Sorry, but that is not how odds are calculated. Whether the combos are valid or not the chances of the combo coming up are only reduced by the number of total combos left to be drawn, not the number of valid combos left to be drawn.
First draw has team with 250 chances to win.
That leaves 999 total combos. It doesn't matter that 249 are invalid. The 10th seed still only has a 11 out of 999 chance of being drawn.
Are you intentionally goofing around on technicalities or do you literally believe what you are saying?
Simple question for you. Let's say the Teams with three worst records are #1 - Bucks (250 combinations), #2 Utah (199 combinations), #3 Orlando (156 combinations).
Lotto starts...
First combination belongs to Orlando. Orlando gets the #1 pick.
Next combination ... Orlando again. They toss it out and try again.
Next combination... Utah. Utah gets the #2 pick.
What are the odds now that the Bucks get the #3 pick?
A - 38%
B - 18%
C - Other
To get into the exact math is fairly complex. It is not simple math. It deals with advanced probability and statistical mathematics. The answer is other.
But before any ball is drawn their chance of getting the 3rd pick is 18%
I think you're over-thinking it, honestly.
Admittedly, I could be wrong. Would be nice to have someone who doesn't suck at math chime in. But as far as I can see, in that scenario... there would be 645 valid combinations left and the Bucks would own 250 of those valid possible combinations. That would mean they would have a 39% chance of ending up with the #3 pick at that point. Meaning... the odds change depending on who ends up with the 1st and 2nd picks.
I hate to bring up "Deal or No Deal" again, but if there are 26 briefcases and one contains a million dollars... You have a 1/26 chance of picking the briefcase with 1 million dollars in it. That's 4% chance of picking the right case.
But if as the game continues and they open 24 briefcases... none of which contain the 1 million dollars. That means there are two cases left... only one contains 1 million dollars. 1/2 chance that you have the briefcase with 1 million dollars in it. That's not "4%" anymore. Unless i'm really REALLY stupid... 1/2 means there is a 50% chance you are holding the case with 1 million dollars in it.
So nick... if I then asked you "Well Nick, there are two cases. One contains 1 million dollars. One doesn't. What's the chances the one you are holding has 1 million in it", your response would be: "To get into the exact math is fairly complex. It is not simple math. It deals with advanced probability and statistical mathematics. The answer is other. " To which I respond, "Well, no Nick. That's foolish. There's a 50% chance you are holding a case with 1 million in it. Simple math, bruh"
Of course, there's another layer here. Maybe you're talking about the fact that certain combinations involve certain ping pong ball numbers... and if some ping pong balls came up over and over during the first 3 draws, they are less likely to come up during the 4th draw... which would impact the probability for the Bucks to land the 3rd pick. But I'm pretty sure that's not what you are suggesting. And even so, it again proves my point that depending on how the first couple picks shake out, the odds fluctuate. Thus... I'm right.