/sci/ - Science & Math

[Anonymous No.6986523]

So /sci/, a di with 420 sides is rolled 420 times, if you added each outcome together, what is the probability this outcome would equate to 88888?

[Anonymous No.6986536]

Dunno, just blaze it

[Anonymous No.6986537]

had2know.com/academics/discrete-uniform-sum-random-variable-generator.html

[Anonymous No.6986544]

>>6986523
If on each face there's written the number 211,6380952380952380952380952381, then the probability is 100%.

[Anonymous No.6986549]

>>6986544
what

[Anonymous No.6986551]

>>6986544
wat no

[Anonymous No.6986552]

>>6986549
if you roll that dice, each time you'll get that number. Try to do a simple 420*(that shit) and you'll get 88888

[Anonymous No.6986555]

>>6986544
No.

420 * 211,6380952380952380952380952381
= 88888,000000000000000000000000002

[Anonymous No.6986567]

>>6986555

Dude what the fuck are you doing. Where are you getting this 211 number from?

[Anonymous No.6986568]

>>6986555
Yeah, periodical bullshit

[Anonymous No.6986573]

>>6986567
88888/420...

[Anonymous No.6986589]

>>6986568
It's 211.6380952 (period 6)

[Anonymous No.6986635]

>>6986573

88888 was the outcome of rolling the di 420 times.

Step 1) Roll di 420 times
Step 2) Add together the numbers you get from rolling the di this many times
Step 3) The outcome was 88888

What's the probability of getting this.

[Anonymous No.6986835]

your question has been asked before:

http://mathforum.org/library/drmath/view/52207.html

[Anonymous No.6988407]

>>6986523
use the normal approximation. will be nearly exact

[Anonymous No.6988423]

>>6986573
888888888888888/Blayze it fggt

[Anonymous No.6989114]

P=1.576129e-04

clear all

N=420; % sides
M=420; % rolls
S=88888; % sum we want

% calculate using normal approximation (fast)
meanS=M*(N+1)/2; % mean sum
stddevS = sqrt( M/12*(N^2-1) ); % standard dev of sum
a=((S-1/2)-meanS)/stddevS; % lower bound for Z
b=((S+1/2)-meanS)/stddevS; % upper bound for Z
P=normcdf(b)-normcdf(a); % the probability estimate

% calculate exactly (slow)
h=ones(1,N)/N;
g=h;
for n=2:M
g=conv(g,h);
end
Q=g(S-M+1);

disp(sprintf('P=%e, Q=%e, diff=%e',P,Q,P-Q))


output:

P=1.576129e-04, Q=1.575608e-04, diff=5.216265e-08

[Anonymous No.6990659]

Isn't there an explicit expression via the multinomial theorem?

[Anonymous No.6990686]

>>6986835
How does Dr Math get this part in red?

[Anonymous No.6990699]

>>6989114
Ok, but how about an analytical solution.

[Anonymous No.6990707]

>>6990699
Ok I spent an extra 30 seconds thinking about how to figure that out and realized that I'd rather use a computer.

[Anonymous No.6990800]

>>6990686
help?

[Anonymous No.6991772]

>>6990800
nobody? anybody?

[Anonymous No.6991804]

>>6990686
Binomial theorem

[Anonymous No.6991855]

>>6991804
with a negative power? how does that work?

[Anonymous No.6991876]

>>6991855
<div class="math">(x+y)^r = \sum_{k=0}^\infty {r \choose k} x^k y^{r-k}</div>

Works for every complex number <span class="math">r[/spoiler].

[Anonymous No.6991934]

>>6991876
Ok, thanks, I can see an infinite sum. For example, you get the standard 1/(1-z) = 1+z+z^2+...

But Dr Math has a finite sum. Perhaps for generating functions you don't care about convergence, but I don't see then how you can do tricks like this.

[Anonymous No.6992347]

>>6986544
I think people are getting confused about the European 211,whatever versus the American 211.whatever

[Anonymous No.6993042]

>>6991934

Dr Math is basically raising (z+z^2+...+z^m) to the n-th power, so he should get a polynomial of degree mn. The second sum should indeed be infinite per the generalized binomial theorem. However, *after* the series are multiplied together to get a single sum, terms corresponding to z^k for k>mn will necessarily be zero. He's a little sloppy, but if you want the coefficient for k<=mn, then Dr Math's formula will work.

You can see that this must work out if you think of f(z)^n as an analytic function on the unit disk, and appeal to the uniqueness of power series.