Deriving the 37-percent rule for dating

The decision to marry or not need only be made when the current result is better than all previous results (otherwise the current choice is definitely suboptimal). The choice, at point $K$ to marry or not (provided the $K$th date is better than all before) does not depend on the previous dates in anay ways (aprt from them being worse than the $K$th) because all relatiev orders are equally likely. Therefore, any optimal strategy must consist of probabilities $p_k$ for $1\le k\le N$ such that when the $k$th date is better than all before, we marry (and stop) with probability $p_k$ and continue otherwise. Let $q_k$ be the probability of picking the best provided we picked none of the first $k$. Then the probability of picking the best provided we picked none of the first $k-1$ is $$ q_{k-1}=\frac{\frac1Np_k+\frac{N-k}{N}\frac1k(1-p_k)q_k+\frac{N-k}{N}(1-\frac1k)q_k}{P(\text{none of the first $k-1$ selected})}=\alpha_kp_k+\beta_k$$ ($k$th is best and we pick, or best is after $k$ and $k$ is best of first and we don't pick it, or best is after $k$ and $k$ is not best of first) and hence is clearly maximized at $p_k=0$ ($\alpha<0$) or $p_k=1$ ($\alpha>0$) or in a rare possibility $p_k$ does not matter at all ($\alpha=0$) and can wlog. be selected to be $0$ or $1$.

You can make yourself clear that haveing $p_k=1$ and $p_{k+1}=0$ cannot be optimal.