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Possible LSU/ Bama Rematch

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    crimson rick

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    macamatic ●

  • Also remember Floridas last NC? After the gators lost to Ole Miss , QB Tim Tebow told the nation that the rest of the year you won't see a team who wants it more, who will fight their way back to the top, or another team play as hard as them the rest of the season and it proved true! 1 loss gators won it all! I hope we can have that kind of leadership and promise out of our players and we could very well put ourselves back into contention! Roll Tide!

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    Bamapoole03

  • bamamba1989 said...

    There are 3 scenarios:
    OkSt loses to Oklahoma (70%) and Standford loses to one of (Oregon, ND, PAC CG) (50%) = 70x50=35% odds
    Arky beats LSU and Arky doesn't lose another game and ranks higher BCS than LSU (20%)
    LSU loses to GA and One of the teams in 1st scenario loses. (20%)

    If you add all this up you get (20+20+35=55%) I'm not a statistics dude but I think that's right. ("and" multiplies and "or" ads) That's how I would handicap our chances of getting back in this thing.

    That's not how it works. If you have two things with a 75% chance of happening, the chances of one of them happening is not 150% for obvious reasons. If we were to go by your probabilities, it would be 58.4%, which you may notice is actually better.

    macamatic ●

  • bamamba1989 said...

    There are 3 scenarios: OkSt loses to Oklahoma (70%) and Standford loses to one of (Oregon, ND, PAC CG) (50%) = 70x50=35% odds Arky beats LSU and Arky doesn't lose another game and ranks higher BCS than LSU (20%) LSU loses to GA and One of the teams in 1st scenario loses. (20%)

    If you add all this up you get (20+20+35=55%) I'm not a statistics dude but I think that's right. ("and" multiplies and "or" ads) That's how I would handicap our chances of getting back in this thing.

    Your stats addition isn't quite correct. You are assuming that all three events are independent and they are not (see event three referring to event 1). To add the probabilities, you'd need to do something like

    P(Event 1, given Event 2 and Event 3 do not happen)+
    P(Event 2, given Event 1 and Event 3 do not happen)+
    P(Event 3, given Event 1 and Event 2 do not happen)

    If these were/are the only 3 scenarios that get bama in, then this would be correct (in a simple way). There is some intersection of the events that you stated, by just adding the probability of all 3 you are double/triple counting some of the event space that puts Bama in the game.

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    menichols74

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