Email a copy of 'MLBTR Free Agent Prediction Contest' to a friend
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By Tim Dierkes | at
Email a copy of 'MLBTR Free Agent Prediction Contest' to a friend
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Austin A.
I’m winning this contest! Yaya!
corey23
Is there anyway to sign up if I don’t have facebook?
oaklandfan22
Lets go Oakland!
Odawg8
im sorta mad cause i got both of the signings so far right before they were announced but now everyone has it… :/
dylanp5030
If there was a person that didn’t have Ortiz going back to the Sox, they should not be participating in this contest.
Austin A.
This contest seems pretty easy. (Most likely not though). Only 15 picks from last year won, and we already have two. Does any one else feel that this is an easy contest?
Karkat
Nope.
Karkat
More to the point: let’s say you were assigning teams at random (because, pretty much…). That gives you a 1 in 30 chance for any given player. Now, we already have two, so the chances of getting at least 5 total right must be in our favor, right?
Nope: Assuming random assignment, we only have a 0.00083% chance of getting three more players correct.
Maybe we loosen things a little bit and assume three more “obvious” players (which don’t really exist, imo). Chances of getting ten out of 50, given 5 freebies: 0.0000011%
phillies1102
Well its not that random. People can analyze other teams, and they know that for instance, the Phillies will not get Zach Greinke, or the Tigers will not get Adam LaRoche. Then there are more judgemental decisions based on how well you know teams (also if you understand the correct goal of the contest). But yes, it looks a lot easeir than it is
Karkat
Yeah, obviously the numbers are a little better than what I presented, but even to get 8 correct at this point is something that comes with a very low (less than .01%, for sure) likelihood
baycommuter 2
I think the odds are better because a player’s old team has much better than a 1 in 30 chance of retaining him, Let’s say it’s 20 percent. Just by picking each player’s old team, on average you would get 10 out of 50 right.