My picks for Sept.
27 were pretty bad but will happen from time to time as as I explain
below.

First, an overview
of my pick outcomes:

$510 (30% of
bankroll) on Indiana -3.5 against Maryland

LOSS Maryland beatsIndiana outright 37 – 15

$510 (30% of
bankroll) on Purdue +8.5 against Iowa

LOSS Iowa beatsPurdue 24 – 10

$407 (24% of
bankroll) on South Carolina -5.5 against Missouri

LOSS Missouri beats South Carolina 21 – 20 outright

$271 (16% of
bankroll, rest of bankroll) on Tennessee +17.5 at Georgia

WIN $517.36 While
Georgia beat Tennessee 35 – 32 Tennessee covered the generous +17.5
spread

Now my bankroll is
$517.36 from $1678.98 or down 71% and I am 2 – 3 for the week.

While the chance of
this exact outcome (losing the first three of my picks, and getting
the last one) is very unlikely, the probability of only getting one
pick right has a low probability, but as unlikely as you may think.
Let me explain.

Calculating the
probability of events that can only have two outcomes in this case is
my pick right or wrong is called binomial. To calculate, the
probability that a particular event will happen, in this case of me
getting one pick right, you have to add together all the scenarios
that causes an event to happen.

So, one of the
scenarios of getting one pick right was what happened yesterday, the
first three picks were wrong and the last pick was right which is
represented by the set, {lllw}.

Where the “l”
represents a lose on my part for that particular pick and “w” is
a win. So you can read {lllw} as picks 1, 2, and 3 are losses and
pick 4 is win. For the event, 3 loses and 1 win, there are four
scenarios that can cause that event the one given above and three
more: {llwl},{lwll}, and {wlll}. The probability of a 1 – 3 pick
day for the probabilities from yesterday's picks would be as follows.

The probability of
{lllw} written in shorthand p({lllw}) given my probability estimates:
game 1 UI -3.5 vs Maryland 71.43%, game 2 Purdue +8.5 vs Iowa 71.43%,
game 3 South Carolina -5.5 vs Missouri 64.00%, and Tennessee +17.5 at
Georgia 60.47% would be

p({lllw}) =
(1-0.7143) X (1-0.7143) X (1-0.64) X 0.6047 = 0.0177 = 1.77%

So, the scenario
{lllw} occurring is very rare, but the probability of only picks one
game correct includes the sum of all scenarios so let calculate the
other three

Finally, the
probability of one win out of three picks this week represnted as
p(three losses) is

p(three losses) =
p({lllw}) + p({llwl}) + p({lwll}) + p({wlll}) = 0.0963 = 9.63%

While these are
probabilities for week 5 of the College Football 2014-2015 season, my
pick probabilities for the 4 games are about the same every week
(between 60% – 70%), so you can expect any given Saturday that
there will be a 10% chance of only getting one pick right (in other words it is most likely to happen a few times a season) and for the
dire probability of picking all the games wrong at 1%, but on the flip side I have a 99% of not loses all my money in one week.

So, I will, literally, probably make the money back, the probability of picking 3 or more games correctly is about 60% (or the sum of getting 4 picks correct, ~20%, and getting 3 picks correct, ~40%, on a any given Saturday). In the long run, I will get my money back and it will keep growing.

So, I will, literally, probably make the money back, the probability of picking 3 or more games correctly is about 60% (or the sum of getting 4 picks correct, ~20%, and getting 3 picks correct, ~40%, on a any given Saturday). In the long run, I will get my money back and it will keep growing.

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