Every condition restricts the solution space, i.e. if any condition is removed the solution space should increase in size.
How Much do Conditions Restrict?
The house in decorum has 12 object slots with 5 possibilities each (all 4 colours and empty) and 4 wall painting slots with 4 options each (all 4 colours). Therefore, the number of total configurations is:
$$5^{12} \times 4^{4} = 6.25 \times 10^{10}$$
Now let’s consider a simple condition: “Condition C_{0}: The House must contain no Red Lamps”. The proportion of configurations that include no red lamps is
$$P(C_{0}) = (4/5)^{4} = 0.4096$$
Here’s another easily calculated condition: “Condition C_{1}: The Kitchen and the Bathroom must be identical, in both objects and wall colour”. The proportion of configurations that satisfy this is:
$$P(C_{1}) = (1/5)^3 \times 1/4 = 0.002$$
These are both conditions I’ve seen in some shape or form, and clearly restrict the space very differently.
The chance that condition C_{0} is satisfied given that condition C_{1} is satisfied is simple:
$$P(C_{0}|C_{1}) = (4/5)^{3} = 0.512$$
Therefore, we have that
$$P(C_{0} \cap C_{1}) = P(C_{0}|C_{1}) \times P(C_{1}) = (5/4) \times P(C_{0})P(C_{1})$$ $$= (5/4)^{0}P(C_{0}) \times (5/4)^{1}P(C_{1})$$ $$= \prod_{n=0}^{1}\biggl((5/4)^{n}P(C_{n})\biggr)$$
Now let’s extrapolate to all conditions:
$$P(\text{Solution})= \prod_{n=0}^{7}\biggl((5/4)^{n}P(C_{n})\biggr)$$
The geometric mean of \(P(C_{0})\) and \(P(C_{1})\) is around 0.03. If we were to say that all 8 of our conditions’ proportions have a geometric mean of 0.03, then we can say that:
$$P(\text{Solution})= \prod_{n=0}^{7}\biggl((5/4)^{n}(0.03)\biggr) \approx 3.4 \times 10^{-10}$$
This matches quite nicely with the number of total configurations, reducing it down to 1-2 possible solutions. This is perhaps too restrictive, but these calculations were rubbish anyway and just validating my choice of having 0.03 as a starting geometric mean for how much the conditions restrict.