It seems to me that $P(d>k)=\tfrac1{k+1}$ for $k\in[0,n]$, so that $$E[d]=\sum_{k\ge 0}P(d>k)=H_{n+1}\simeq \ln n.$$ The reason is that the last $k+1$ terms of the sequence are in random order, so that the last one is the smallest among the last $k+1$ terms of the sequence with probability $P(d>k)=\tfrac1{k+1}$. When the last is the smallest of all I assume that $d$ is set equal to $n+1$. Does it make sense ?

For this to be valid, you need to assume that the distribution has no atoms, which is true for instance if it has a density of probability.

It seems to me that $P(d>k)=\tfrac1{k+1}$ for $k\in[0,n]$, so that $$E[d]=\sum_{k\ge 0}P(d>k)=H_{n+1}\simeq \ln n.$$ The reason is that the last $k+1$ terms of the sequence are in random order, so that the last one is the smallest among the last $k+1$ terms of the sequence with probability $P(d>k)=\tfrac1{k+1}$. When the last is the smallest of all I assume that $d$ is set equal to $n+1$. Does it make sense ?

For this to be valid, you need to assume that the probability distribution has no atoms, which is true for instance if it has a density of probability.