Math/Logic Question

**Data** · 01-30-09, 10:13 PM

For each pair of handicappers, you must know a given handicapper's winning percentage for the plays the other handicapper has lost.

**solobass** · 01-30-09, 10:31 PM

also what the handicappers specialties and areas of expertise are. bottom line, you are the one making the bet.

**Ganchrow** · 01-31-09, 12:34 AM

Originally posted by TotallyTilt

Say handicapper A wins 54% of his picks, handicapper B wins 58%, and handicapper C wins 62%. Let's say they all pick the Bulls to win tonight, what are your chances of winning? What if only A and B pick the Bulls? What if A and C pick the Bulls?

Were one to assume the mutual statistical independence of each handicapper picking a winner then:

Pr(Team X wins | A, B, C pick Team X)
= Pr(A picks winner) * Pr(B) * Pr(C) / (Pr(A) * Pr(B) * Pr(C) + (1-Pr(A)) * (1-Pr(B)) * (1-Pr(C)))
= 54% * 58% * 62% / ( 54% * 58% * 62% + (1-54%) * (1-58%) * (1-62%) )
≈ 72.5650%

Similarly:

Pr(Team X wins | A, B pick Team X, C picks Team X's opp.)
= Pr(A) * Pr(B) * (1-Pr(C)) / (Pr(A) * Pr(B) * (1-Pr(C)) + (1-Pr(A)) * (1-Pr(B)) * Pr(C))
= 54% * 58% * (1-62%) / ( 54% * 58% * (1-62%) + (1-54%) * (1-58%) * 62% )
≈ 49.8392%

And:

Pr(Team X wins | A, C pick Team X, B picks Team X's opp.)
= Pr(A) * (1-Pr(B)) * Pr(C) / (Pr(A) * (1-Pr(B)) * Pr(C) + (1-Pr(A)) * Pr(B) * (1-Pr(C)))
= 54% * (1-58%) * 62% / ( 54% * (1-58%) * 62% + (1-54%) * 58% * (1-62%) )
≈ 58.1058%

Note, however, that within this context statistical independence is no more than an assumption of convenience and one that we'd only expect to have any realistic chance of being valid were each handicapper basing his picks off disparate selection criteria (and probably not even then).

To truly answer this question accurately without such a limiting assumption one would need further information as to the probabilities of conditional occurrences (i.e., what Data said).

**TotallyTilt** · 01-31-09, 02:36 AM

This is exactly what I was looking for. Thank you very much for taking the time to respond and for being so helpful.

**Indecent** · 09-08-09, 11:01 PM

Thanks Ganch! (For this and all your other threads)

Any chance you can explain further how you could calculate this given we know the predictions for each game? Here's a 3 game example that might be easy to illustrate how to calculate.

For example, assume A is 33.33% with (W, L, L), B is 66.66% with (W, L, W) and C is 100%. Ignoring the obvious that 3 games is too small of a sample size to be of much use, how could we calculate the winning chances if all three pick the Bulls? A and B ?

**Ominous** · 09-09-09, 09:53 AM

The concept of looking on covers for instnace and then picking based on what other "sucessful" handicappers are picking is not a valid strategy imo.

First of all, if all cappers are on a team its prolly a bad signal rather than a good one because they prolly bet on the same information which in that case most likely is overrated.

Second, most handicappers who post on forums and who seem to be doing well are just on a lucky spree, and thus to take thier previous 60-40 record and assume he has 60% winchance is probably a BIIIG mistake. Some of these cappers may even be creating multiple accounts until they get a lucky spree and a therefore high % (mostly based on luck).

IMO this mathematics is useless for practical purposes and while the problem may be logically interesting it does not say anything about your actual winchances in real life

**Indecent** · 09-09-09, 03:43 PM

Originally posted by Ominous

The concept of looking on covers for instnace and then picking based on what other "sucessful" handicappers are picking is not a valid strategy imo.

First of all, if all cappers are on a team its prolly a bad signal rather than a good one because they prolly bet on the same information which in that case most likely is overrated.

Second, most handicappers who post on forums and who seem to be doing well are just on a lucky spree, and thus to take thier previous 60-40 record and assume he has 60% winchance is probably a BIIIG mistake. Some of these cappers may even be creating multiple accounts until they get a lucky spree and a therefore high % (mostly based on luck).

IMO this mathematics is useless for practical purposes and while the problem may be logically interesting it does not say anything about your actual winchances in real life

What if someone were designing a prediction program using computer machine learning techniques? I'm automating the production of hundreds of prediction systems using various stats collected and am trying to start using multiple systems to make one prediction. I won't go into too many of the nitty details but I think using this approach sshould not fall into either of your traps...

1) The systems are given different sets of stats and learns without intervention from me. Although some of the systems use similar stats (all of them use rushing yards per minute, pass yards per play, etc) there are various other stats used that change the way the network learns. Each system is created using random values so the production of each system is both random and independent.

2) These systems can be back tested over thousands of games, producing a very good idea of how accurate each system expects to be.

So, there's no cause-effect relationship between each system's picks, and we do know the historical accuracy of the system.

Does my reasoning still seem flawed?