Quote:
Originally Posted by Ganchrow
There's nothing wrong with estimating p-values using Monte Carlo simulations. There is, however, something wrong with estimating p-values using Monte Carlo simulations in Excel. It's hard on the soul.
Here's a very simple Monte Carlo script coded in Perl. (You can download a free copy of Perl from http://www.activeperl.com/.) On a decent machine you should easily be able to run 2,000,000 55-bet trials in a minute or less. You should modify the EDGE, TRIALS, and BOGEY constants to suit your needs.
Assuming all bets are independent of one another, then the simplest method would just be to appeal to the Central Limit Theorem. Take the sums of the variances of each bet (betting to win n unit at decimal odds d and edge E, variance would be (1+E)*(d-E-1) * n^2 / (d-1)^2) and then take the square root of that sum to obtain the standard deviation and a z-score. (So for zero-edge and betting to win 1 unit, variance would just be d-1.)
In your example, assuming no edge, we get a standard deviation of about 15.47%. This means your results of ~15.72% is about 1.016 standard devs from breakeven for a p-value of about 15.48% (=1-NORMSDIST(1.016)). |
Monte Carlo method eh? Well, I feel pretty good that there is an official name for what I was doing.
I thought I had invented it.
You do have the correct picture of me sitting for 15 minutes waiting for Excel to chug through 10,000 simulations. Thank you for the link to the program code. I haven't downloaded it yet as I am just reading this post now but I would also be interested in the randomizing code you wrote about. Seems to me if I was running 100,000,000 trials to test a model that I would want to stay away from any repeating pattern.
Regarding your simplest suggestion, there seems to be one parenthesis missing and I can not make the equation work.
If I have:
E=0 (Edge)
d=.6 (American Odds of +150)
n=1 (Units to win)
Then I get:
(1+0)*(.6-0-1) * 1^2 / (.6-1)^2)
From my understanding, I assume I would calculate the results of the above equation for each of the 55 independent samples and take the square root which will give me the standard deviation. I believe the above example of +150 would evaluate to 3.75 if I ignore the last parenthesis. The square root of many such large numbers will not come close to 15.48%, so I am lost. Also, do the z-test and standard deviation both evaluate to 15.48% in your example and this is why you are able to get both numbers from one?
I do have a question for you about adhering to the Central Limit Theorem. Not so much from this post but from other posts you have made. I get the feeling that when you say things like "as long as your comfortable appealing to the Central Limit Theorem" --> (not a direct quote as I am going from memory on this) that maybe I shouldn't be appealing to it and maybe I should be thinking along the lines of Bayesian. Here are 2 simple questions that I have often wondered. Relevant to sports betting, do you personally appeal to the Central Limit Theorem for calculating probabilities and significance? Are there situations that you do not?
Also, if it were you and you had the choice of running 100,000,000 Monte Carlo simulations (with the better randomizer in place of course) or using the Central Limit Theorem example you proposed, which would you choose?
Finally, the following equation did not work in Excel as it is missing parameters for the function. I am sure that they are probably assumed numbers like 1's and 0's but it would be helpful if you could fill them in for me.
15.48% (=1-NORMSDIST(1.016))
I promised lots of smiley faces for the Excel type formulas included with the answers so here they are. Thanks for coming through, again, for me Ganchrow.
