Let A(n) be the event 'exactly n red ones are caught'. Let B(n) be the event 'there are exactly n red ones in the ponds'. We are looking for P(B(10)|A(10)), ie the probability that ten rocks are red given that ten red rocks are drawn. So, by something called Bayes' theorem,
P(A(10)|B(10)) = P(B(10)|A(10))*P(A(10))/P(B(10))
which simplifies things, because the probability that we get 10 red ones given that there are ten in the pond (the bit on the left of the equation) is obviously just 1. So now we've got:
1 = P(B(10)|A(10))*P(A(10))/P(B(10))
Furthermore, since the probability that we throw a red rock in on any given trial is 1/2 (flip of a coin), P(B(10)) = 2^-10 = 1/1024. Thus:
1/1024 = P(B(10)|A(10))*P(A(10))
Now, P(A(10)), the probability that ten red ones are picked, will be equal to the average probability that this happens, specifically:
P(A(10)) = SUM(n=1 to 10) P(B(n))P(A(10)B(n))
= (1024^-1)*SUM(n=1 to 10) (10 choose n) (n/10)^10
If we substitute this in the other equation, the 1/1024s cancel, and by evaluating that sum we find that P(B(10)|A(10)) = 1/14.26 or about 7%.
The answer to the title is in this post.