A Serious Question for You Geniuses

[Mjulian]

Here is what I would like to know from the stat guys in here:

I am trying to determine whether a guy's performance is statistically significant or is simply a positive abberration culled from a much larger timeframe.

He is 100% mechanical, or so he says.

His record in bases last year from July 1 on was:

507 - 425 with a hold of 4.8%

Using slightly modified selection criteria, his record in pucks this year is:

203 - 181 with a hold of 0.6%

And in Hoops it is:

146 - 120 with a hold of 5.5%

All told from July 1 to yesterday he is:

856 - 726 with a hold of....not sure how to figure out the combined hold but its gotta be somewhere in the 3 - 4% range.

My question then is are these enough observations to conclude that this guy's methodology is sound or could this just simply be a decent run.

The key is of course that he says he selects games 100% mechanically based on a few key variables adjusted sport to sport, variables that he claims are not subject to change, ie: rules changes, juiced ball, etc.

[count zero]

A good quickie measure for the reliability of a given set of results is z-score, which takes into account both the win percentage and the number of trials, each of which bears on the question of reliability. The calculation is simple, though there is some room for interpretation in evaluating the results.

The calculation first: figuring the z-score normally involves a more complicated procedure than the following, but in the case of sportsbetting results, this shortcut can be used: divide the difference between wins and losses (wins minus losses) by the square root of the number of trials (wins plus losses). That's it. Typically, you'll come up with a number between zero and 2 or 3, sometimes (rarely) higher. Doing this calc for the examples you mention gives the following:

1) 507-425 in bases, z = 2.7 (507 + 425 = 932; SQR(932) = 30.53; 507 - 425 = 82; 82 / 30.53 = 2.7).

2) 203-181 in pucks, z = 1.1 (203 + 181 = 384; SQR(384) = 19.6; 203 - 181 = 22; 22 / 19.6 = 1.1).

3) 146-120 in hoops, z = 1.6 (146 + 120 = 266; SQR(266) = 16.3; 146 - 120 = 26; 26 / 16.3 = 1.6).

4) 856-726 overall, z = 3.3 (856 + 726 = 1582; SQR(1582) = 39.8; 856 - 726 = 130; 130 / 39.8 = 3.3).

The usual way of interpreting the z-score is that z=1 means there's about a 65% likelihood that the record is meaningful and about a 35% likelihood that the record was achieved by chance; z=2 means there's about a 95% likelihood that the record is meaningful and about a 5% likelihood that the record was achieved by chance; z=3 means there's about a 99.5% likelihood that the record is meaningful and about a 0.5% likelihood that the record was achieved by chance. However, and this is very important, that's in the unchanging world of the physical sciences, where you can be 100% sure that every trial, both past and future, is exactly the same. You don't come close to that in real-world activities like betting -- players change, coaching strategies change, the rules change, you have the human element, etc. The statistics texts I've read (all two of them) suggest that, as a rule of thumb in "sociological" applications, you should roughly double the requirements for what constitutes each level of reliability. Thus, you'd need a z-score of 2 to establish 65% reliability, a z of 4 to establish 95% reliability, and a z of 6 to establish 99.5% reliability. This agrees reasonably well with my experience, where I've found that betting concepts or personal W-L records with a z of 1 or 2 subsequently perform essentially randomly, while the few cases where I've found something with a z of 4 or 5 continue to perform well year after year.

So on the basis of z-score, I'd say that this bettor's overall record is encouraging, but not godlike, and that his individual sport records, especially in hockey and basketball, are not particularly meaningful.

[PMcK]

Count zero, how about a 500-500 record with a +120 average moneylines in baseball or hockey? z = 0?

Doesn’t the z-score calculation above assume that the W-L outcomes are equally probable?

You can convert the moneyline sport records to those “flip-a-coin” sport records like this:

Calculate the “hold” using (decimal line)*win%. A 50% record with +120 lines (or 2.20 decimal line) gives:

(decimal line)*win% = 2.20*0.50 = 1.10 or 10% “hold”.

An equal -110 or 1.91 decimal line ATS win% would be:

win% = 1.10/1.91 = 0.576 or 57.6%.

So the 500-500 record with +120 lines is equal to a 576-424 record with -110 ATS lines if you consider just the $.

[count zero]

Oops, forgot about that. Right you are, PMcK, thanks for the info.

So, how about this: using Mjulian's figure for the baseball segment of 507-425 with a hold of 4.8%, we'd use 1.048 / 1.91 = .549 or 54.9% as the adjusted win % with both outcomes equally likely. Over the 932 plays in Mjulian's sample, 54.9% would be roughly equivalent to 511 wins and 421 losses. Calculating the z-score as above on this record gets z = 2.9 rather than the 2.7 I found. Is that legal?

[Mjulian]

CZ and PMcK,

You guys are great. That is exactly the type of information I was looking for. I appreciate it.

And thanks for walking a dope like me through the calculations too.

MJ

[GottaWinToday]

THANK YOU for the information here.....I'll use this in my future tries to find the perfect guru......

Thanks again.....


GottaWinToday

[rabbit]

Regarding the Z-score, you can get the same result in Excel with the following formula:

=(W-(0.5*(W+L))*((W+L)*0.25)^-0.5

where W = wins in your sample, and L = losses

[Jack_straw]

aren't you really figuring the "distributio" of the occurances, standard deviation?