Seeing the 3rd tallest person doesn't mean you can see 3 people or more. You might only see the 3rd tallest and the tallest, with the rest obscured behind the tallest. Also, the probability of the 3rd tallest being visible is 1/3 not 1/4. The probability of the nth tallest person being visible is (n-1)!/n! = 1/n.


With these corrections the problem is actually eaiser. Notice that if we just add the probabilities for seeing the tallest, 2nd tallest, etc. people, we will get the desired average since we will be counting all the times we see 2 people exactly twice, all the times we see 3 people 3 exactly 3 times, etc. So the series is:


1 + 1/2 + 1/3 + ....+ 1/10 = 2.93


So almost 3 people would be visible on average.