I read an MLB.com story about the Rangers’ game 5 win, and it cited this statistic:
The team that won Game 5 prevailed in 26 of the previous 39 World Series that were tied at 2 after four games.
This sounds like a good omen for Texas. But wait a minute. Suppose each team has a 50% chance to win any given game. Then we’d expect the game 5 winner to win the series 3/4 of the time, not 2/3. After all, the only way to lose a 7 game series is to lose twice in a row, a 25% chance.
Hmmm… Now the chances of winning any given game aren’t 50%. Home teams tend to win more often than visiting teams. So I used the baseball-reference play-finder to search for postseason games won by home and away, and found the home team won 720 games, and lost 605, or a .543 winning percentage. So now if I assume the home team won game 5, then since they’ll be visiting in games 6 and 7, they’d have a 54.3% chance of losing any given game, and so a 29.5% chance of losing 2 in a row. Even with this pessimistic assumption, though, I’d expect the game 5 winner to win the series 70% of the time, not 67%.
If it’s the visiting team wins game 5, the chances of them losing 2 home games in a row is just 21%. When I blend these, assuming the chance of a home team winning is always 54.3%, and thus 54.3% of game 5 winners are home teams, then I get the chance of the game 5 winner winning the series working out to 74.4%. So accounting for home/road doesn’t really make a significant difference over simply assuming it’s a 50/50 coin flip.
Now there have been only 39 previous World Series where the teams were tied at 2. That’s a tiny sample size, so I shouldn’t really be surprised that just 2 out of 3 of them won the series, when I’d expect it to be closer to 3 out of 4. But it is amusing, and also somewhat counter-intuitive, that so far game 5 winners have actually won the World Series less often than you’d expect by random chance alone!