[ssanders82]

Hi, I'm new here (hello all). I've been betting on Matchbook for about a month based on patterns I've discovered in historical NBA and NCAABB data. I'm doing well so far but wanted to calculate my z-score to see what the chance is that I've just been lucky.

I'm familiar with the standard z-score formula of (raw score - pop. mean) / std deviation but I'm not sure if there's a way to apply that to binary data (win or lose), instead of using a continuum such as calculating the average # of points I've beaten the spread by, and dividing by std dev to find my edge over the spread.

Does this make sense? Let's say I've won 58 out of 100 bets. How do I calculate the % chance that my system is better than 50%, 55%, or 60%?

Or, since I'm mainly concerned with a day-to-day equity curve (I use an amalgamation of several systems to hopefully smooth the ups & downs), does it make more sense just to perform a z-score analysis with daily P&L?

(I'm coming from a stock market background so the terminology here of measuring everything in units is foreign to me, I use ROI and $'s.) FWIW I'm trading at .5-.6 Kelly using a conservative win% estimate.

I couldn't find anything using the search so feel free to point me to a previous thread if it exists. Thanks.

Edit: To clarify/simplify, assume all bets are made at +100

[Ganchrow]

See http://forum.sbrforum.com/players-ta...tml#post859159.

[ssanders82]

Originally Posted by Ganchrow

See http://forum.sbrforum.com/players-ta...tml#post859159.

Awesome, thank you. I may be missing something from that post, but can you explain how you calculated the std dev of 1000 coin flips as 15.81 heads? Without going into a bet-by-bet variance analysis I'm trying to calculate % chances that I have a winning system after n bets and x wins.

[Ganchrow]

Originally Posted by ssanders82

Awesome, thank you. I may be missing something from that post, but can you explain how you calculated the std dev of 1000 coin flips as 15.81 heads? Without going into a bet-by-bet variance analysis I'm trying to calculate % chances that I have a winning system after n bets and x wins.

After n Bernoulli trials (i.e., trials of a binary outcome event -- in this case heads/tails) with success rate p, variance is given by:

σ² = n * p * (1-p)

and hence standard deviation:

σ = √n * p * (1-p)

So in the case of 1,000 trials of a coin flip:

σ = √1,000 * 0.5 * (1-0.5) ≈ 15.81

But if all your bets have roughly the same breakeven probability anyway, then you're probably best served determining the Type I error of the hypothesis that you're no better than a 0 EV handicapper directly from the binomial distribution.

So if you've placed N bets at breakeven probability p and have won W of them, the probability that you'd see these results or better were you in fact only a 0 EV bettor would be given in Excel by:

=1-BINOMDIST(W-1,N,p,1)

So let's say you've placed 200 bets at -110 and have won 109 of them, the probability that these results would be seen by chance alone had you in fact no advantage would be given in Excel by:

=1-BINOMDIST(109-1,200,110/210,1) ≈ 29.87%

meaning that you could not reject the null hypothesis at α = 0.05 that you were no better than 0 EV.

Performing the same analysis using the Normal distribution:

σ = √200 * 110/210 * (1-110/210) ≈ 7.06
Z_{one tailed} ≈ (109-200*110/210)/7.06 ≈ 0.60
p(Z ≥ 0.60) ≈ 27.42%

which due to the flat tails of the normal distribution implies slightly higher significance than that of reality.

[ssanders82]

Aha, Bernoulli trials, I knew there were terms from college statistics rattling around in my brain but couldn't pull them out. Thanks man, this is exactly what I was looking for.