Here you go:

probability of success in first attempt, P(s1)=8/10
probability of failure in first attempt, P(f1)=2/10

probability of success in second attempt, P(s2)=5/10
probability of failure in second attempt, P(f2)=5/10

The user goes for a second attempt only if he fails at the first. You can use the conditional probability of success at second attempt given the first is a failure, P(s2|f1)=P(s2^f1)/P(f1)= P(s2)*P(f1)/p(f1) = p(s2)=5/10. (ie. same as P(s2)...other way of looking at this is because the second success probability is INDEPENDENT of the failure at the first)

But the user will go to the second attempt ONLY when he fails at the first. Therefore success at the second attempt after failure from the first will be, P(s2|f1 ^f1)= P(s2|f1)*P(f1)=5/10*2/10=0.1

Hence the total success probability will be P(s)=P(s1+ P(s2|f1 ^f1)=8/10+0.1 = 0.9

(Note: ^ means AND operator)

However, be informed that it's been EONS I haven 't played with probabilities...so a double check wouldn't hurt.

Hope it helps..