Author Topic: Interesting, illegal immigration and the House of Representatives composition.  (Read 548 times)


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Had this pointed out on social media, and decided to do my own little bit of research, couldn't find the projection they cited for 2020, but I did find a Pew study for 2014 I decided to use.

As of 2010, the average House Seat represents about 710,000 people.
States with 375,000 or more "unauthorized immigrants" in 2014.

California - 2.35 Million (netting them 3 House seats)
Texas - 1.65 Million (netting them 2 House seats)
Florida - 850,000 (1 house seat)
New York - 775,000 (1 house seat)
New Jersey - 500,000
Illinois - 450,000
Georgia - 375,000


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I'm unsure what the point is here. 

Yes. Illegal immigrants are counted as part of the census and are part of apportionment.  Not sure if you can say that California or Texas actually owe 2 or 3 House seats to illegal immigrants, however.  If they don't count, they don't count at all for anybody.  That means you have to subtract the entire illegal immigrant population from apportionment, which is somewhat impossible, because Democrats made it so you can't really tell. 

If you take the 2010 census numbers, then subtract the 11 million illegal immigrants from the total that the pew report posits, you get 297,645,538. 
Take California's population, then subtract the 2.35 million the pew report states, you get 34,903,056.  This would be approximately 11% of your updated US population without illegals.  11% of 435 would get you 51.  That would be 2 less house seats instead of 3. 

This of course only works if you think all the illegals actually took part in the census. 
« Last Edit: January 15, 2019, 06:47:05 PM by Grant »


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That's correct about the math, it isn't so simplistic. If California doubled the us population with illegals, they'd still not have 100% of the seats.

That math also assumes that all those illegals are returning census forms, or more accurately that they are returning them at a rate equivalent to other residents. I would guess it isn't, but I would have a hard time proving it.