What are the odds?

[Silvercast90]

I'm currently in a dice rolling contest, the prizes include 3 mil gold, ninja headband, 2 katanas, and a gaia tan.

The goal of this contest is to roll a certain number set out of 20 20 sided die.

What are the odds that i will roll less than 100 from rolling 20, 20 sided dice?

Also, what are the odds of me rolling higher then a 390?

Seeing as i do not know the equations to do so, it would be a great learning experience for me if you posted the equations as well as the answers.

Or just post the equations and I'll work them out smile

Yay statistics smile  

[VorpalNeko]

Each die has an expected value of 21/2 and variance of (20^2-1)/12. Hence the mean of the sum is 20*21/2 = 210 and the variance σ² = 665--the dice are independent, so the variances just add. Since the number of dice is fairly high, their sum can be closely approximated by a normal distribution of the same mean and variance.

In particular, the z-score of 99.5 is z = (99.5-μ)/σ = -4.285. You can look it up on z-score tables, or if you've a computer to evaluate error functions, P(Z
So how many ways are there to get, say, 25 out of twenty dice? Well, each die rolls at least one, and at most 20, so it's got to be the same as the coefficient of x^25 in
F = (x + x² + x³ + ... + x^{20})^{20}
That's ugly, but what's inside is just a truncated geometric sum, so it's easy to write as x(1-x^{20})/(1-x). From basic calculus, the coefficient of x^n should be F^n(0)/n!, where the ^n denotes the nth derivative rather than nth power. That's convenient for low n, but pretty useless for high n without a computer. I know it's possible to get a pretty summation based of multinomial coefficients and some , but I'd rather not derive it. For low numbers, the first nonzero terms are the binomial coefficients C(19+n,19), but the whole sequence is not that simple.

Crunching this out, the total for 20-29 is 10015005. Hence the probability of getting less than 30 is 10015005/20^20 = 9.55E-20, or 1 in 10 billion billion. By symmetry, that's also the probability of getting over 390. For each possible state of 20 dice with a sum over 390, reflect the sum by subtracting from 21: you get a dice state with a sum under 30 (20*21 - 390), and vice versa.

Incidentally, the number of ways to get a sum under 100 turns out to be exactly 376287842588079066876. Hence the probability of such is 3.59E-6, or 1 in 280 million thousand.  

[Silvercast90]

VorpalNeko

Each die has an expected value of 21/2 and variance of (20^2-1)/12. Hence the mean of the sum is 20*21/2 = 210 and the variance σ² = 665--the dice are independent, so the variances just add. Since the number of dice is fairly high, their sum can be closely approximated by a normal distribution of the same mean and variance.

In particular, the z-score of 99.5 is z = (99.5-μ)/σ = -4.285. You can look it up on z-score tables, or if you've a computer to evaluate error functions, P(Z
So how many ways are there to get, say, 25 out of twenty dice? Well, each die rolls at least one, and at most 20, so it's got to be the same as the coefficient of x^25 in
F = (x + x² + x³ + ... + x^{20})^{20}
That's ugly, but what's inside is just a truncated geometric sum, so it's easy to write as x(1-x^{20})/(1-x). From basic calculus, the coefficient of x^n should be F^n(0)/n!, where the ^n denotes the nth derivative rather than nth power. That's convenient for low n, but pretty useless for high n without a computer. I know it's possible to get a pretty summation based of multinomial coefficients and some , but I'd rather not derive it. For low numbers, the first nonzero terms are the binomial coefficients C(19+n,19), but the whole sequence is not that simple.

Crunching this out, the total for 20-29 is 10015005. Hence the probability of getting less than 30 is 10015005/20^20 = 9.55E-20, or 1 in 10 billion billion. By symmetry, that's also the probability of getting over 390. For each possible state of 20 dice with a sum over 390, reflect the sum by subtracting from 21: you get a dice state with a sum under 30 (20*21 - 390), and vice versa.

Incidentally, the number of ways to get a sum under 100 turns out to be exactly 376287842588079066876. Hence the probability of such is 3.59E-6, or 1 in 280 million.

Oh, okay, thanks much. I didn't graduate high school so i never took statistics or calc/pre-calc. Man, So, in your opinion, is there a point in trying to get under 100? And i know that there is no way in hell i can get above 390 in this century. But seeing that 100 is closer to 200 (the average roll) i might have some hope.  

[VorpalNeko]

For some reason when I flipped the last probability 3.59E-6, I used started thinking of E-9 instead, so the odds are 1:280,000. The other ones are correct. (Note to self: check posts more often.)

Silvercast90

Oh, okay, thanks much. I didn't graduate high school so i never took statistics or calc/pre-calc. Man, So, in your opinion, is there a point in trying to get under 100? And i know that there is no way in hell i can get above 390 in this century. But seeing that 100 is closer to 200 (the average roll) i might have some hope.

If the individual success probability is p, then the probability of having at least one success in n tries is 1-(1-p)^n. Well, what are acceptable odds for you? If p is very low and n is not too high, then 1-(1-p)^n is about np, so you'd need about 280 tries to bring up your chance of winning to in a thousand.

The exact solution to to solving 1-(1-p)^n = P is actually n = log(1-P)/log(1-p), where P is the probability you'd like to have (P = 0.001 = 1:1000 above). So, for example, you'd need over 193,154 thousand tries to make your chance of winning over one-half. That might actually be more than the total number of tries of everyone who participates, combined (moral: chances are, no one will get this prize, unless I'm severely underestimating the popularity of the contest, or it just goes on indefinitely without time limits).  

[Silvercast90]

It already has almost 100,000 replies, and the lowest roll was 101. That was me biggrin

Edit: i have posted over 200 times in this topic.