removed.  In this universe where the Sixers win, the Celtics would have a 20.8% chance of picking second (156 chances in 750 winning combinations).  However, the Sixers only win in 25% of the draws.  If the Lakers win the first overall, the Celtics would have a 19.5% chance of picking second (156 chances in 801 winning combinations).  But the Laters only win in 19.9% of draws.  So you have to go through what the Celtics' odds would be to pick second given all the various outcomes of the first draw, and weight them by the likelihood of that occurring.  When you do this, the Celtics have a 0.157389 chance of picking second.

It's not the NBA trying to give better or worse odds.  It's just math.

Ooooooooookay, I get it now. I wasn't factoring in the actual probabilities of each team getting the first pick, which makes sense now. TP!

So it makes sense that they put it down that way PRIOR to the lottery, but once those picks actually do happen, our chances actually still do increase for each pick in real time.

Exactly.  If we don't pick first, we have between a 15.7% and 20.8% chance to pick second, depending who gets to pick first.  However, if we win the first pick, we have a 0% chance to pick second.  As that happens 15.6% of the time, it brings down the overall odds of picking second.

[nickagneta]

You can not take into effect just one scenario in the math. You have to account for all scenarios. Therein lies your math problems

[Csfan1984]

The old odds vs probability dilemma

But that's merely a rhetorical issue, isn't it? It's like the difference between saying the C's record was 48 AND 34 versus 48 out of 72. How would that dilemma apply here?

Odds are the chances.
Probability is projecting the outcome.

Odds is mostly presented as a ratio. Probability is always percentage and mostly only used to define the highest or lowest case.

Probability is always combined numbers. While odds however only concentrate on a single chance OR in this case pick.

So odds= What are the chances C's get #1 pick?
Probability=What is the most likely spot C's draft?

Though similar they aren't exactly the same. And when you confuse the question itself you can go crazy over the numbers.

[Vermont Green]

If they don't win #1, their odds for #2 should increase.  They still have 250 combinations, but the total pool of eligible combinations has decreased somewhere between 199 (if Lakers win) and 5 (if Chicago wins).  So 250 divided by 801 or 995 would equal either 31.2% or 25.1% odds.  This is not reflected in the chart everyone uses though which shows decreased odds of 21.5% for Philly.

You are missing one key aspect.  The 25.1% to 31.2% odds apply only if they don't get the first pick.  So their odds of getting the second pick is 28.1% (roughly) of 75% which works out to be roughly 21%.  That is the probability before the first pick is done.  After the first pick, all the odds change as you suggest (their odds would increase at that point but not until) but because the first pick is not done, you have to subtract out the odds Philly gets the first pick.

Clear as mud?