Ok so maybe not "improbabilities" so much as differing from the average.

Let's say you have an N-sided, equally weighted dice. You roll the dice K times, and average the results. The average must approach the expectation value for the dice, namely: (N+1)/2. My question is, what is the probability of any given K rolls of the dice, that current average differs from expectation value?

Example: a 6-sided dice is rolled 5 times and the results are: {3, 3, 6, 4, 5} which averages to 4.2. The expectation value is (6+1)/2 = 3.5. What is the probability for this difference between 3.5 and 4.2?

## Probability of improbabilities

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### Re: Probability of improbabilities

badwiz wrote:Ok so maybe not "improbabilities" so much as differing from the average.

Let's say you have an N-sided, equally weighted dice. You roll the dice K times, and average the results. The average must approach the expectation value for the dice, namely: (N+1)/2. My question is, what is the probability of any given K rolls of the dice, that current average differs from expectation value?

Example: a 6-sided dice is rolled 5 times and the results are: {3, 3, 6, 4, 5} which averages to 4.2. The expectation value is (6+1)/2 = 3.5. What is the probability for this difference between 3.5 and 4.2?

I think you're asking about "for any given difference from the expectation value (e.g. ±0.7?), what is the probability?".

I think the answer will be based upon the fact that for every roll of the dice, a value is given of a given difference from the expected mean. 1/6th chance, each, of -2.5, -1.5, -0.5, +0.5, +1.5 or +2.5, for a fair D6. You'd sum the possibilities that relate to the finally required difference multiplied by the number of dice rolls made. Or, in other words, take the sum of each variation and divide by the dice roll, and count all results that are within your specification. (That specification might be between (inclusively) a given value and zero, beyond (also inclusively) a value or specifically a given value, and might or might not be concerned with magnitude (respectively -x≤result≤+x, not(-x<result<_x), result=±x) or of a matching sign. Hard to tell what you're looking for, though.

In the unlikely event that you mean any non-zero difference from the expectation, five rolls have 100% chance of coming out that way. For reasons that are obvious when you think about it.

### Re: Probability of improbabilities

If you are rolling sufficiently many dice, then there's an easy trick to know how likely you are to get the rolls within ±q of the average. You apply the Central Limit Theorem, which states that the average of a bunch of iid* random variables starts looking like a normal distribution once you have enough of them**. So the expected value of the average is 3.5, and since the variance*** of a single die roll is 35/12 so the variance of the average of n dice is 35/12n, and hence you can take z = (x - 3.5)/sqrt(35/12n) to be approximately a standard normal variable and use that to convert between Prob(x is within q of 3.5) and Prob(z is between -blah and +blah) and look up the latter on a table.

* Independent, identically distributed, i.e. they're draws from the same distribution like rolling a fair die lots of times, or counting the number of times the 11,582nd word in a book is "the" at your local library.

** Technically, it's "in the limit as the number of variables tends to infinity", but anyway.

*** Variance = Expected value of the square of the difference between the random variable and its expected value, i.e. E[(X - E(X))²].

* Independent, identically distributed, i.e. they're draws from the same distribution like rolling a fair die lots of times, or counting the number of times the 11,582nd word in a book is "the" at your local library.

** Technically, it's "in the limit as the number of variables tends to infinity", but anyway.

*** Variance = Expected value of the square of the difference between the random variable and its expected value, i.e. E[(X - E(X))²].

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