Just to clarify the debated odds of a 10th worst record ('seed') landing ANY of the top 3 picks:
Nick is correct as to the odds of getting any ONE of the top three picks are all small, prior to the lottery. They are:
1st: 11 in 1000
2nd: 13 in 1000
3rd: 16 in 1000
Larry is correct that the odds as you progress to the next pick depend on who picked before you. This is because they are dependent events. Even though the combinations for a picked team remain in the tumbler, the fact that they are ignored means they are not part of the result domain.
However that doesn't mean the odds are any different than expressed above. The above numbers already fold that in.
Let's walk through the actual calculation.
The odds of the 10th seed getting the first pick are .011. That one is simple.
Once the first pick is made, the maximum chance the 10th seed would have would in the case where the first pick went to the 1st seed, because that would remove 250 combinations from the result domain, which would mean the 10th seed would have 11 in 750 chances of getting the 2nd pick, or .0146667. But that scenario only occurs if the first seed got the first pick - which is only 1 in 4 chance. That means that scenario only has a .0366667 chance of occurring.
If the 2nd seed got the 1st pick, then the 10th pick would have odds of 11 in 801, or .01373 of getting the 2nd pick. But the odds of the 2nd seed getting the first pick are 199 in 1000, so the total odds of that scenario occurring are just .0027328.
Similarly, if the 3rd seed got the 1st pick, the 10th pick would have just a .002033 chance of getting the 2nd pick. And if the 4th seed got the 1st pick, the 10th seed would have just a .0014858 chance at the 2nd pick. And so on, if the 5th seed got the first pick, the 10th seed would have a .0010614 chance of getting the 2nd pick.
There are 13 different scenarios where the 10th seed ends up with the 2nd pick, all based on one of the other 13 teams getting the 1st pick. When you add the odds for any one of these 13 scenarios occurring all up together you get a total probability for the 10th seed getting the 2nd pick of (writing a quick little beanshell script):
int[] C2 = new int[]{250, 199, 156, 119, 88, 63, 43, 28, 17, 8, 7, 6, 5};
P2 = 0.0;
for(int combinations : C2){
P2+=(11*combinations)/((1000.0-combinations)*1000.0);
}
P2;
0.013008772045002403
So, the total odds of getting any of the 13th scenarios that result in the 10th seed getting the 2nd pick are, indeed, .013. Just like the original tables said.
Similarly (requires an extra level of iteration), the total odds of getting any of the 156 (13*12) scenarios that result in the 10th seed getting the 3rd pick are .016.
Thus the total odds of any of the 1+13+156 scenarios that lead to the 10th seed getting ANY of the top 3 picks is .011+.013+.016 = .040.
So, the odds of the 10th seed getting a Top 3 pick is just 4%.
This is really just basic common sense. The table indicates the odds of a seed getting each pick and for each seed, all those chances must add up to 1.0.
Every seed has the largest chance of getting it's own pick. I.E., the 10th seed has the largest chance of getting the 10th pick. In fact it is an 87% chance that happens. That means that the chances of getting any of the other 5 allowed picks must add up to no more than 13%.
In the interest of competitive balance, the lottery is slotted to favor the lower seeds. So the bulk of that extra 13% (9%, in fact) is actually loaded towards a chance of ending up with a _worse_ pick.
Basically, the 10th seed has a greater chance of ending up with the 11th pick than it does any of the top 3 picks.