Why doesn't my house get misty?
It's a matter of temperature and dew point. If it's foggy outside, that means the temperature outside is at or below the dew point. Chances are it's warmer inside your house. If so, no fog.
On the other hand, if the temperature inside is the same as outside, it may actually be foggy inside, but you don't notice because sight lines inside may not be long enough to notice the fog.
P.S. Dew point is the temperature at which the rate of condensation of water vapor into droplets (releasing heat) equals the rate of evaporation of droplets into vapor (absorbing heat). It depends, of course, on the amount of water vapor in the air.
The fog evaporates when it enters your house. If you want fog in your house, you should try filling it with a couple of inches of water so that the interior is at 100% humidity.
Your house, and most other places, are unlikely to be at 100% humidity because they're not soaking wet. If not, dew will condense on anything that's cold, and the water will run off, probably to somewhere inaccessible. When the temperature rises again, that water will no longer be available to evaporate, so the humidity will be below 100%.
So,suppose your humidity is below 100%. When fog blows in, you will mix the 100% humidity air at Toutside around the fog with the X% humidity air at Tinside in your house. If your house is enough colder than the air outside, it's possible that the humidity will be over 100% at Tmix and even more fog will condense out of the air. Chances are, though, that your house is either the same temperature or warmer than the air outside.
So, now you have fog droplets at (50 + X/2)% humidity, let's say, where equal parts of inside and outside air have mixed. The fog will evaporate into the air, or at least some of it will. But if these fog droplets last for long enough, you'll still see them in your house. How long will they last?
Fixed the part below about how long it will last.
Droplets turn out to have an evaporation rate that is nearly constant:
dD^2 / dt = -K
where K depends linearly on the partial pressure of water that you could still add to the air before it's at saturation, and D is the diameter. In "Kinetics and evaporation
of water drops in air" by H.J. Holterman (IMAG report 2003 – 12, Wageningen UR, July 2003), Table 4, we see that K is a constant of about 100 μm2/s at 90% humidity and 15C. (And even if that were after mixing that would mean your house is incredibly humid.)
Since typical cloud/fog droplets are only 10-20 microns in diameter, even under these very humid conditions, most droplets will last no more than about 1-4 seconds. With more typical humidity, they'd last a fraction of a second, which means that the lifetime of the droplets would be dominated by the time taken for the outside and inside air to mix, not the evaporation of droplets in mixed air. So to get a noticeable amount of fog inside a livable house, you'd have to have a decent breeze, and then mostly what you'd see is droplets in air that had not yet warmed up or mixed with the air inside. Fog normally forms in calmer conditions, but maybe in San Francisco where fog blows in with a decent breeze you'd have a chance at seeing some indoor fog.
Fog / clouds form when the air temperature drops below dew point. The dew point is the temperature at which 100 % relative humidity is reached. At and below dew point, condensation exceeds evaporation at the condensation nuclei, and droplets can survive.
There are several ways to decrease the temperature of a given air mass: Decrease air pressure by vertical motion to create clouds, move air into a colder environment to create advective fog (or go outside in winter and breathe), let moist air stay still in a depressed area over night to create radiative fog.
On the other hand, and to answer the question, if an air mass is heated (e.g. by entering the inside of a house), evaporation of the droplets exceeds condensation and the fog will disappear.