The Riddler was sort of feeling the rising boredom of being under lockdown when proposing the following variant to the Monty Hall puzzle:

There are zero to three goats, with a probability ¼ each, and they are allocated to different doors uniformly among the three doors of the show. After the player chooses a door, Monty opens another door hidding a goat or signals this is impossible. Given that he did open a door, what is the probability that the player’s door does not hide a goat?

Indeed, a straightforward conditional probability computation considering all eight possible cases with the four cases corresponding to Monty opening a door leads to a probability of ^{3}/_{8} for the player’s door. As confirmed by the following R code:

s=sample m=c(0,0) for(t in 1:1e6)m=m+(range(s(1:3,s(1:3,1)))>1)