With results for all its precincts now in, it emerges that Brighton Beach (aka “Little Odessa”) voted 84.1% for Trump – his sixth highest result in all of NY’s ~300 neighborhoods.
The also heavily Sovok Jewish neighborhood of Seagate-Coney Island voted 81.4% for Trump.
(The big gray area on the map above is a single neighbood and only had 8 votes in total, which were evenly split between Trump and Cruz).
As you theorized: Brighton goes for Trump (mostly), evangelical ^H^H^H^H^H^H^H Orthodox Jews for Cruz. The richies in Manhattan go for Kasich. You can actually see the less upscale precincts east of Third Avenue going for Trump, whereas the superrich between Third and Fifth go for Kasich.
Cruz did pretty poorly outside of Orthodox Jews, which isn’t surprising when you insult the city.
These are huge outliers even relative to Trump’s outstanding performance in NY as a whole; as I write this, 96% of the votes have been counted and Trump is set to end up at 60% and with almost all of its delegates.
The only major region in NY state which Trump didn’t win was Manhattan which went for Kasich.
T even got more votes in BB and adjoining West Brighton than either D candidate, which in the rest of NYC happened only in SI neighbourhoods.
In 2010 Brighton Beach had 110,000 people. Trump’s 84% win there was the result of about 430 votes for him (he got over 69,000 votes in the city). Largely meaningless result.
The southern half of that big grey area is the beach itself, i.e. sand. There are some Soviet Jews in Coney Island’s (Fred) Trump Towers, and lots in Sheepshead Bay (where I used to live) and in neighborhoods north of it.
Trump got 708 votes (82.7%) in West Brighton. The census category of Brighton Beach might have included that area.
In Sheepshead Bay Trump got 1316 votes (78.6%).
In comparison, Howard Beach had 26,000 people and gave Trump about 900 votes.
Gravesand cast 425 votes for Trump (81.9%). Homecrest gave Trump 612 votes (74.5%). Lots of Soviet Jews there, though not as many as in Brighton.
Most basic interpretation of BB:
Older Sovok Jews = Trump
Younger Sovok Jews/half-assimilated Jewish-Americans = Bernie
NAMs = Hillary
?
Yes, that sounds right. One part of Brighton (the single-family homes north of Brighton Beach Ave.) is Hispanic. One part (the relatively-new Oceana development) is upscale – many kinds of upper middle class people live there. But most of Brighton is actual Little Odessa, the thing that everyone imagines BB to be. That’s the area that must have voted for Trump.
Can you explain why you are drawing conclusions based on the votes of 430 people in an area with a 110,000 population?
I live in an upscale apartment complex in Newton, MA. Many of the tenants are older Russian/Ukrainian Jews living with their kids (STEM workers in Cambridge, etc.). I work from home so I’ve had a few conversations with them about world events and politics. I can’t imagine many of them voted for someone other than Trump.
It doesn’t surprise me actually. Many years ago I knew a lot of Jewish taxi drivers – smart but not egghead smart and rough boys at heart; Trump is a lot like them in mannerism and style.
Because it fits in with my own impressions as well as apparently those of most other commentators here and as such is worthy of interest?
Turnout might be relatively low for the Republicans in BB, but it is also low for the Democrats as well. This might say something about the level of civic involvement in this demographic but it is probably an adequate enough representation of their political views.
“These are huge outliers even relative to Trump’s outstanding performance in NY as a whole”
Trump did very well in areas with a lot of Russian speakers, but its not a “huge outlier” when compared to Staten Island where Trump got 82% of the vote. Trump received over 70% in South-West Brooklyn neighborhoods. So yes Sovok Jews voted for Trump in large numbers, but so did other outer borough white Republicans in southern Brooklyn/Staten Island. (The more religious Jewish neighborhoods did vote less for Trump and more Cruz) So to what extent is it Sovok-ness versus their current location?
Neighborhoods with lots of Russian speakers-
Brighton Beach 84.1%
West Brighton 82.7%
Seagate-Coney Island 81.4%
Sheepshead Bay Area 78.6% (Sheepshead bay has more Russian speakers than Brighton at this point)
Homecrest 74.5%
Other Staten Island & Southern Brooklyn neighborhoods (these also have lots of Russian speakers, but they must be a relatively small percentage of Republican primary voters)
Staten Island 82.1%
Gravesend 81.9%
Bath Beach 78.6%
Bensonhurst East 77.9%
Bensonhurst West 76.5%
GeorgeTown-Marine Park 76.1%
Dyker Heights 76.3%
Bay Ridge 69.2%
It’s just so funny to me to see this huge map of NY state all red with a tiny little sliver of green on it – Kasich’s Manhattan. In every debate he bored us with his “son of a mailman” schtick. He’s the plain-spoken guy from the heartland with common-sense solutions. And the only places he could win here were Wall St, the Upper East Side, SoHo, Greenwich Village, etc.
Large % isn’t large numbers. Only 430 of Brighton Beach’s 110,000 people voted for Trump. The percentage of voters choosing Trump was high, but very, very few people voted at all.
In contrast, Italian Howard Beach with 26,000 people gave Trump 900 votes.
430 votes out of a population of 110,000 doesn’t seem representative of much. One can conclude that of the tiny number of BB people who bothered to vote, Trump was quite popular.
I know several Russian Jews who would choose Trump over Clinton (or especially Sanders) but none who like him. A few liked Rubio and Christie. BB results may reflect this type of attitude.
It’s not 110,000 people it’s 35,244.
The vote numbers are for neighborhood tabulation areas. The population numbers for NYC neighborhood tabulation areas can be found here.
https://data.cityofnewyork.us/City-Government/New-York-City-Population-By-Neighborhood-Tabulatio/swpk-hqdp
See also this map of primary results. Same numbers as the NYT, but they actually show you the neighborhood boundaries. http://www.wnyc.org/story/map-ny-primary-vote-nyc-2016/
Total republican primary votes
Brighton Beach – 2.12% of population
Lindenwood-Howard Beach -5% of population
(Even this overstates the difference because, per 2009-2013 Census ACS profile, BB has approximately 25K citizens, and Howard Beach has 27K. So they have roughly the same number of people eligible to vote.)
Also, remember that large numbers of people in this neighborhood are immigrants who are NOT from the former Soviet Union.
Per ACS profile foreign born Europe 13.7K, Latin America 2.5K, Asia 6K (Pakistanis, Arabs)
I imagine relatively few of the Mexicans and Arabs are registered Republicans. I think its safe to assume that of those people who voted in the Republican primary in that neighborhood, many are Jews from the former Soviet Union, even if they are only say 20% of the population overall. (Btw, I am not even sure that “Sovok Jews” are a majority of the Russian speaking population in at this point. There are large numbers of Russians, Ukranians, Georgians, Uzbeks etc. They don’t really fit neatly into any of Anatoly’s 5 types of Russian Americans. Many of them must be citizens by now. )
“The also heavily Sovok Jewish neighborhood of Seagate-Coney Island voted 81.4% for Trump.”
For what it’s worth here are the Russian + Ukrainian ancestry numbers, as a percent of overall population, from the 2009-2013 ACS Profile. Seagate is 9th in Brooklyn.
http://maps.nyc.gov/census/
West Brighton – 42.1%
Brighton Beach – 25.9%
Sheepshead Bay Area – 19.1%
Madison – 17.4%
Gravesend – 16%
Midwood – 13.9%
Homecrest – 13.6%
Bensonhurst East – 12.6%
Seagate-Coney Island – 11.2%
But the more meaningful number is Russian+Ukranian ancestry divided by the number of Non hispanic Whites. Seagate/Coney Island has a lot of Asians and Blacks few of them vote in republican primaries.
West Brighton 43%
Brighton Beach 37%
Seagate-Coney Island 37%
Gravesend 30%
Sheepshead Bay Area 27%
Bensonhurst East 26%
Madison 25%
Midwood 18%
Homecrest 17%
Thanks for the clarification. Multiple other sources place Brighton Beach’s population at 110,000. West Brighton’s population is elsewhere listed as 64,000 – but the table you linked to claims only 17,750. This underestimate seems to be evident across the city – Chinatown’s population is estimated at 150,000 but the table lists about 50,000. I wonder how the discrepancy is explained. Does MTA count citizens only? The table’s figures are from 2000 but the population would not have increased that much in 10 years.
The difference in population is due to different sources referring to different things when they say “Brithton Beach.” As the WNYC notes point out “[f]rom a conversational standpoint, it makes sense to count votes by neighborhood. From a computational standpoint, that’s tricky to do, because the precise boundaries of New York City neighborhoods are notoriously difficult to define.
We settled on New York City’s map of Neighborhood Tabulation Areas, which is imperfect but divides the city into roughly equal population chunks.”
http://www.wnyc.org/story/where-heck-hells-kitchen/
When you google Brighton Beach population you get one a map of one area and a population number from wikipedia for a different area. The wikipedia article had population of 111,063
for “Brighton Beach and Coney Island, combined.” The wikipedia source link is this http://www.osc.state.ny.us/osdc/rpt8-2012.pdf
Which says “[n]eighborhood data used in this report are either for the greater Coney Island area, covered by ZIP codes 11214, 11223, and 11224; or for a U.S. Census
neighborhood area that covers Coney Island and Brighton Beach.” Basically they threw a bunch of neighborhoods together, one of which is Brighton Beach. Then wikipedia used the 110,000 number and then a bunch of crappy websites took the number off Wikipedia. So those are your multiple sources. The combined population of neighborhood tabulation areas Brighton Beach, West Brighton, Seagate-Coney Island is 86,881. Throw in pieces of neighboring neighborhoods like Homecrest and Gravesend and eventually you’ll get close to 110,000.
“West Brighton’s population is elsewhere listed as 64,000” I doubt it, where did you see this? The total population from the 2010 Census for the West Brighton NTA is 17,750 according to this http://maps.nyc.gov/census/ (The data.cityofnewyork.us link I gave before actually lists 2000 census numbers)
In the real world nobody says “West Brighton” its just part of Brighton. Just use neighborhood tabulation areas, that’s what the vote numbers are for and there is census population and survey data for these units.
The Chinatown NTA 2010 census population is 47,844. Again depends where you draw the borders of “Chinatown.” Also matters who is doing the estimating. Some Chinatown tourist organization PDF says “Chinatown is home to a resident population estimated at 150,000.” Yea I wouldn’t pay any attention to sources like that. The Census may be under counting the population in areas with a lot of immigrants like Brighton Beach and Chinatown but its a better startign point than some random number from wikipedia or a tourist brochure.
By the way the 2009-2013 ACS profile ancestry survey, used above for the neighborhoods (NTAs), shows a total of 286,053 for Russian + Ukrainian ancestry in New York city, and 114,042 for Brooklyn.
The 2014 American community survey of languages spoken at home counted 185,759 people 5 years and over who speak Russian at home in New York City. The number for Brooklyn is 125,877. The same survey has reported ancestry of Russian + Ukrainian of 270,976 for New York City, of which 116,070 is in Brooklyn.
So while I haven’t seen any data for language spoken at home by NTAs the Russian + Ukrainian ancestry total for Brooklyn of 114K is close to the 125K Russian language spoken at home. So I think it is a pretty good proxy in Brooklyn. (Manhattan is different, 89K for Russian+Ukrainian ancestry versus only 11K speaking Russian at home.)
http://www1.nyc.gov/site/planning/data-maps/nyc-population/american-community-survey.page
http://maps.nyc.gov/census/
Thank you, again, for clarifying things. Your effort is much appreciated.