Question about the lottery odds

[bdm860]

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?

Oh, I get it.

I just think most people think of it in more in real time for their team only, as opposed to overall before any picks have been made (especially at the top of the draft).  As in, "We want #1, but if we don't get #1, then what are the odds of us getting #2" and the answer is always better.

[droopdog7]

It's real simple fellas.  Say we have one ball in 1000.  Odd of being selected first is 1/1000 or 0.1%.  If we don't get picked first the odds of getting picked second are 1/999 or   0.1%.  If we aren't picked first or second then our odds are 1/998 or  to get picked third or 0.1%.

So as you can see, the denominator only goes down by 1 (out of a thousand) for each of the first three picks, meaning we have essentially the same odds at each of the first three picks.

Easy peasy.

[Chris22]

The percentages change once the picking starts.

[max215]

It's real simple fellas.  Say we have one ball in 1000.  Odd of being selected first is 1/1000 or 0.1%.  If we don't get picked first the odds of getting picked second are 1/999 or   0.1%.  If we aren't picked first or second then our odds are 1/998 or  to get picked third or 0.1%.

So as you can see, the denominator only goes down by 1 (out of a thousand) for each of the first three picks, meaning we have essentially the same odds at each of the first three picks.

Easy peasy.

Except you're not removing 1 ball, but as many as 250, which radically changes the percentages. Saltlover's explanation is spot on.

[celticmania]

My brain hurts

[kozlodoev]

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).

[fairweatherfan]

It's real simple fellas.  Say we have one ball in 1000.  Odd of being selected first is 1/1000 or 0.1%.  If we don't get picked first the odds of getting picked second are 1/999 or   0.1%.  If we aren't picked first or second then our odds are 1/998 or  to get picked third or 0.1%.

So as you can see, the denominator only goes down by 1 (out of a thousand) for each of the first three picks, meaning we have essentially the same odds at each of the first three picks.

Easy peasy.

Except you're not removing 1 ball, but as many as 250, which radically changes the percentages. Saltlover's explanation is spot on.

Basically the reason why our probability for #2 is almost the same as #1 is because the average increase in our chances when the various other teams get #1 is almost perfectly offset by our chances of getting #1 ourselves (in which case our odds of #2 are of course 0).


...it's also combinations of balls, not individual balls, but that's probably needlessly complicating things.

[Greyman]

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.

[SHAQATTACK]

If your the  Lakers your  odds are always greater than they have any right to be.

[guava_wrench]

It's real simple fellas.  Say we have one ball in 1000.  Odd of being selected first is 1/1000 or 0.1%.  If we don't get picked first the odds of getting picked second are 1/999 or   0.1%.  If we aren't picked first or second then our odds are 1/998 or  to get picked third or 0.1%.

So as you can see, the denominator only goes down by 1 (out of a thousand) for each of the first three picks, meaning we have essentially the same odds at each of the first three picks.

Easy peasy.

I don't think that is correct.

4 balls each numbered 1-14 are placed in a machine. 4 are drawn to generate a 4 digit combination (order doesn't matter).

Each team has a set # of combinations assigned to them based on record. For example, Philly has 250 for their own pick.

For the initial draw, the denominator is 1000. If the combination matches Philly's own pick, the denominator will drop by 250 because all of the combinations assigned to Philly's pick will no longer be valid for the second pick.

That's my understanding.

[guava_wrench]

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%.

[guava_wrench]

It's real simple fellas.  Say we have one ball in 1000.  Odd of being selected first is 1/1000 or 0.1%.  If we don't get picked first the odds of getting picked second are 1/999 or   0.1%.  If we aren't picked first or second then our odds are 1/998 or  to get picked third or 0.1%.

So as you can see, the denominator only goes down by 1 (out of a thousand) for each of the first three picks, meaning we have essentially the same odds at each of the first three picks.

Easy peasy.

Except you're not removing 1 ball, but as many as 250, which radically changes the percentages. Saltlover's explanation is spot on.

Basically the reason why our probability for #2 is almost the same as #1 is because the average increase in our chances when the various other teams get #1 is almost perfectly offset by our chances of getting #1 ourselves (in which case our odds of #2 are of course 0).


...it's also combinations of balls, not individual balls, but that's probably needlessly complicating things.

There is a limited set of possible outcomes. The probability can be determined by listing out all outcomes and seeing what percentage of outcomes have us at a particular pick.

The probability is only true before they start drawing combinations. As other have indicated, the odds once #1 is set change because a large number of possible outcomes are removed.

Similarly, as results are revealed on TV, the odds chance as outcomes are removed.

But all that really matters is the overall percentages since we won't be there to reconsider during the drawing of balls and because everything is already determined when the reveal happens.

[Greyman]

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.

[fairweatherfan]

It's real simple fellas.  Say we have one ball in 1000.  Odd of being selected first is 1/1000 or 0.1%.  If we don't get picked first the odds of getting picked second are 1/999 or   0.1%.  If we aren't picked first or second then our odds are 1/998 or  to get picked third or 0.1%.

So as you can see, the denominator only goes down by 1 (out of a thousand) for each of the first three picks, meaning we have essentially the same odds at each of the first three picks.

Easy peasy.

Except you're not removing 1 ball, but as many as 250, which radically changes the percentages. Saltlover's explanation is spot on.

Basically the reason why our probability for #2 is almost the same as #1 is because the average increase in our chances when the various other teams get #1 is almost perfectly offset by our chances of getting #1 ourselves (in which case our odds of #2 are of course 0).


...it's also combinations of balls, not individual balls, but that's probably needlessly complicating things.

There is a limited set of possible outcomes. The probability can be determined by listing out all outcomes and seeing what percentage of outcomes have us at a particular pick.

The probability is only true before they start drawing combinations. As other have indicated, the odds once #1 is set change because a large number of possible outcomes are removed.

Similarly, as results are revealed on TV, the odds chance as outcomes are removed.

But all that really matters is the overall percentages since we won't be there to reconsider during the drawing of balls and because everything is already determined when the reveal happens.

Yup, that's probability.  I think the original issue is that some are thinking in terms of a conditional probability - the likelihood we draw #2, but only in cases where we don't get #1.   It would be intuitive to expect the probability of #2 to go up if that was the case.  But the overall probabilities are including all the outcomes where we win #1 and #2 is moot.  That's where I think the confusion has been coming in.

[guava_wrench]

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.