How many dice rolls until things even out?

How many times would you have to roll 2 dice (2d6) until they started to match their probability averages? In other words, I could roll snake eyes (total 2) five times in a row, but how long until the rolls begin to get closer to the probability averages?

2d6 total	probability occurring
2	2.78
3	5.56
4	8.33
5	11.11
6	13.89
7	16.67
8	13.89
9	11.11
10	8.33
11	5.56
12	2.78

Based on the Wizard's response to previous "dice influence test" questions, I suspect 3,600 rolls should produce a relatively uniform distribution most of the time.

Read the article on the WoO site here (scroll down to the question with the large graph.)

Quote: Ayecarumba

Based on the Wizard's response to previous "dice influence test" questions, I suspect 3,600 rolls should produce a relatively uniform distribution most of the time.

Read the article on the WoO site here (scroll down to the question with the large graph.)

Just ran a quick simulation, and 3600 didn't always get very close. The Chi-square test frequently produced percentages well over 50%.

Quote: Joseph

How many times would you have to roll 2 dice (2d6) until they started to match their probability averages? In other words, I could roll snake eyes (total 2) five times in a row, but how long until the rolls begin to get closer to the probability averages?]

Bold emphasis added.

I haven't thought this through, but is the key issue the wording "started to match" and "begin to get closer"? Doesn't this tendency start right away? No, it doesn't get closer with every roll from roll #1. Some rolls take the data farther from the theoretical. Even after a million rolls, each roll makes the observed percentage of each number either increase or decrease, taking it either closer or farther from the theoretical, not always closer, but that's the tendency from the very beginning, isn't it?

Did I misunderstand the question?

removed
silly

Sally

Ok so one thing I don't think was really addressed was the example situation you gave. If you do something so far from the normal (like rolling five straight snake eyes) then it's going to take much longer to reach a normal distribution.

In my random opinion, at the minimum you have to reach a number of rolls where 5 snake eyes is 1% of the expected. In order to get 500 of them you need to roll 18000 times. So in my opinion-without-facts you'll need around 20k rolls to "make up" such crazy variance.

The OP kind of asked two different questions.

In the video I linked to, around roll 243 the number 11 had rolled almost 2X more than expected.
In another 240 rolls, it falls back, percentage-wise.
By the 800th roll, it still is ahead, but the percentages are now getting closer and closer.

It is the percentages that can quickly converge, relatively speaking, way faster than the absolute values.

Watching the "law of large numbers" in action

Fair question to ponder.

As long as the OP realizes there is no value in keeping the stats and then thinking there will be a trend to correct the numbers that he can then bet on. An anomaly will be corrected by burying small numbers by large numbers, not by the dice feeling a need to skew some of the next rolls the "other way"

This is a little like those questions about random number generators.

Its always going to be a long time before you can satisfactorily prove its randomness.
Its going to be a much shorter time before all but the purists are satisfied its random.
Quibbling over non-expected results ain't gonna get you anywhere.

My answer is 36^e = 17000 rolls.

I think it depends what you mean by "even out." You kind of have to look at it on a log scale, and still in probabilistic terms. Essentially, the most volatile are going to be the 2 and 12, so let's call it "how many rolls are needed before a 3 sigma radius is less than 5% of the rolls for the frequency of snake eyes." (That is, more or less, when 99.76% of all trials will give you a result within 5% of the expected.) Treating it as a Bernoulli trial with a 1/36 chance of success (with a variance, therefore, of n*35/1296, i.e., n*(1/36-1/36^2)), it becomes a question of getting 3*sqrt(n*5/1296) < n/(36*20)
sqrt(n*35/1296) < n/2160
n*35/1296 < (n^2)/4665600
n > 126000
Or, for two sigma, about a 95% probability of being within 5%, 56000 will do.

But "even out" is really the wrong way to think about it. In particular, consider that with every roll you make, the volatility in raw number of rolls increases, decreasing only as a fraction of the total rolls. So betting a lot in the hope that your bets will regress to an affordable mean is a losing strategy even if it's not quite the gambler's fallacy, since the error bars on the expected return grow in terms of dollars, even as they shrink in proportion to the bet - it's just that they grow more slowly. Basically, the standard deviation is proportional to the square root of the number of random events - bets, rolls, etc. - and the probability of it being in a certain distance of the mean is pretty much a function of the standard deviation (number of trials matters a little, but this vanishes quickly), and this creates regression to the mean, even as the raw breadth of the error increases.

So I guess the short answer is, "they don't."