1999 AL Winning Percentage by Team
Every team's total, added up from its players' lines. Wins per decision
The New York Yankees led the AL in winning percentage in 1999, with .605, .006 ahead of the Cleveland Indians. The average across 14 AL teams is .496.
Led the AL in 1999
New York Yankees
162 games
.605
Winning Percentage
.006 clear of Cleveland Indians
| # | Team | G | W-L% |
|---|---|---|---|
| 1 |
New York Yankees
|
162 | .605 |
| 2 |
 Cleveland Indians
|
162 | .599 |
| 3 |
 Texas Rangers
|
162 | .586 |
| 4 |
 Boston Red Sox
|
162 | .580 |
| 5 |
 Oakland Athletics
|
162 | .537 |
| 6 |
 Toronto Blue Jays
|
162 | .519 |
| AL average | .496 | ||
| 7 |
 Seattle Mariners
|
162 | .488 |
| 8 |
 Baltimore Orioles
|
162 | .482 |
| 9 |
 Chicago White Sox
|
162 | .466 |
| 10 |
 Anaheim Angels
|
162 | .432 |
| 11 |
 Detroit Tigers
|
161 | .429 |
| 12 |
 Tampa Bay Devil Rays
|
162 | .426 |
| 13 |
 Kansas City Royals
|
161 | .398 |
| 14 |
 Minnesota Twins
|
161 | .394 |
A team's total is its players' numbers for the club added up, so a player traded mid-season counts for each team only what he did there. Rates such as ERA and batting average are worked out from the team's summed totals, not averaged across players.