MikeJones said:
Guess you've never heard about the "frozen envelope" in the eighties and the #1 pick to, conveniently enough, the team that had just been taken over two weeks earlier by the NBA's biggest superstar of all time...
Just because it's on TV doesn't mean the fix wouldn't be in. The NBA does it regularly. I wonder how Cleveland just happened to get the #1 pick when hometown hero Lebron was coming up, hmm?
I don’t follow basketball so I don’t know that story, sorry.
As for lottery on TV being fixed, yes it‘s still possible but it’s not that easy and sure.
In this case on TV we could see that all the 48 balls had the right property.
We could also see if some teams’ balls come out all at first.
And if they wanted, they could make the lottery in a way that’s very difficult to think it’s fixed.
Just think about throwing 48 balls with the same colour in a big basket that can mechanically mix them. Then give someone a long shoveler, he mixes again with it and then picks up a ball. They are going to fix it? Good luck!
HockeyCritter said:
From a purely statistical standpoint, the chances would remain the same if you removed a “dead†ball if it were selected. However, the probability changes.
Same chances because of same (sum of) probability.
ronald26 said:
Insightfull. Please elaborate.
Yes they do have to remove them, as the sample, if you even understand statistics, would be invalid.
No, they don’t.
Let’s elaborate….
Suppose the Rangers with one of their 3 balls (3/48) are picked first.
Now consider another team, for example Columbus (3 balls as well).
-1st scenario-
For the second drawing, if you take out the two left NYR (dead)balls, Columbus has 3/45 to be picked second. 3/45=0.06666 -->
6,666%.
-2nd scenario-
Now reconsider the second drawing without taking away NYR dead balls: 47 balls are still in the basket. That gives Columbus a 3/47 probability to be picked straight away.
BUT that’s not all, you have to add Columbus’ chances resulting from the eventuality that one of the two dead balls is picked instead, so that the second drawing has to be repeated. There are 2/47 chances that a dead ball is the one picked in the second drawing. In the consequent repetition, 46 balls are left and Columbus has 3/46 chances to see one of their balls picked.
So that means Columbus has (2/47)X(3/46) more chances to be picked second.
Again, it’s not all. In the first repetition (2/47 to be taken) there is the remote eventuality (1/46) that the one NYR left ball is picked. If that’s the case, a second repetition is taken with 45 balls left and this time Columbus has 3/45 chances to see one of their balls picked.
That means Columbus has (2/47)X(1/46)X(3/45) more chances to be picked second.
Now add together all the possibilities Columbus has to be picked second:
3/47 + (2/47)X(3/46) + (2/47)X(1/46)X(3/45)=
= 0.06383 + 0.00277 + 0.00006 = 0.06666 -->
6.666%
Insight-full?