Sometimes, our desire for luxury gets a little ahead of our actual budget. There’s no shame in wanting a five-star experience; who wouldn’t want a fabulous stay with incredible dining and fabulous hotels? Yet maybe you lack the cold hard cash (or your credit card doesn’t have room), but that’s okay because a recent study by Wethrift has figured out the cheapest cities where you can have a five-star vacation. Seven of the top 10 cities in this study are in California, which makes sense for the state with an easy-going nature covered by a varnish of luxury.
The top city for an affordable five-star experience? It’s Sonoma, the heart of wine country in northern California. It has an average restaurant price of just $44 and an average hotel price of $518, earning it a low-cost luxury rating of 1.83.
In second place is Chicago, which has an average restaurant price of $53 and a hotel price of $411, with a low-cost luxury rating of 1.63. In third place is San Diego, with a dining average of $49, a hotel average of $579, and a luxury rating of 1.51. Another takeaway? Sonoma, San Diego, and Chicago are the three most affordable cities for luxury dining, while Chicago, Paso Robles (California) and San Francisco are the three most affordable cities for luxury hotels.
You’re likely wondering about the methodology behind this. The study started by identifying 956 Michelin-recognized restaurants (with stars and Bib Gourmand ratings) and using TripAdvisor and Booking.com to identify 348 luxury three-to-five-star rated hotels across the U.S. The prices were then standardized by applying min-max normalization to arrive at the results.
Here are the top 10 best cities for an affordable five-star vacation:
1. Sonoma, CA
2. Chicago, IL
3. San Diego, CA
4. Santa Barbara, CA
5. Los Angeles, CA
6. San Francisco, CA
7. Miami Beach, FL
8. Washington, D.C.
9. Paso Robles, CA
10. Healdsburg, CA