Every fish has at least 8 stripes, so if we had 10 fish, there would be 80 stripes with 6 stripes not yet counted.

As a male fish has one extra stripe, these 6 stripes must belong to 6 male fish.

So there are 6 males and 4 females.

Therefore the ratio of male to female fish is 3:2.

Alternatively, if there are f females and m males then $86=8f+9m=8(f+m)+m$.

Now $86=8\times 10+6=8 \times 9+14...$

Only the first of these is feasible so m=6 and f=4.

Therefore the ratio of male to female fish is $3:2$.

As a male fish has one extra stripe, these 6 stripes must belong to 6 male fish.

So there are 6 males and 4 females.

Therefore the ratio of male to female fish is 3:2.

Alternatively, if there are f females and m males then $86=8f+9m=8(f+m)+m$.

Now $86=8\times 10+6=8 \times 9+14...$

Only the first of these is feasible so m=6 and f=4.

Therefore the ratio of male to female fish is $3:2$.

*This problem is taken from the UKMT Mathematical Challenges.*