I Am Taller Than Everyone I See

It may not be obvious at first but is possible to show that;

${ P }_{ N }={ P }_{ N-1 }+\frac { 1 }{ N }$

Let us consider the last person, this person has $\frac { 1 }{ N }$ of being the tallest. Which gives you $1+{ P }_{ N-1 }$ people who are able to give this statement. If we consider everyone except the last person, that is $N-1$ , you get ${ P }_{ N-1 }$ who are able to give this statement.So in fact we get,

${ P }_{ N }={ P }_{ N-1 }+\frac { 1 }{ N }$

also for ${ P }_{ N-1 }$ by the same method

${ P }_{ N-1 }={ P }_{ N-2 }+\frac { 1 }{ N-1 }$

Inductively we get

${ P }_{ N }=1+\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +\frac { 1 }{ 4 } .............\frac { 1 }{ N }$

which is essentially a finite harmonic series in the form

$\sum _{ k=1 }^{ N }{ \frac { 1 }{ k } }$

$\left\lfloor \left( 1000 \right) \sum _{ k=1 }^{ 200 }{ \frac { 1 }{ k } } -\sum _{ k=1 }^{ 198 }{ \frac { 1 }{ k } } \right\rfloor =\left\lfloor 1000\left( \frac { 1 }{ 199 } +\frac { 1 }{ 200 } \right) \right\rfloor =10$

Let the people be $\alpha_1 \cdots \alpha_N$ , where $\alpha_1$ is the first person in line and $\alpha_N$ is the last person in line. For each $1 \leq i \leq n$ , let $X_i$ be equal to 1 if person $\alpha_i$ is taller than everyone in front of him, and 0 otherwise. Then $P_N = X_1 + \cdots X_n$ . The average value of $P_N$ is the same as the expected value of $P_N$ . The expected value of any $X_i$ is just $\frac{1}{i}$ because it is just the probability that in the first $i$ people, $\alpha_i$ is the tallest, and any person in this set of people is equally likely to be the tallest. Thus, by the linearity of expectation, we see that the expected value, or average, of $P_N$ is just the sum of the first $n$ terms of the harmonic series, and now it is very easy to numerically compute that answer, which comes out to 10.