Normal Winners

Recently, 1,158 participants selected the winner, against the spread, of 19 NCAA College Football bowl games. I, being one of those participants, selected 8 correct winners – not my best effort. Let’s see how everyone else did.

And the data

Wins Players Normal
0 0 0
1 0 0
2 0 0
3 1 2
4 6 7
5 31 22
6 77 55
7 145 108
8 196 170
9 217 213
10 209 213
11 160 170
12 68 108
13 38 55
14 6 22
15 4 7
16 0 2
17 0 0
18 0 0
19 0 0
1158 1154

The formula in column C is

=ROUND(NORMDIST(A2,AVERAGE($A$2:$A$21),$F$1,FALSE)*(AVERAGE($B$11:$B$12)/NORMDIST($A$11,AVERAGE($A$2:$A$21),$F$1,FALSE)),0)

I really don’t know much about normal distributions. I scaled up the result of NORMDIST so that it’s highest point would match the highest point of the data. I started with 1 standard deviation and it was a much narrower curve. I ended up putting the standard deviation in F1 and adjusting it until the number of participants was about equal to the actual number of participants. I probably broke some statistical commandment, so I’ll see you in statistics hell.

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4 thoughts on “Normal Winners

  1. Given the nature of the problem – 19 trials in a row – you could use a Binomial distribution rather than a Normal. That way, no approximation needed! =BINOMDIST(8, 19, 0.5, FALSE)*1158 should give you the expected number of people correctly calling 8 games, assuming they have a 50% chance of calling it right.
    Otherwise, matching a normal to a sample is usually done computing the sample average and standard deviation.

  2. […] 8 correct winners – not my best effort. Let’s see how everyone else did. And the data… [full post] Dick Kusleika Daily Dose of Excel charting 0 0 0 0 […]

  3. @Mathias:
    I like your suggestion of using the binomial distribution. I would also add that instead of assuming 0.5 chance of calling it right, we use the sample average for a best fit. For this data-set, the average # of wins is 9.05, for an event probability of 9.05/19 = 0.477.

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