On this blog, I am also going to record a hypothetical betting for this college football season. The rules for the betting are:

- I will start the season with a bankroll of $1000 on Sept 19
^{th}, 2014, I know this is a day late - There will be no maximum to the amount I can bet
- The minimum bet amount is $5
- If the bankroll goes below $100, it plus up back to $100
- The following day bankroll will be the based on the amount of the bet from winnings and loses from the previous day, so even though I could have a compounded bets multiple times on Saturdays due to the stagger of game times, I will consider all games starting at the same time for simplicity
- All bets will be in whole dollar amount; where $100 would be allowed but not $100.49
- I will cap the Kelly bet to 30% of the bank roll, I do this for a few of reasons

1. I’m a conservative better and the idea of betting 60% or 70% on one game makes me squeamish

2. This allows me to have at least four bets a day therefore hedging my bets a little more

3. My testing on previous season shows this was a good number for maximum winning

- I will do no parlays, as all explain on later blogs parlays are not good bets individual game betting has an overall better pay back

Based on this criteria my betting for Sept 19

^{th}would have been:
$200
on UConn or 20% of my bankroll based on my picks. UConn lost 14 to 17
to South Florida the spread was UConn +2.5, UConn could not cover the
spread and I lost my bet. I will have $800 to bet on Sept 20

^{th}
My bets for Sept 20

^{th}are based on my algorithmic picks will $800 in the bankroll:
$240 (Kelly 30%) on Utah State +2.0 against Arkansas State

$240 (Kelly 30%) on Mississippi State +9.0 against LSU

$216 (Kelly 27%) on Utah +3.5 against Michigan

$104 (rest of the bankroll) on Maryland + 2.5 against Syracuse

It
might be considered risky to bet the whole bankroll, but statistically
if my algorithms estimates are correct there is a very small chance I
will lose all the bets. The probability of losing all the bets are

(1-0.8333) X (1-0.6667)X(1-0.6531)X(1-0.6190) = 0.0073

So, I only have a 0.7% of losing all the bets. This formula is similar to a binomial distribution where I would have 4 outcomes where none of them are true, but the value of p is different for each pick so the above equation work.

While I believe that algorithm works based on testing with real data and Monte Carlo testing on simulated outcomes, even if my predictions were totally wrong and the picks are totally by chance the probability of all four being wrong is 6.25% or 0.50 X 0.50 X 0.50 X 0.50 where random picks have a 50% of being correct.

I get my data from the Sunshine Forecast, and this weeks stats were not updated so my confidence in my predictions is not as high as I would like.

I get my data from the Sunshine Forecast, and this weeks stats were not updated so my confidence in my predictions is not as high as I would like.

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