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Idiot with a question about probabilities

Anonymous Sun 29 Mar 03:20:54 2020 No.11511140 View Reply Original Report

Quoted By: >>11511195

I am currently trying to program a random number generator with a total of 6 possible outcomes to start. The way I want the logic to work is to draw from a number of 6 possibilities, then from the remaining 5, then the remaining 4, and so on until it is forced to pull the last unmatched possibility. My question is what the percentage of each possibility would roughly be. I think I remember hearing these types of probabilities work in a curve with the middle being the most frequent, but I could be completely wrong about that.

Anonymous

Anonymous Sun 29 Mar 2020 03:44:49 No.11511195 Report

Quoted By:

>>11511140
https://en.wikipedia.org/wiki/Hypergeometric_distribution
>My question is what the percentage of each possibility would roughly be
This question is ambiguous, so I'm not entirely sure what exactly you want, but if sounds like the basic thing you're asking for is how to model "sampling without replacement", which is what the hypergeometric distribution is used for. (For very large populations, "sampling without replacement" is very close to "sampling with replacement", which is where the "normal distribution" is used because it's easier to compute, but here your population is only 6, so just use the real deal.)

Anonymous

Anonymous Sun 29 Mar 2020 04:52:31 No.11511333 Report

Quoted By: >>11513535 >>11513701 >>11513896

I think that's what I'm looking for, but just to clarify my application a bit further.

Say you have 6 balls in a jar, 5 red and 1 blue. At first you have a 1/6th change of pulling the blue ball, and if you take out one red, it becomes 1/5th then 1/4th and so on. Once the blue ball is pulled, all of them are placed back in the jar and the process is started over. What I'm trying to calculate is if that trial were repeated over set intervals (5 times, 10 times, 25 times, etc) what the probability each of the 6 potential draws would have of containing the blue ball, and how those numbers shift in relation to the sample size. I can work with more static equations but still like this was always my weakness with code.

I'm pretty sure I remember it not being equal to 1/6th chance for any given place, but I don't remember how to determine how it gets weighted over the size of the sample set. Most of the examples combine multiples for each color which I think is part of whats throwing off my dumbass conceptualization of it

Anonymous

Anonymous Sun 29 Mar 2020 23:31:16 No.11513535 Report

Quoted By:

>>11511333
Yeah, you'll want the hypergeometric distribution.
In the broadest terms, if you have a number $N$ of objects, a number $K$ of them have a specified property, and you draw from this population $n$ times WITHOUT replacing what you drew, the probability that you succeed exactly $k$ times in drawing objects with the specified property is given by the probability mass function $f$ : $f(k)=\frac{K\choose k {N-K}\choose{n-k}}{N\choose n}$
Let me know if you have further questions.

Anonymous

Anonymous Sun 29 Mar 2020 23:53:19 No.11513595 Report

Quoted By:

Sorry, $f(k)=\frac{{K\choose k}{{N-K}\choose{n-k}}}{{N\choose n}}$
This does have some shortcomings. $k$ must be between $\max(0,n+K-N)$ and $\min(K,n)$ for this equation to make any sense.

Oh, I'm looking at this problem a bit deeper and it's a lot more complicated than I thought. Basically, at the outset, before any draws, the probability of drawing the blue ball is 1/6. But once you draw, the probability of blue is either 1/5 (if you drew a red) or 0 (if you drew the blue). Basically, your parameter is also random, so this is some mixture. I am clueless about probability at this point.

Anonymous

Anonymous Mon 30 Mar 2020 00:28:35 No.11513701 Report

Quoted By: >>11513707 >>11513896

>>11511333

I don't think it's as difficult as you make it out to be.

Position of the blue ball being pulled:
1) 1/6 - get blue ball first try
2) 5/6 * 1/5 - get red ball first try, get blue ball second try
3) 5/6 * 4/5 * 1/4 - get red ball first two tries, get blue ball third try
... and so on

Anonymous

Anonymous Mon 30 Mar 2020 00:30:46 No.11513707 Report

Quoted By:

>>11513701
PS. I'm on mobile so can't check, but if these all add up to one (positions 1 to 6, following the pattern I laid out), then you can be even more sure that this solution is correct.

Anonymous

Anonymous Mon 30 Mar 2020 02:15:32 No.11513896 Report

Quoted By:

>>11513701
> 1) 1/6 - get blue ball first try
> 2) 5/6 * 1/5
= 1/6
> 3) 5/6 * 4/5 * 1/4
= 1/6
And so on. I.e. each of the 6 places has a 1/6th chance of getting the blue ball.

To look at it another way, suppose that you number the balls from 1-6. There are 6! = 720 possible permutations (orderings). For a given ball in a given position, there are 6!/6=5!=120 permutations which have that ball in that position. The choice of which of the 6 balls is the blue ball doesn't affect anything.

>>11511333
> I'm pretty sure I remember it not being equal to 1/6th chance for any given place,
It is 1/6.

OTOH, if you're interested in the probability of the Nth draw yielding the blue ball *given that that previous N-1 draws didn't*, it's 1/(6-N+1), i.e. 1/6, 1/5, 1/4, 1/3, 1/2, 1 (i.e. if the first 5 were all red, the 6th draw has to be blue ball).

Anonymous

Anonymous Mon 30 Mar 2020 04:54:40 No.11514232 Report

Quoted By: >>11514869

lol I love kula

so easy to play and does that whack combo (tatsu into dp), makes all the spics on fightcade rage, always goes first on my team in 2k2

Anonymous

Anonymous Mon 30 Mar 2020 11:53:16 No.11514869 Report

Quoted By:

>>11514232
kula is gay, whip is cuter

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