Someone that knows probability, tell me what I'm doing wrong.

So if we have 500 fish in the lake, and 50 are already tagged, then the probability of catching a tagged fish is 50/500.
Then catching another tagged fish is 49/499 and a third would be 48/498. So in general would it just be 50*49*48/(N)*(N-1)*(N-2)?

Or would it be N choose 50 / 50 choose 3?

(* is multiplication)
First of all, your answer is not between 0 and 1, it is not a probability.
We have 50 tagged (white)fishes and N-50 untagged fishes. Probability of catching a tagged fish is 50/N and not catching a tagged fish is N-50/N. In the second attempt, catching a tagged fish if we have already catched one is 49/N-1 and catching a untagged fish is N-50/N-1. And catching a tagged fish in the second run, if we haven't catched in the first attemp, is 50/N-1 and catching an untagged fish is N-51/N-1. So we should continue like that but we have an information which says we end up with only 3 tagged fishes. So this means, we see 50, 49 and 48 on numerator for tagged fishes and N-50, N-51 to the N-96 for the untagged fishes and we see N to the N-49 on the denominator. So we have something like this:
50*49*48*(N-50)*(N-51)*...*(N-96)
divided by
N*(N-1)*...*(N-49)
If we try to write this in a simpler way:
Answer: P(50,3)*P(N-50,47) / P(N,50)

N fish in the lake
choose 50
50 possible successes
we have 3 successes

k = 3
n = 50
K = 50
N = N

Hypergeomtric distribution google it

>>8348426
Ok so ((50 choose 3)*(N choose 47)) / (N choose 50)?

>>8348445
(C(x,y) means (x choose y))
No, but I made a mistake by using permutation instead of combination.
[I just simply count what is the probabilty of catching 3 tagged fishes in the first 3 runs and then catching 47 untagged fishes and this was a mistake, if we add other posiblities, we'll get what you say: (C(50,3)*C(N-50,47)) / C(N,50)]

>>8348497
Well thanks for the help. Definitely in a better spot than I was. Now time to make a function for this in R and plot it. Fun stuff.

>>8348511
NP, but if you didn't fully understand my explanation (and most probably I am the one to blame for that), you should google hypergeometric distribution.

>>8348361
50 marked fish , N-50 non marked fish

find probability that the first 3 fish caught are marked and next 47 fish caught are non-marked

50/N * 49/(N-1) * 48/(N-2) * (N-50)/(N-3) * (N-51)/(N-4) ... (N- 99)/(N-49)

Which is equal to [( 50! / 47!) * ((N-50)! / (N-100)!) ] / (N! / (N-50)!)

Then you take that probability and you multiply it by the number of possible permutations there are for catching 50 fish and finding 3 are marked fish and 47 are non-marked (because instead having the 3 marked fish at the beginning you could have them at the end, or in the middle , or in any order). which is given by the binomial coefficient 50!/(3!*47!)
So the answer is (50!/(3!*47!)) * [( 50! / 47!) * ((N-50)! / (N-100)!) ] / (N! / (N-50)!)

which equals
(50!/(3!*47!)) * ( 50! / 47!) * ((N-50)! / (N-100)!) * ((N-50)! / N!)


which does not simplify neatly

Woops I've just spotted thanks to >>8348426
that It should say

50/N * 49/(N-1) * 48/(N-2) * (N-50)/(N-3) * (N-51)/(N-4) ... (N- 96)/(N-49)

I forgot tha when you start catching unmarked fishes, thre are only 47 of them, got carried away and thought that there are 50.

so it should be
[( 50! / 47!) * ((N-50)! / (N-97)!) ] / (N! / (N-50)!)

then
(50!/(3!*47!)) * ( 50! / 47!) * ((N-50)! / (N-97)!) * ((N-50)! / N!)

I think the final answer of >>8348426 is wrong though because it forgot to multiply by binomial coefficients.

For example if it was not fish and instead the question were "what is the probability of rolling a dice 50 times and getting within your 50 roles 3 sixes and 47 non-sixes , you would of course remember to multiply by the binomial coefficient.

>>8348497
> (C(50,3)*C(N-50,47)) / C(N,50)
This is the only correct answer ITT, everything else is autism.

>>8348497
(C(50,3)*C(N-50,47)) / C(N,50)
equals
[(50!/(3!*47!))*((N-50!)/(47!(N-97)!)] / (N!/50!(N-50!))
equals
(50!/(3!*47!)) * ((N-50)! / (N-97)!) * (50!/47!) * ((N-50!)/N!)

which is my answer >>8348621

guess it does have a simpler expression haha

>>8348628
this answer >>8348621
>(50!/(3!*47!)) * ( 50! / 47!) * ((N-50)! / (N-97)!) * ((N-50)! / N!)
is correct too

>>8348636
[math](N-97)!\neq3![/math]
So no, it's not, and writing it down like that is autistic as hell.