# 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.

## 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. grossman says:

Here’s a chart of my interest in college bowl games this year:

3. […] 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 […]

4. @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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