Why does relative humidity drop if I open the window in winter? What about the temperature drop?
Let's do this with accurate equations. We know how to calculate the saturated water vapour pressure for temperature $T$:
$$e_s(T)=e_{s0}e^{\frac{h_i}{R_w}(\frac{1}{T_{s0}}-\frac{1}{T})}=610\text{ Pa} \times e^{5423(\frac{1}{273\text{ K}}-\frac{1}{T})}$$
$e_s$ is saturated water vapour pressure as function of $T$ in Pascals
$T$ is the temperature in Kelvins
So, the saturated water vapour pressure at $T=0°C=273K$ is equal to $e_s(273K)=610 Pa$. Similarly, $e_s(293K)=2360Pa$ and $e_s(283K)=1230Pa$.
So we have two parcels of air: 1. 20°C, 40% and 2. 0°C, 40%. What are the masses of the water vapour in both parcels?
$$e=f\times e_s(T)$$
$f$ is relative humidity, $e$ is the water vapour pressure
$$m(T)=\frac{eV}{R_w T}=\frac{f\times e_s(T)\times V}{R_w\times T}=\frac{fe_s(T)V}{461\frac{J}{kg}T}$$
$$m(293K)=0.0070kg$$
$$m(273K)=0.0019kg$$
The next step is to combine these two masses:
$$M=m(293K)+m(273K)=0.0089kg$$
So the density is $\rho=\frac{M}{V}=0.0045\frac{kg}{m^3}$
The water vapour pressure is given by:
$$e=\rho RT=587Pa$$
The relative humidity is then:
$$f=\frac{e}{e_s(283K)}=0.477$$
So this is around 48% of the relative humidity.
Why did I put here so many equations? Simple! These equations hold on high degree of accuracy, so they can't be wrong. Why would the relative humidity drop in reality?

Also, for Skew-T Log-P lovers:

This solution is even more concise and less error prone.