Dice Analysis (using a z-test)

[Lionz]

How often are there 3v1 losses back to back for people?

Also, how about someone compare dice of a general with dice of a cook? Who has not wondered if dice are weighted with score?

[Snowgun]

OMG you people are KILLING me! LOL
First of all power is NOT what you are saying it is. Second the t distribution approaches the Z distribution at high sample size. (I wanna say 30 or 40, but i'm tired right now). And nobunga has gotten into a shitcan of worms breaking out an ANCOVA

You all are confusing one another because no one has described exactly what they are comparing! Are you comparing how many times a die comes up a number in a certain amount of rolls? Binomial! Are you determining probability of different types of attacks with multiple dice? Are you trying to fit a histogram of discrete dice outcomes to a normal curve?

Basic Z and T tests compare sets of data to see if they are the same. (i'm trying to make this pretty simple so I'm not going into technical specifics, so don't be calling me out on definition semantics)

Example: American Males have heights with an average of 5'10'', and a STD (standard deviation) of 4". We are going to measure the heights of CC males. We get an average of 5'6" and STD of 2". Question: are CC males statistically different in height from the american male population?

At first you would say "yea" because 5'6" is smaller than 5'10". But the spread (or STD) of the american male population could cover this.

(What is Standard deviation? In our example it's 4". that means that ~65% of the american male population is 5'10" +- 4" (or 5'6" to 6'2"). it gives an idea as to spread of the values. 1 STD is 65%, x2 STD is 96%, x3 is 99% fall in this range.)

So even though CC height mean falls into the spread of american male mean, we use the calculations in the z and t tests to see what the chances are of us getting these values randomly. IN OTHER WORDS:, you assume that CC players are representive of american males (in height), and you want to see what the chance is of these (theoretically) randomly picked players being short. (like picking M&M's out of a bag with your eyes closed and happening to get all browns)

That z or T score (like 1.96) is how many standard deviations the two data sets are away. (remember x2 is 96%, so x1.96 is ~95%, and 100 - 95 is 0.05%, which is the common alpha someone was talking about)

Long story short, if the two sets have scores bigger than 1.96, then you have a 1 in 20 chance of accidentally picking these randomly (assuming they are in the same set). Most people call bullshit on this and then say they are different populations (e.g. reject the null set).

-Ok, done dropping the science, i'm off to bed. Good luck guys, first pick a question, then look for a statistical method to answer. Not the reverse.

[BadMoFu]

play more hands

[jrh_cardinal]

OMG you people are KILLING me! LOL
First of all power is NOT what you are saying it is. Second the t distribution approaches the Z distribution at high sample size. (I wanna say 30 or 40, but i'm tired right now). And nobunga has gotten into a shitcan of worms breaking out an ANCOVA

You all are confusing one another because no one has described exactly what they are comparing! Are you comparing how many times a die comes up a number in a certain amount of rolls? Binomial! Are you determining probability of different types of attacks with multiple dice? Are you trying to fit a histogram of discrete dice outcomes to a normal curve?

Basic Z and T tests compare sets of data to see if they are the same. (i'm trying to make this pretty simple so I'm not going into technical specifics, so don't be calling me out on definition semantics)

Example: American Males have heights with an average of 5'10'', and a STD (standard deviation) of 4". We are going to measure the heights of CC males. We get an average of 5'6" and STD of 2". Question: are CC males statistically different in height from the american male population?

At first you would say "yea" because 5'6" is smaller than 5'10". But the spread (or STD) of the american male population could cover this.

(What is Standard deviation? In our example it's 4". that means that ~65% of the american male population is 5'10" +- 4" (or 5'6" to 6'2"). it gives an idea as to spread of the values. 1 STD is 65%, x2 STD is 96%, x3 is 99% fall in this range.)

So even though CC height mean falls into the spread of american male mean, we use the calculations in the z and t tests to see what the chances are of us getting these values randomly. IN OTHER WORDS:, you assume that CC players are representive of american males (in height), and you want to see what the chance is of these (theoretically) randomly picked players being short. (like picking M&M's out of a bag with your eyes closed and happening to get all browns)

That z or T score (like 1.96) is how many standard deviations the two data sets are away. (remember x2 is 96%, so x1.96 is ~95%, and 100 - 95 is 0.05%, which is the common alpha someone was talking about)

Long story short, if the two sets have scores bigger than 1.96, then you have a 1 in 20 chance of accidentally picking these randomly (assuming they are in the same set). Most people call bullshit on this and then say they are different populations (e.g. reject the null set).

-Ok, done dropping the science, i'm off to bed. Good luck guys, first pick a question, then look for a statistical method to answer. Not the reverse.

we're comparing his stats (the sample) to the "ideal stats" (the population) given by the Dice Analyzer Script. This is a simple z-test (I still say it is a t-test but with a sample this size it really makes no difference) The formula is:

z=sample value - pop value/sample standard dev.

sample value would be .553404
pop value would be .571065
standard dev. is unknown (which makes it a t-test), but is really easy to calculate and I trust he did it right, but that's not the point, the point is that we are doing it correctly, and we are not confusing ourselves (at least I'm not confused)

if you think this is wrong you had better be a statistics professor or something like that because I currently have a 99.91% in an AP Stats class where it is impossible to get over 100, so fire away

[jrh_cardinal]

but you're right about the power

[natty dread]

You guys speak like another language here...

[tscott]

So I've added a few more dice rolls to the mix, the quite the opposite of "evening out" is happening. My new calculated z-score is -2.79 which corresponds to a probability of .0026. I also did a chi-squared test of the distribution of my dice rolls. Basically it compares my distribution of 1's through 6's with what should be expected with random dice. my chi-square value comes in at 14.78. With 5 degrees of freedom, this FAILS at the .025 probability threshold, though is still above the .001 threshold.

So what does all of this really mean? First, if the dice really are random, I am just really unlucky, which is certainly possible. How unlucky? Well, I've thrown 7471 that have been recorded, if we were to take any random set of this many rolls, only 26 out 10,000 would be this unlucky. If we look at the chi-squared results, it is telling me that if roll dice 7471 times, less than 2.5% of the time will the dice be this "far" from random. (I know the concept of far from random is a misnomer since I am assuming the dice are random for the moment, but I am just trying to make it make the most sense to the most amount of people). Now here is the thing that really gets me. With these crappy dice I should be LOSING way more than I am winning, however this just isn't the case. It seems as though everybody is having less than average dice. Although this may seem fair, I feel as if it takes some skill out of the game, as an understanding of probability and how dice behave is essential to success at the game. When the dice don't meet those expectation OVER THE LONG TERM, those that based their strategy on that behavior are at a disadvantage.

[army of nobunaga]

You update the excel sheet yet man? Id like to look. You might be right. I dunno. I always keep an open mind. My test was much more simple and came out perfect in favor of the dice. Yours is a little more in depth and if right shows that mine did not take in account the various permutations of a two dice total... ie

you can get 5 by 1 and 4... and by 2 and 3 ...

My test looked at totals only.

But I cannot believe any coder would go to the trouble of adding some sort of multiplyer to make the permutations less than random. Im digressing, lets see your excel.

[tscott]

OK, the spreadsheet is updated. I added the chi-squared test at the bottom left. The numbers differ slightly from my last post as I added in my rolls from the last few hours. Z-score of -2.82. Chi-squared of 15.02, with df=5. In my opinion, these dice are crap.

[army of nobunaga]

if you run this EXCLUDING the 1vs 2 and and 1vs 1... I think you will see the dice are fine... You just need a higher level of the small throws.. 2v2 down.. imo

[BrutalBob]

OK, the spreadsheet is updated. I added the chi-squared test at the bottom left. The numbers differ slightly from my last post as I added in my rolls from the last few hours. Z-score of -2.82. Chi-squared of 15.02, with df=5. In my opinion, these dice are crap.

Ok so maybe you have "proven" it "statistically" but my mind wont be made up until we get a greater body of anecdotal evidence.

Can we now focus a bit of time compiling a list of people who claim to have be wronged- perhaps we could rank them by the degree to which they were robbed of a win.

That would really seal the argument for me.

[army of nobunaga]

Can we now focus a bit of time compiling a list of people who claim to have be wronged- perhaps we could rank them by the degree to which they were robbed of a win.

That would really seal the argument for me.

Wow mr wizard, thats pretty scientific..

I think you should start that special thread yourself.