Math question to Ganch (if you have time)

Dark Horse · 09-28-2007, 04:11 AM

A method objectively divides 7 points between two teams in a contest. When these 7 points are divided as 4-3, the results are:

Home favorite 4-3: 9-29 ATS
Home dog 4-3: 13-3 ATS
Road dog 4-3: 3-12 ATS

What is the chance that these ATS results are the result of luck?

And if they're not the result of luck, what type of strength (or bet size) would you assign to such a method?

Many thanks!

Ganchrow · 09-29-2007, 08:07 AM

Quote:

Originally Posted by Dark Horse

A method objectively divides 7 points between two teams in a contest. When these 7 points are divided as 4-3, the results are:

Home favorite 4-3: 9-29 ATS
Home dog 4-3: 13-3 ATS
Road dog 4-3: 3-12 ATS

What is the chance that these ATS results are the result of luck?

And if they're not the result of luck, what type of strength (or bet size) would you assign to such a method?

Haven't we had this discussion before? As I recall you never seem to be satisfied with my answer.

Anyway, just as last time the, most important issue is of data sampling methodology. Did you go into this expecting these types of results, looking at only this one scoring method? Or did you look at many different scoring method, with this one happening to be the one (out of sever candidates) that worked? The former is indicative of a model substantially more likely than the latter to possess predictive power. The latter represents what's known as "data mining" and is to be avoided. (The idea being that if you look at a randomly generated set of data long enough, you'll eventually find patters, which although they might describe set itself, have no predictive ability whatsoever.)

I also can't help but ask why you're only presenting scores of 4-3. If points are generally a "bad" thing for a team, then in general wouldn't we expect that more points would be indicative of a greater probability of defeat? Why would an even attribution of points (4-3) ostensibly represent a higher probability of loss (or victory) than a more lopsided one (such as 7-0, or 0-7)? Now obviously there might be a good explanation for this, but at first blush my initial reaction would be that this is indicative of data mining.

Ganchrow · 09-29-2007, 08:16 AM

But anyway, assuming your data were exclusively obtained out-of-sample, and that this represented the only strategy which you were considering, the probability of a true 50/50 picker going better than 44-25 or worse than 25-44 over 69 trials would be 2.949%.

Dark Horse · 09-29-2007, 07:06 PM

Not data mining. Nor going into it with any expectation. Just a neutral spectator, scratching his head at the early sample. The obvious potential being that this sample is not 25-44, but 54-15. Time will have to tell. Thanks for the input.

Ganchrow · 09-29-2007, 08:23 PM

Quote:

Originally Posted by Dark Horse

Not data mining. ... The obvious potential being that this sample is not 25-44, but 54-15.

You see that right there ... that's data mining. You've looked at your results and realized how much better they'd be if you had reversed the meaning of your scoring system for home dogs versus home faves and road dogs (and where, by the way, are road faves on the list ... were they not included because the results were close to 50/50?) Because you're allowing your results to dictate the terms of your model you shouldn't be surprised when those results confirm that model's predictions.

As we've discussed before, it's a question of properly segmenting your data set into in-sample and an out-of-sample partitions. One should first first formulate his model based his in-sample data set, but then only consider his model confirmed when the model's predictions are verified within the out-of-sample data set.

But anyway. All that said. If indeed the above results were properly obtained out-of-sample, there would be a roughly 0.000261% probability of it being the result of pure chance. (That's a 0.000131% probability of results of 54-15 or better and an equivalent probability of results of 15-54 or worse.)

Dark Horse · 09-29-2007, 09:36 PM

The point behind my question was that the results were so obviously different for HF-HD-RD. I wanted to know what the chances of that were.

Also note that I merely observed the 4-3 division of pts. I didn't even go so far as to assign strength to it. How could I be more objective than by being open to the possibility that it could be a strength to a HD and a weakness to a HF?

If you just take dogs the result is 16-15 ATS. Nothing there. But divide it into HD and RD and the results are 13-3 and 3-12. Coincidence?

CrazyL · 09-29-2007, 10:15 PM

Alright enough numbers talk fellas my brain is starting to hurt. Take it to private! =)

m3vr6 · 09-29-2007, 11:20 PM

isn't road dog the same as home favorite? also your question can be stated as "What is the chance that these ATS results are the result of NOT luck?" and your answer would be 1-(What is the chance that these ATS results are the result of luck?")

Ganchrow · 09-30-2007, 01:04 AM

Quote:

Originally Posted by m3vr6

isn't road dog the same as home favorite?

Ostensibly, we're looking only at teams which qualify for 4-3 scoring. Otherwise, as you've pointed out, we'd be double counting.

And, by the way, just to make it perfectly clear, while I disagree with roughly 99⁹³/₁₀₀% of Dark Horse "quantitative" conclusions, I still think he qualifies as one of those sharpest guys at this place ... just so long as we aren't talking math. ;-)

Dark Horse · 09-30-2007, 01:58 AM

A man can't ask a simple question anymore, without being accused of 'quantitative conclusions'.

On the subject of sharpness, what if the 4-3 division line were not an angle, but a knife? Does it make any difference if it is used to cut a birthday cake or slice a throat? None whatsoever. It's still just a f*cking knife.

Happy birthday!