Explanation
USSR vs. Rest of the World 1970 was a team match held in Belgrade from March 29 to April 5. A Soviet selection faced a team assembled from leading players of other countries. The main lineups contained ten boards, ordered roughly by playing strength, with reserves available to replace starters.
Four rounds were played. On each board, the regular representatives could meet up to four times while alternating colors. Every game contributed one point to the winning team, half a point to each side for a draw, and zero to the loser. After all forty games were added, the USSR won by the narrowest margin, 20.5-19.5.
The Soviet team included five world champions: Botvinnik, Smyslov, Tal, Petrosian, and Spassky. Bent Larsen played first board for the Rest of the World, with Bobby Fischer on second board. Fischer faced Tigran Petrosian and scored 3 points from 4, one of the match's leading individual results.
The event carried strong Cold War symbolism because it tested the depth of the Soviet chess system against an international selection. It was nevertheless a specific representative match, not an edition of the , which is a different recurring institutional competition.
Common confusions
World Team Chess Championship
The Belgrade contest was a special match between two selections. It was not an edition of the recurring institutional world team championship.
View termOne game on each board
The match lasted four rounds, and regular opponents generally met several times. The final score combined forty games.
Sources
- 1.A look back at the 1970 USSR vs. Rest of the World, FIDE
- 2.Belgrade: Narrow win for the USSR, ChessBase
