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ATP: Probability Question

Screamindemon3

Screamindemon3

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It's been a while since I took Statistics 1/2, so I want to make sure I'm getting the correct numbers here. Here's my question:

If you play in a FF league for three season, the first season having 10 owners, the second season having 12 owners, and the third season having 12 owners. What is the probability that the same 2 owners will finish 9th/10th the first season and 11th/12th the next 2 seasons? (doesn't matter which one they finish, just that they're in the bottom 2 of all 3 seasons)

Were assuming everything is equal, so individual manager knowledge doesn't come into play. Just pure probability of the situation occurring.
 
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My guess was:

1/10 x 1/9 = 1/90

1/12 x 1/11 = 1/132

1/12 x 1/11 = 1/132

1/90 x 1/132 x 1/132 = 1/1568160

OR

1/5 x 1/9 = 1/45

1/6 x 1/11 = 1/66

1/6 x 1/11 = 1/66

1/45 x 1/66 x x1/66 = 1/196020

Couldn't decide which one it would be. Since it doesn't matter which position the first owner finishes in, I was leaning towards 1/196020 as well...
 
2(8!)/(10!) * (2(10!)/(12!))^2=5.102*10^(-6)=1/196020

Got the same answer
 
thanks guys

Rule #1 of being a FF commish, always make sure your numbers are correct before you put out an article
 
It depends if you're asking if the two specific people finish at the bottom three years in a row or if two people finish at the bottom three years in a row.
 
JDawg said:
It depends if you're asking if the two specific people finish at the bottom three years in a row or if two people finish at the bottom three years in a row.
Click to expand...

Um, two people will always finish at the bottom of any fantasy league unless you have a league of only two people.
 
I meant "Jack and John are in the league, what are the odds they finish at the bottom three years in a row" vs "what are the odds the last two spots are occupied by the same two people for three years in a row" because in the second question, the first year doesn't really matter because like you said someone will finish in the last two spots regardless. You don't care who.
 
also, the odds are identical to the odds that the same two people would finish 1st and 2nd.
 
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