[Bullajami]

I am trying to figure the statistical relevance and reliability of a bet selection methodology, but I need some help.

All of the bets are on moneylines.

60 bets were made using 1% as the betting unit, and 33 bets were made using 2% as the betting unit. After 93 bets I am up 20%.

What, if any, conclusions can be made at his point about this strategy?

Thanks in advance for your help.

[Ganchrow]

Originally Posted by Bullajami

I am trying to figure the statistical relevance and reliability of a bet selection methodology, but I need some help.

All of the bets are on moneylines.

60 bets were made using 1% as the betting unit, and 33 bets were made using 2% as the betting unit. After 93 bets I am up 20%.

What, if any, conclusions can be made at his point about this strategy?

Thanks in advance for your help.

At what odds were each of the 93 bets placed?

When you refer to 1% as the "betting unit" is that a "to-risk" amount or a "to-win" amount?

Are you compounding your bets so that had you started with 100 units and then found yourself up 10 units, would 1% then imply a wager of 1 unit or of 1.1 units?

[Bullajami]

Originally Posted by Ganchrow

At what odds were each of the 93 bets placed?

When you refer to 1% as the "betting unit" is that a "to-risk" amount or a "to-win" amount?

Are you compounding your bets so that had you started with 100 units and then found yourself up 10 units, would 1% then imply a wager of 1 unit or of 1.1 units?

From -105 to +1000. Mostly around +130. Do you really need all 93 odds?

Betting unit is risked amount.

Yes, betting unit is based on compounding results.

[Ganchrow]

Originally Posted by Bullajami

Do you really need all 93 odds?

That depends ... do you really need an answer?

[Bullajami]

120
120
160
140
120
145
110
130
115
150
160
121
145
105
135
150
140
145
120
135
-105
120
140
165
165
130
100
130
130
110
114
170
155
115
155
155
145
380
140
132
140
165
180
145
128
165
105
115
145
160
190
-105
-105
135
130
120
175
170
135
210
105
120
125
145
140
135
210
135
200
125
210
125
120
100
150
150
125
105
125
150
135
135
120
140
240
130
150
195
105
1000
140
170
140

[Ganchrow]

Originally Posted by Bullajami

-snip-

Which bets were at 1% and which were at 2%?

[Bullajami]

Bottom 60 at 1%.

[Ganchrow]

I'm coming up with a p-value of about 11.8%. This corresponds to the probability that a player not paying vig and with no edge would have achieved the same or better results strictly by chance.

[Bullajami]

0.882 probability that my method is effective? Am I understanding you correctly?

Can you show me how it's calculated so that I can run the numbers again when I have more data?

Thank you for your assistance!

[Ganchrow]

Originally Posted by Bullajami

0.882 probability that my method is effective? Am I understanding you correctly?

There's an 88.2% probability that a bettor with no advantage and paying no vig would have achieved results worse than that you obtained.

That's slightly different than simply stating that there's an 88.2% probability that your method was effective.

Also ths conclusion takes your results in a vacuum,, meaning that it's assuming that this is the only strategy you're investigating. If these simply corresponds to but a single strategy out of many that just happened to be effective than the 88.2% figure would be too high. Perhaps drastically.

By tradition, 95% is generally considered the minimum desired level for statistical significance. Hence, even if the above were true these results would not be considered statistically significant (meaning that we could not reject the hypothesis were due to "luck". This does not imply the results were due to luck, just that a statistician would be unable to reject that possibility at the 95% confidence level.

Originally Posted by Bullajami

Can you show me how it's calculated so that I can run the numbers again when I have more data?

Because of the compounding effect I simply ran a 5,000,000 trial Monte Carlo simulation, which is how I obtained the 11.8% value.

I need to run off to wife's birthday party now, but at some later point I'll write a post detailing the process involved. It's rather straightforward.

[Bullajami]

I thank God that I have access to your insane genius.

[Ganchrow]

Originally Posted by Bullajami

I thank God that I have access to your insane genius.

God has exactly zero to do it -- I made a pact with the devil years ago.

(That's why I have such bad breath.)

[Bullajami]

Didn't Pascal call making a pact with the Devil a bad bet?

I should have studied more when I was younger, but I was always too horny.

I appreciate you sharing your scholarly knowledge, your Monte Carlo simulation software, and for using Altoids when you post.

[Ganchrow]

We're assuming zero-vig so the expectation on each bet is zero.

The variance of a bet of size x at decimal odds of d is given by (d-1)*x². So for example the variance of a 1-unit bet at +130 would be 1² * (-1) = 1.3, while the variance of a 2-unit bet at -105 would be 2² * (-1) ≈ 3.8095.

If we ignore compounding the variance of a linear combination of independent bets is the sum of the individual variances. In the provided data set the total variance works out to be about 272.86 units². The standard deviation is the square root of the variance, which in this case is (272.86 units²)^½ ≈ 16.519 units.

The realized return is ²⁰/_16.510 ≈ +1.2108 standard deviations better than expected, implying a p-value of =NORMSDIST(1.2108) ≈ 88.701% from the normal distribution and =1-TDIST(1.2108,93-1,1) ≈ 88.545% from the t-distribution with 92 (i.e., 93 bets - 1) degrees of freedom (where NORMSDIST and TDIST refer to MS Excel's cumulative standard normal and Student t-distribution functions, respectively).

One problem with the above analysis, however, is that it completely discounts the effect of compounding. Nevertheless, because we're dealing with a relatively small number of bets, each at a relatively small percentage of bankroll, this effect is fairly small.

Taking compounding into account, the only real procedural difference would be in the calculation of variance. Because the result of previous bets impact the results of future bets (insofar as if we win an earlier bet we'd be betting more on a later bet it's no longer correct that the total variance equals the sum of the variances.

Rather than go into some long derivation, I'll just state the result here. The total variance is given by the sum of the variances plus the sum of the product of the variances taken 2 at a time plus the sum of the product of the variances taken 3 at a time ... plus the product of all N variances.

This obviously represents a huge number of terms (2^N - 1 terms for N bets, 9,903,520,314,283,042,199,192,993,792 terms for 93 bets), but if bets are sufficiently small we can approximate by only looking at terms up to only the second or third order. Given the 93 bets of the original problem, the total variance (up to the second order) would then be:

Total Variance ≈ Σ^N_i=1[σ²_i] + Σ^N_i=1[σ²_i * Σ^N_j=i+1[σ²_j ]]

which works out to a standard deviation of 16.629%, implying a t-distribution (with 92 degrees of freedom) p-value of 88.391%. Were we to go to the third order we'd of course find a slightly higher σ (~16.630%) and hence a slightly lower p-value (~88.390%).

This is faily close to the p-value of 88.215% that I obtained from my 40,000,000-trial Monte Carlo run. The difference stems from having approximated the distribution of results with the t-distribution.