Confidence that feeds on itself

Idea. Every sequence of results with the same number of goals has the same probability. Counting them shows that the number of goals is uniform on $1,\dots ,99$.

Step 1: one sequence. After the first two kicks the player has 1 goal and 1 miss in 2 kicks. Look at the remaining 98 kicks, with $j$ more goals and $98-j$ more misses in a given order. When the $m$-th of the new goals is scored, he has already scored $m$ times, so the numerators of the goal probabilities are $1,2,\dots ,j$, whatever the order. Likewise the numerators for the misses are $1,2,\dots ,98-j$. The denominators are always $2,3,\dots ,99$. So every such sequence has probability $$\frac{j!(98-j)!}{2\cdot 3\cdots 99}=\frac{j!(98-j)!}{99!}.$$

Step 2: count. There are $\left(\genfrac{}{}{0ex}{}{98}{j}\right)$ orders, so $$\begin{array}{rl}\mathbb{P}(j\text{ more goals})& =\left(\genfrac{}{}{0ex}{}{98}{j}\right)\frac{j!(98-j)!}{99!}\\ & =\frac{98!}{99!}=\frac{1}{99}.\end{array}$$

Step 3. So the total number of goals, $1+j$, is uniform on $1,2,\dots ,99$, and $$\mathbb{P}(\text{exactly 40 goals})=\frac{1}{99}.$$

Remark. Any count from 1 to 99 is equally likely: early luck is locked in. As the number of kicks grows, his scoring rate converges to a random limit that is uniform on $[0,1]$.