Comedian David Spade has found a buyer for a Beverly Hills mansion that was his home for two decades.
The four-bedroom, seven-bath property, located at 1120 Wallace Ridge in Trousdale Estates, went into contract on January 12, six days after it was put on the market. Its last asking price was nearly $20 million, according to Zillow, and the deal closed at $19.5 million. The buyer is currently unknown.
Architect Edward Fickett designed the 6,426-square-foot mansion. It has features such as a backyard pool, a tennis court and a fire pit. Spade bought the property in 2001 for $4 million, property records show. The mansion was updated several times under his ownership, getting a metal parapet in place of a Spanish-style roof, a new kitchen and a TV room.
Spade recently moved to the Hollywood Hills. In October, he paid $13.9 million for a newly-built spec home above the Sunset Strip. His new house is twice as large as the Beverly Hills property. The six-bed, eight-bath property, designed by architect George Kurczyn, spans 11,300 square feet and has features such as an infinity pool, a bar and club room and a panic room.
The deal adds to Spade’s real estate deal sheet. He previously owned a beach house on Malibu’s La Costa Beach. He sold that property to Netflix co-CEO Ted Sarandos for $10.2 million in 2013. In 2020, he parted with a Spanish-style bungalow in West Hollywood in a $2.3 million off-market deal.
Hilton & Hyland’s Stuart Vetterick, the broker in charge of the listing, did not respond to a request for comment on Spade’s most recent sale.