So Any Forums, show your real IQ:

so Any Forums, show your real IQ:
>yesterday, session of d&d with friends and we went into a rabbit hole to calculate this:
>d20 dice, what would be the probablility to roll the dice 18 times and have the same number 5 times as result (doesn't matter which one), not consecutive.

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2.62144000E+23
I think?

Oh sorry it should
I think

1/2.62144000E+23

100 percent the dice are weighted

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0.00000625

Op here, today with fresh brain, i came up with:
(20 5) * (1/20)^5 * (19/20)^13
but i'm not sure though

which is:
45038745288077310189/32768000000000000000000

and the result is 0.001374473427980874944732666015625

Wait, is the number specificied at the beginning or can it be *any* number that shows up 5 times?
Because if it's any number that gets harder to calculate.

One in 3,200,000

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i think youre close
so, its (1*20)^4, since the first roll doesnt matter-- were not looking for a specific number yet

as to whether or not we need (10/20)^12 i think is determined if those with over 5 of the same number do not count, i calculated it as if they do-- since they DO have 5x the same number strictly speaking, they just also have more

wa thinking the same, it can be either way, first roll is the number which has to be rolled another 4 times or a most difficult calc would be *any* number which is rolled 5 times

No not 5 Times
18 times
I had the same mistake

>(20 5) * (1/20)^5 * (19/20)^13
i made a mistake on my previous formula.. it should be:
(17 4) * (1/20)^4 * (19/20)^13 - if we take the first roll result as seed

>first roll is the number which has to be rolled another 4 times or a most difficult calc would be *any* number which is rolled 5 times
it doesnt matter, permutations can be rearranged. so if a number is rolled 5x, the original array just needs reordered such that the number appears first

so something like this?

(1/20) * (17 4) * (1/20)^4 * (19/20)^13

The binomial is (17 4) and not (18 5), and 1/20^5 because the part of the search for the 4 useful results must be multiplied by the probability of having the expected result in position 1 (so we multiply everything by an additional 1/20).

bump while i do somethin

thinking if we are bound to the first roll or not.. if not my formula is wrong

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based.
look at this nigger, you won for me dude!

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derp, made it roll 20x instead of 18
big diff

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Did you dig out R for this?
Incredibly based.

so i was wrong, with my formula above:
(17 4) * (1/20)^4 * (19/20)^13
my result is way different than yours
result is 0.00763

cleaned up logic a bit so its easier to read

Attached: 2022-06-21-065712_1920x1080_scrot.png (1920x1080, 570.77K)

What I got was:
((1 / S)^N) x (R / N)

Where:
S = Sides of the die
N = Amount of equal numbers
R = The number of rolls

So the equation would be:
((1 / 20)^5) x (18 / 5)

(I like unnecessary brackets)

They make things more readable, so they're necessary to me user.

have you applied Bernoulli process and trials?

because it doesn't seem correct to me