/sci/ - Science & Math » Thread #12077086

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Anonymous Wed 02 Sep 16:10:05 2020 No.12077086 View Reply Original Report

Quoted By: >>12077095 >>12077938 >>12078291 >>12078375

I'm really shit at maths, but trying to figure out something with regards to chance and dont know where else to ask really - it's not my homework or anything like that, just a discussion i had with someone that i couldn't answer

If an event has a certain % chance to trigger a positive outcome - and you are given an item that grants another 'roll' at that outcome if the first results in failure - is there a sort of combined outcome likelihood that can be calculated? One guy said it doesnt change since its say, a 1 in 5, then a 1 in 5 again. One guy said it's doubled, but i dont think that can be right since a 50% chance to occur would then become 100% and always occur which is also not the case.

Or is there a name for what i'm describing so i can figure this out?

Anonymous

Anonymous Wed 02 Sep 2020 16:14:52 No.12077095 Report

Quoted By:

>>12077086
Its the same as the coin thing. No mater how many times you toss it its always 50/50

Anonymous

Anonymous Wed 02 Sep 2020 19:46:03 No.12077938 Report

Quoted By:

>>12077086
Yup. It's union of 2 outcomes success and failure/reroll/success.
These outcomes are disjoint, so we can add them up. If success is p, the success with a reroll is p+(1-p)p
1/5 success with a reroll is 9/25.
1/2 success with a reroll is 3/4

Anonymous

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Anonymous Wed 02 Sep 2020 21:01:41 No.12078285 Report

Quoted By:

assume you have a fair $N$ -sided die. the probability of rolling $n \in [1,N]\subset\mathbb{Z}$ is
$ P_{dN}(n) = \frac{1}{N} $
assume you want to meet or exceed challenge $C$ . the probability of success = 1 minus the probability of failure. the probability of failure is the sum of the probabilities of results below a threshold, which is represented by the cumulative probability function
$ X_{dN}(n) = \sum_{m=1}^{n} \frac{1}{N} = \frac{1}{N} \sum_{m=1}^{n} 1 = \frac{n}{N} $
and thus probability of beating challenge $C$ with a dN is
$ P_{dNpass}(C) = 1 - X_{dN}(C - 1) = 1 - \frac{C - 1}{N} $

for your scenario you want to know the chance of success given the special condition that you can reroll once if you fail. effectively you take the max value of two rolls. we need to compute the probability of getting each result and then compute the cumulative probability function to compute the success chance.

a concrete example helps us derive the probability. enumerating the possible max value of two rolls we find
$ \begin{array}{c|cccc} & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 2 & 3 & 4 \\ 3 & 3 & 3 & 3 & 4 \\ 4 & 4 & 4 & 4 & 4 \\ \end{array} $
so $P_{dNmax}(1) = \frac{1}{4^2}$ , $P_{dNmax}(2) = \frac{3}{4^2}$ , $P_{dNmax}(3) = \frac{5}{4^2}$ , $P_{dNmax}(4) = \frac{7}{4^2}$ . the pattern is
$ P_{dNmax}(n) = \frac{2 n - 1}{N^2} $

summing this
$ X_{dNmax}(n) = \sum_{m = 1}^n \frac{2 m - 1}{N^2} = \frac{1}{N^2} \left( 2 \sum_{m = 1}^n m - \sum_{m=1}^n 1 \right) = \frac{1}{N^2} \left( 2 \frac{n(n + 1)}{2} - n \right) = \frac{n^2}{N^2} $

thus the chance of success with a reroll is
$ P_{dNmaxpass}(C) = 1 - X_{dNmax}(C - 1) = 1 - \frac{(C-1)^2}{N^2} $

check the graphs in the picture for your chance to succeed. you want the reroll

Anonymous

Anonymous Wed 02 Sep 2020 21:02:39 No.12078291 Report

Quoted By: >>12080265

>>12077086
You need to calculate the probability of succeeding on the first attempt + the probability of failing the first attempt and then succeeding on the second.

Presuming that the two attempts are independent* (i.e. attempt 2 has exactly the sane odds as attempted 1), you can calculate it as follows:

$\frac{1}{5} + \frac{4}{5} \cdot \frac{1}{5} = \frac{1}{5} + \frac{4}{25} = \frac{5}{25} + \frac{4}{25} = \frac{9}{25}$

*this allows us to multiply the probabilities together to get the probability of both events occuring

Anonymous

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Anonymous Wed 02 Sep 2020 21:20:34 No.12078375 Report

Quoted By: >>12078557

>>12077086

Anonymous

Anonymous Wed 02 Sep 2020 22:05:21 No.12078557 Report

Quoted By:

>>12078375
Why do you make inputs to your spreadsheet so ugly blue/green?

Anonymous

Anonymous Thu 03 Sep 2020 11:52:59 No.12080215 Report

Quoted By:

hello op here thanks for the responses i understand a bit better now

Anonymous

Anonymous Thu 03 Sep 2020 12:21:05 No.12080265 Report

Quoted By:

>>12078291
This doesn't make sense. It's not clear what you are answering. You are assuming that events 1 and 2 are independent when 2 depends on whether 1 fails?

Anonymous

Anonymous Thu 03 Sep 2020 21:52:27 No.12082158 Report

Quoted By:

are there any other interesting rolling mechanics that /sci/ can analyze?

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