Let [tex]K_i[/tex] denote the number of encounters required to see a new Pokemon after having encountered [tex]i-1[/tex] of them. Then [tex]K_1=1[/tex].
If Gary has already seen [tex]i-1[/tex] Pokemon, that leaves [tex]m-(i-1)[/tex] still to be encountered, and the next new encounter has a probability of [tex]p_i=\frac{m-(i-1)}m[/tex] of occurring. Then [tex]K_i[/tex] is geometric, with PMF
[tex]P(K_i=k)=\begin{cases}(1-p_i)^{k-1}p_i&\text{for }k\ge1\\0&\text{otherwise}\end{cases}[/tex]
Then the expectation and variance of [tex]K_i[/tex] are
[tex]E[K_i]=\dfrac1{p_i}=\dfrac m{m-(i-1)}[/tex]
[tex]V[K_i]=\dfrac{1-p_i}{{p_i}^2}=\dfrac{m(i-1)}{(m-(i-1))^2}[/tex]
[tex]K[/tex] is the total number of encounters needed to record all [tex]m[/tex] Pokemon, so
[tex]K=\displaystyle\sum_{i=1}^mK_i[/tex]
By linearity of expectation,
[tex]E[K]=\displaystyle\sum_{i=1}^mE[K_i]=\sum_{i=1}^m\frac m{m-(i-1)}=\frac mm+\frac m{m-1}+\frac m{m-2}+\cdots+\frac m1[/tex]
[tex]E[K]=m\displaystyle\sum_{i=1}^m\frac1i[/tex]
The [tex]K_i[/tex] are independent, so the variance is
[tex]V[K]=\displaystyle\sum_{i=1}^mV[K_i]=\sum_{i=1}^m\frac{m(i-1)}{(m-(i-1))^2}[/tex]
[tex]V[K]=m\displaystyle\sum_{i=0}^{m-1}\frac i{(m-i)^2}[/tex]
i got this completely wrong cause im just gonna say the answer is pikachu
but really quick gary never sees all of the pokemon
