Question about calculating Z-Scores

**Juret** · 03-27-14, 02:56 AM

Then you have to calculate the variance for each bet: Stake^2 * (decimal odds - 1). In the simplified formula the variance is the same for every bet.

Sum the variance for all bets and take the square root to get the standard deviation. Divide the amount won in your currency by the standard deviation and you have the z-score. If you divide the amount won by the sum of stakes for all bets, you get the ROI.

However, if the odds in your sample are widespread, the z-score could be a poor estimate.

**JohnGalt2341** · 03-27-14, 03:13 PM

Thanks. Let me see if I did this correctly. Let's say I have a record of 470-375 and ALL of these bets were $110 to win $100. Would that give me a Z-Score of 3.116?

**Miz** · 03-27-14, 08:15 PM

if you search this forum you'll find the answer. It's been asked a few times and answered pretty well already... no need to rephrase it.

**JohnGalt2341** · 03-28-14, 09:29 PM

Originally posted by Miz

if you search this forum you'll find the answer. It's been asked a few times and answered pretty well already... no need to rephrase it.

I did find a few threads about it. I was just looking for a simple equation that would incorporate standard juice of -110. I didn't find a specific equation for this but this is what I came up with: Z Score=(Wins-Losses)/sqrt(sample size*1.1). Would this be correct? Assuming all of your bets are 1 unit and with -110 juice.

**agendaman** · 03-29-14, 04:04 AM

equations wont work with gambling in the long run-trust me on that

**Miz** · 03-29-14, 06:13 PM

assuming this is a point spread sport where the expected win pct is 50% if you flipped coins...

470-375....

(470 - (470+375)/2) / ( ((470+375)/4)^0.5 ) = 3.26

Thank me later for this one: in excel .... Probability that luck created the results = =1-NORMSDIST(z)

https://www.fourmilab.ch/rpkp/experiments/analysis/zCalc.html

**Miz** · 03-29-14, 06:16 PM

Originally posted by agendaman

equations wont work with gambling in the long run-trust me on that

Nah, of course not. Why would something so hokey pokey like math work...

**JohnGalt2341** · 03-29-14, 07:27 PM

Originally posted by Miz

assuming this is a point spread sport where the expected win pct is 50% if you flipped coins...

470-375....

(470 - (470+375)/2) / ( ((470+375)/4)^0.5 ) = 3.26

Thank me later for this one: in excel .... Probability that luck created the results = =1-NORMSDIST(z)

https://www.fourmilab.ch/rpkp/experiments/analysis/zCalc.html

How is the formula you gave any different than (Wins-Losses/sqrt(sample size)? Plug (470-375)/sqrt(845) into excel and you will get the exact same result. This is the correct formula if all bets are +100(no juice on either side). Because of the -110 juice the Z-Score is no longer 3.268. If I calculated it correctly it is now 3.116.

**Miz** · 03-29-14, 08:11 PM

it isn't wins - losses.

It is wins - (total number of games)/2

assuming a 50% chance of winning a game, your expected number of wins is the total number of games divided by 2.

**JohnGalt2341** · 03-29-14, 08:36 PM

Originally posted by Miz

it isn't wins - losses.

It is wins - (total number of games)/2

assuming a 50% chance of winning a game, your expected number of wins is the total number of games divided by 2.

The formula you gave (470 - (470+375)/2) / ( ((470+375)/4)^0.5 ) = 3.268099.
The formula I gave (470-375)/sqrt(845)=3.268099. Coincidence?

The whole point of this thread is that neither formula is correct for -110 juice. Lets assume that there are 2 people playing the same exact plays at 2 different Sportsbooks. Person A, is playing at a fictitious Sportsbook where all the prices are +100. And person B, is playing at a terrible Sportsbook where all the prices are -120. They make the same exact plays and they both end up with a record of 470-375. Are you telling me that they will both have the same Z-Score?

**Miz** · 03-29-14, 10:52 PM

I use the simplified version (only applied to 50/50 propositions) I went through it once a long time ago and have always used this simplified approach. There is a more complete version that you have to apply if you try to do it for events where the win probability isn't 50/50. By applying juice the total probability is greater than 1, right? Z-score is not about juice I don't think, rather your performance against the implied probability. I use it to loosely gauge the validity of a model, and that's about it. Good luck with your calculations.

**Miz** · 03-29-14, 10:58 PM

to be honest i just thought you needed a hand coming up with your z-score, and was trying to be helpful. i didn't really read the post thoroughly

... just thinking out loud here... if you include juice in your z-score calcs then you will have an implied probability greater than 1 when you consider both sides of a given event ... I think z-score is independent of any vigorish, and really is about the performance versus what was expected with no edge/model and assumes a total probability that adds up to 1.

the simplified version above works when it is 50/50... but for baseball MLs for example, it gets trickier. Either way you have to have a prob that adds to 1 I believe.... even if it is Team A has a 75% chance to win and Team B is 25%. I think someone went through this exact situation in here once (Ganchrow probably)