Statistical Homework for the weekend

I think I have found an answer to a question that has been bugging me regarding sports wagering. I hope you all can find this statistical treatise useful for next week.

Given that you have $X amount available to bet on sports, and you are picking N games over a finite time period, and you accept R% risk that your cash may be lost on these game, then what is the optimal bet size that you should make?

First, we need to establish the probability p of a successful bet. What should p be:
p =
0.5 for normal uses
0.57 for Eric's NFL picks, 2005-2006
0.41 for Eric's NFL picks, 2007
0.37 if your name is Leland
0.75 if you are Kermit and betting against any games where Wiz and Eric agree

Each bet is a sample from the binomial distribution.
The expected wins are Np.
The standard deviation for the # of wins is the square root of (N)(p)(1-p)

So if you are someone like Eric in 2003, N would be 60 and p=0.5, the expected # of wins is 30 with a std deviation of sqrt(15), or about 4.

OK - now as the number of games increase, this approaches the bell curve distribution. That's the key here. You can now use the bell curve to have a confidence that you won't lose more than a certain number of games.

You have a 3% or less chance to lose more than (Np)-2 std deviations. You only have a 15% chance to lose more than (Np)-1 std deviation. We'll call D to be the # of deviations you are willing to accept. A 10% chance is 1.3 std deviations.

Given the D you choose, you now have that confidence in going having (Np-D*Std Dev) wins and (Np+D*Std Dev) losses.

You then use this to figure out your maximum loss. Your maximum loss given your D is:
(Np - D*(Std Dev))*B - (Np + D(Std Dev))*(B*1.10), where Std Dev is the standard deviation you figured out above.

Now you set this maximum loss to your bankroll and solve for B - voila! Your optimum unit bet size!

For example, Layup has $500 to work with and wants to bet 16 games (5 mens, 10 womens, with a double unit on the Lady Boilers). He is willing to take 15% risk.

We assume he does not listen to Leland and his p for each game is 50%.

Expected wins: Np = 8
Std deviation: sqrt(N*p*(1-p)) = sqrt(16*0.25) = 2

Now, his D for 15% risk is 1. So his worse case is (8-2) wins and (8+2) losses, with 15% risk. In other words, he is taking a 15% chance that he will not do worse than 6-10.

What would Layup bet to be 6-10 and lose his bankroll, $500?
6B - 11B = 500; B = 100.

So therefore he will make $100 bets.

It can be shown that if Layup was more conservative and took only 3% risk, his unit bet would then be only $54.34. This will be an exercise for the reader. (I've always wanted to say that.....)

"Well", you say, "he may go 1-6 and be out of money by 2:00 pm on Thursday!". That's the 15% risk. Or that's the alcohol making the picks. Or Don is whispering "Big Ten" in his ear all weekend.