[MadCapper]

If someone wins at a 60% clip for 1000 plays, is it safe to assume that he will likely win no less than 55% percent of his next 1000 plays?

[MadCapper]

Or do you need need a larger sample size?

[mofome]

no one will ever hit 60% over 1000 plays.. it's virtually impossible. unless they're betting into opinionated books who play around with their lines, or beating very slow moving books who don't move lines quickly enough and are getting 2 to 3 pts better on every game. but playing into real lines, no one will hit 60% over 1000 games. It's a pretty large enough sample size.

but when betting into real lines, i would say 1 in 25,000 bettors will go 600-400 over 1,000 games. and over 20,000 games, not one bettor will go 12,000-8000 (60%).. they would have a better chance at winning the power ball lottery 3 times over.

What about pavy?

[MadCapper]

Nick,

Thanx for the speech

But you didn't answer my question.

All I asked was IF someone does..........

Thanx in advance.

[MadCapper]

maddy,

of course if you can hit 60% over 1000 games, then you should have no problem hitting 60% consistently, but the problem is, it's virtually impossible. it's like me asking, if i can pick up 25 hot girls in one day and bang each one of them, will i be a playboy.. of course, but i can never do this, so why mention it. it's just a fantasy.. 60% over 1000 games is impossible, like i said, unless you are beating lines left and right.


why don't you try for 54%.. you can make a great living hitting 54%.. all you have to do is if Bos is -4.5, try to find a bos -3.5.. it's much easier to do this than to try to hit 60%.

or if you want to hit 56%, if bos is -5.5, bet the game at bos -4.. if you can beat the closing line by 1.5 pts, you can easily hit 56%.. each 1 pt in nba is worth very close to 4%.

Thanx NICK.

I know I will never hit 600 out of 1000. But I am sure someone can.

I dont have to hit 60% or even 55% to make money because I play a lot of mlb and tennis underdogs.

[durito]

I can guarantee you I can hit 60% on a 1,000 plays.



* you all might want to mention what type of odds you are talking about.

[MadCapper]

durito, good one. i know exactly what you are saying.



if all your picks are -160.. then you should hit 60% with no problem

if all of your 1000 picks are -210/+190, then you will hit 66% over 1000 games..


if all of your 1000 games are -330/+270, then you will hit 75% on your 1000 games.




so as you see maddy, you can hit 60% over 1000 games..

moneyline dogs, sides, and totals at -107.

No bigger moneyline favs.

[MadCapper]

Nick,

No. I asked IF someone does hit 60% WITHOUT betting on big moneyline favorites over 1000 games, can we assume he will hit at that pace for the next 1000.

Or will he (however impossible it was to do) drop down to around 50%?

[durito]

Nick,

No. I asked IF someone does hit 60% WITHOUT betting on big moneyline favorites over 1000 games, can we assume he will hit at that pace for the next 1000.

Or will he (however impossible it was to do) drop down to around 50%?

You'd be safe to assume nothing.

Nevertheless, the probability of hitting 600 or more wins out of 1,000 bets (assuming coin flips, ie 50% win chance) is .0000000136423%

All this says, however, is that his past results were very likely not the result of chance. It says nothing about what will happen in the future.

[LT Profits]

MadCapper,

I don't think ANYONE can hit 60% over 1000 plays without betting big ML favorites. It is physically impossible.

[MadCapper]

MadCapper,

I don't think ANYONE can hit 60% over 1000 plays without betting big ML favorites. It is physically impossible.

I agree its near impossible.

But if someone SOMEHOW does....

What can we expect from his next 1000 plays?

I would think this guy has some special talent to do something like this. It cant be all luck.

[katstale]

If he did it--subscribe!!

[tacomax]

MadCapper,

I don't think ANYONE can hit 60% over 1000 plays without betting big ML favorites. It is physically impossible.

Robzilla hits 75%

Dodif hits 108%

[imgv94]

I agree its near impossible.

But if someone SOMEHOW does....

What can we expect from his next 1000 plays?

I would think this guy has some special talent to do something like this. It cant be all luck.

Start fading him.. Seriously if someone was hitting 60% over 1,000 plays then it's obvious he didn't many bad beats.. A lot of that was luck.

It's really hard to answer this cause we don't know how experienced he is or if he just has beginners luck.

[beetman]

Given the -110 odds or lower caveat, 1000 picks is a rather large sample and unlikely to be obtained due to luck.

[raiders72002]

http://faculty.vassar.edu/lowry/binomialX.html

n=1000
k=600
p= .5

impossible

[tacomax]

http://faculty.vassar.edu/lowry/binomialX.html

n=1000
k=600
p= .5

impossible

What's impossible?

[raiders72002]

Look at the binomial distribution chart and see what the odds of hitting 600 or more.

Something could be wrong with the chart because I can see you hitting 400 or less.

[tacomax]

Look at the binomial distribution chart and see what the odds of hitting 600 or more.

Note a couple things:

1) Impossible and statistically improbable are not the same thing

2) You assumed p=0.5

[Ganchrow]

This is a traditional exercise in what’s known as “Bayesian inference”. You have a number of observations of a random variable (1,000), which you use to infer knowledge about a statistical parameter (the handicapper’s “true” success rate).

As the question has been phrased, however, the answer is not well defined. This is because we still need to make an assumption regarding the distribution of the parameter within the population. To put it in simple terms, we need to know how likely it would be that a randomly selected handicapper would have a true success rate of X (this is known as the “prior distribution” of X as it reflects our knowledge of X prior to any observations of the particular handicapper).

The reason for this requirement should be clear on reflection. Imagine, for example, if we knew with 100% certainty that because of the randomness inherent in sports betting no handicapper could ever be a > 58% true picker. If this were the case then we could say with certainty that at least some of the handicapper’s observed success would be attributable purely to luck.

One common simplifying assumption is that the prior distribution of our parameter is “uniform” meaning that all values are equally likely within the population. Understand of course that we know this to be untrue. We know, for example, that 50% pickers are infinitely more common within the population than 100% pickers. But that said, assuming a uniform distribution does provide for computational ease. First, I’m going to demonstrate how to use Excel to calculate a solution to this problem under the working assumption of a uniform distribution pickers within the population, then I’m doing to demonstrate the approximate solution to the same problem using a slightly more realistic prior distribution.

The problem: Over 1,000 trials a handicapper has demonstrated a pick rate of 60%. We wish to determine the posterior distribution (i.e., after considering the available evidence) of the handicapper’s true pick rate. We’ll then use the posterior distribution to estimate the handicapper’s probability of picking 55% or greater over his next 1,000 picks.

Case 1: A uniform prior distribution of true pick rates within the population.

Bayes’ theorem states:

P(H|E) = P(E|H) * P(H) / P(E)

Where P() is the probability operator, H is a given hypothesis, and E is the observed evidence. In this case E corresponds to the observed 600 wins over 1,000 trials, and H could correspond to any given hypothesis regarding the handicapper’s success rate (for example, the hypothesis that the handicapper is actually a 60% picker, or the hypothesis that the handicapper is actually a 50% picker). What we’re seeking is the probability distribution of H given the observation E.

Because P(E) represents the probability of observing E under all hypotheses, P(E) can be rewritten as follows:

P(E) = ∫ P(E|H_x) * P(H_x) dx

However, because we’re assuming a uniform prior distribution of all hypotheses we have P(H_x) = P(H_y) for all x, y. This allows us to bring the term P(H_x) outside the integral and then cancel it out with the P(H) term in the numerator.

So this gives us:

P(H|E) = P(E|H) / ∫ P(E|H_x) dx

We can approximate the above integral using the discrete sum:

P(E) ≈ Σ_i P(E|H_i)

over some countable set of possible hypotheses regarding the handicapper's true pick rate.

If you open up the attached spreadsheet you'll see in column A the set of different hypotheses considered. Each cell corresponds to the hypothesis that "the picker's true pick rate is within 0.05% of the value within the cell".

In column B we have the probability of P(E|H) which is the conditional probability of observing the evidence (600 winners out of 1,000) given the hypothesis in the corresponding cell in column A. This is simply the p-value from the binomial distribution (caveat: we're using a linear interpolation here which is only approximately correct).

Taking the sum of values in column B gives us Σ P(E|H_i), which in column C we use to divide each value in column B. This gives us the normalized likelihood of the stated hypothesis being true given the observed evidence.

Column D is the (linearly interpolated) probability of winning 550 of the next 1,000 picks assuming the given hypothesis is true.

Multiplying together columns C and D (results in column E) and then taking the sum (cell E1002) gives us the (approximate) expected value of the true pick rate, 98.9%

This means that if the distribution of pick rates within the population were uniform there would be about a 98.9% probability of the handicapper going at least 55% over his next 1,000 plays.

Case 2: A Gaussian prior distribution of true pick rates within the population. We'll assume a mean of 50% and a standard deviation of 1.5% (this implies that 0.0000000013% of the population is truly a 60% handicapper or better, while 0.0429% of the population is a 55% handicapper or better).

We start again with Bayes' Theorem:

P(H|E) = P(E|H) * P(H) / P(E)

and the discrete sum approximation of P(E):

P(E) ≈ Σ P(E|H_i) * P(H_i)

(Note that because values of P(H_i)'s are in general not the same for different hypothesis we can no longer cancel P(H) in the numerator and denominator.)

In Excel columns F & G we calculate the standard Z-scores for each hypothesis. Z_l corresponds to the hypothesis that the true pick rate is more than 0.05% lower than the hypothesized value in column A, while Z_u corresponds to the hypothesis that the true pick rate is more than 0.05% higher than the hypothesized value in column A.

Taking N(Z_u) - N(Z_l) (column H) gives us the prior probability of randomly selecting a handicapper from the population with a true pick rate within 0.05% of column A. This corresponds to P(H_i), the prior distribution of H_i (which we've assumed to be normal).

Multiplying the prior probability on column H by the conditional probability of observing 600 wins out of 1,000 picks (column B) gives us column I, which is the numerator of Bayes' theorem. Taking the sum (cell I1002) and dividing through each value in I gives us column J, the normalized likelihood of the stated hypothesis given the evidence.

Multiplying column J by the probability hitting of 550/1000 given the hypothesis yields column K.

Taking the sum of column K gives us the probability of the observed handicapper going at least 55% over his next 1,000 plays, which is about 46.1% (cell K1002).

Conclusion:
What we see here is that our estimates of the handicapper's posterior probability are highly dependent on our assumptions about the prior probability distribution. This is quite common in problems of Bayesian inference.

Bayesian inference allows us refine our beliefs regarding the truth of a specific hypothesis based on the availability of additional information. Specifically, Bayes' Theorem tells us the degree to which our beliefs should change as new information comes to light. In the two examples above we had certain prior beliefs regarding the probability of a randomly selected handicapper winning 550 out of his next 1,000 plays (it would work out to a 45.1% probability in the case of the uniform distribution and a 1.152% probability in the case the Gaussian distribution). However, based on the observed evidence (winning 600 out of his previous 1,000 plays) in each situation our estimate of his future probability increased.

[MadCapper]

Thank you Einst....err...Ganchrow.

I appreciate the time and effort tou put in to answer my question!

[raiders72002]

Note a couple things:

1) Impossible and statistically improbable are not the same thing

2) You assumed p=0.5

use poisson. Plug in p=.55 for a 55% capper and it's still almost impossible to hit 60%.

[mathdotcom]

Madcapper
If N = 1000 you will get significant t-statistic even if you test > 59% on next 1000 plays.

But this assumes independent samples... maybe your success over first 1000 plays will cloud your judgment for next 1000, etc.

[SlickFazzer]

At some point that 60% is going to come down. You win 55% to 56% at -110
for a lifetime with the right bankroll, and you will be one happy SOB.

Its harder than it looks.