# Probability question : What gender are my fish?

This is NOT a school homework question; I did my last one of those three decades ago!

We were sitting in the sunshine, watching our six golden rudd swimming around the garden pond and my wife asked me "how many do you think are females?" Well, lacking a detailed veterinary examination (of the fish!) I found it hard to give any sort of answer. All I could suggest was I'd tell her what the likelihood of each mix was . . . later.

Truth is, I've found it hard to come up with an answer and get varying results depending on the reasoning I apply. So, can someone tell me the correct method for this and the probability of each possible mix of males and females. For the sake of argument we'll assume these six fish were drawn from a large population of equal males and females and all were equally hard to catch.

I'm sure my wife is on tenterhooks, awaiting my answer to her question ;)

We were sitting in the sunshine, watching our six golden rudd swimming around the garden pond and my wife asked me "how many do you think are females?" Well, lacking a detailed veterinary examination (of the fish!) I found it hard to give any sort of answer. All I could suggest was I'd tell her what the likelihood of each mix was . . . later.

Truth is, I've found it hard to come up with an answer and get varying results depending on the reasoning I apply. So, can someone tell me the correct method for this and the probability of each possible mix of males and females. For the sake of argument we'll assume these six fish were drawn from a large population of equal males and females and all were equally hard to catch.

I'm sure my wife is on tenterhooks, awaiting my answer to her question ;)

active| newest | oldestIn that case, each individual fish has an independent 50% chance to be chosen male, or chosen female (there's no correlations). So the 64 permutations all have exactly the same probability (1/64 = 1/2

^{6}), and the only issue is to count the number of permutations for each combination:0F 6M ==> N=1 permutation, so P = 1/64

1F 5M ==> N=6, P = 6/64

2F 4M ==> N=5+4+3+2+1=15, P= 15/64

3F 3M ==> N=4+3+2+1, P = 10/64

4F 2M = 2F 4M ==> P = 15/64

5F 1M = 1F 5M ==> P = 6/64

6F 0M = 0F 6M ==> P = 1/64

So you have 10/64 = 15.625% chance of 3 of each. The most likely case is two of one and four of the other.

They are all female ........ and ..... when needed

one will transform to a masculine gender called .(protogyny)..

A

According to a USGS page, "Reproduction occurs from April to August. Adhesive eggs are laid among vegetation. Fecundity ranges from 3500–232000 eggs. Maturation is at 2–3 years, 90–150 mm TL. Males are smaller with fine tubercles [nodules] on head and body."

I can't find any references that suggest rudd are likely to switch gender, nor of any gender imbalance in breeding populations, so I'd be willing to assume that the probability of any particular fish being male or female is 50/50.

We only got them last summer at 2" long, so not yet breeding age yet, but that's another way . . . Wait a year or two then count the fat ones!

(I'll have a look for spotty ones with nodules, though.)