The Astros have a 53% chance of winning 83 games in 2009
I was inspired to get fancy by BtB and plugged the Astros WAR Projection Project into a binomial distribution function to determine what the probability was that the Astros actually hit the mark our input projected. The actual probability was .531, so I understated the probability, slightly, in the title.
I know this isn't entirely accurate, but I thought it would be fun none the less.
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::head explodes::
When statistical analysis goes too far….
Ceci n'est pas une signature.
by DbacksSkins on
Jan 11, 2009 8:56 AM CST
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Joe Morgan voice:
YOU DON’T PLAY BASEBALL WITH CALCULATORS, DAMMIT!
Seriously though, good job. Can you figure out what the probability is for 90 wins?
by Only_A_Lad on
Jan 11, 2009 10:47 AM CST
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I got 18%, using DQ's sheet.
Here’s the total:
115 0.02
110 0.08
105 0.69
100 2.22
95 9.09
90 18.7
85 41.2
80 58.8
75 81.3
70 90.9
65 97.7
60 99.3
55 99.9
by R.J. Anderson on
Jan 11, 2009 11:18 AM CST
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Hmmmmm
I wonder what i did terribly wrong?
The Crawfishboxes
A good friend of mine used to say, "This is a very simple game. You throw the ball, you catch the ball, you hit the ball. Sometimes you win, sometimes you lose, sometimes it rains." Think about that for a while.
by DyingQuail on
Jan 11, 2009 11:21 AM CST
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Nevermind
Like a jackass I put in the cumulative probability. I got 6.26% (I still can’t figure out where the disconnect is, I used 83, 162, .512345679, and 1).
I should really not be blogging at 5:00 AM at the peak of insomnia.
The Crawfishboxes
A good friend of mine used to say, "This is a very simple game. You throw the ball, you catch the ball, you hit the ball. Sometimes you win, sometimes you lose, sometimes it rains." Think about that for a while.
by DyingQuail on
Jan 11, 2009 11:30 AM CST
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My formula is:
=100-(100*BINOMDIST(O23/2,81,0.51234568,1))
With O23 set to 83, or whatever wins total you want.
by R.J. Anderson on
Jan 11, 2009 11:43 AM CST
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So.. there is a chance we can win 115 games?
Sweet!
by entropic soul on
Jan 11, 2009 4:45 PM CST
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