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By Root 549 0

they could have lost. Two games of the series had been played, so there were 5 possible games yet to play. And since each of those games had 2 possible outcomes—a Yankee win (Y) or a Braves win (B)—there were 25, or 32, possible outcomes. For instance, the Yankees could have won 3, then lost 2: YYYBB; or they could have alternated victories: YBYBY. (In the latter case, since the Braves would have won 4 games with the 6th game, the last game would never have been played, but we’ll get to that in a minute.) The probability that the Yankees would come back to win the series was equal to the number of sequences in which they would win at least 4 games divided by the total number of sequences, 32; the chance that the Braves would win was equal to the number of sequences in which they would win at least 2 more games also divided by 32.

This calculation may seem odd, because as I mentioned, it includes scenarios (such as YBYBY) in which the teams keep playing even after the Braves have won the required 4 games. The teams would certainly not play a 7th game once the Braves had won 4. But mathematics is independent of human whim, and whether or not the players play the games does not affect the fact that such sequences exist. For example, suppose you’re playing a coin-toss game in which you win if at any time heads come up. There are 22, or 4, possible two-toss sequences: HT, HH, TH, and TT. In the first two of these, you would not bother tossing the coin again because you would already have won. Still, your chances of winning are 3 in 4 because 3 of the 4 complete sequences include an H.

So in order to calculate the Yankees’ and the Braves’ chances of victory, we simply make an accounting of the possible 5-game sequences for the remainder of the series. First, the Yankees would have been victorious if they had won 4 of the 5 possible remaining games. That could have happened in 1 of 5 ways: BYYYY, YBYYY, YYBYY, YYYBY, or YYYYB. Alternatively, the Yankees would have triumphed if they had won all 5 of the remaining games, which could have happened in only 1 way: YYYYY. Now for the Braves: they would have become champions if the Yankees had won only 3 games, which could have happened in 10 ways (BBYYY, BYBYY, and so on), or if the Yankees had won only 2 games (which again could have happened in 10 ways), or if the Yankees had won only 1 game (which could have happened in 5 ways), or if they had won none (which could have happened in only 1 way). Adding these possible outcomes together, we find that the chance of a Yankees victory was 6 in 32, or about 19 percent, versus 26 in 32, or about 81 percent for the Braves. According to Pascal and Fermat, if the series had abruptly been terminated, that’s how they should have split the bonus pot, and those are the odds that should have been set if a bet was to be made after the first 2 games. For the record, the Yankees did come back to win the next 4 games, and they were crowned champion.

The same reasoning could also be applied to the start of the series—that is, before any game has been played. If the two teams have equal chances of winning each game, you will find, of course, that they have an equal chance of winning the series. But similar reasoning works if they don’t have an equal chance, except that the simple accounting I just employed would have to be altered slightly: each outcome would have to be weighted by a factor describing its relative probability. If you do that and analyze the situation at the start of the series, you will discover that in a 7-game series there is a sizable chance that the inferior team will be crowned champion. For instance, if one team is good enough to warrant beating another in 55 percent of its games, the weaker team will nevertheless win a 7-game series about 4 times out of 10. And if the superior team could be expected to beat its opponent, on average, 2 out of each 3 times they meet, the inferior team will still win a 7-game series about once every 5 matchups. There is really no way for sports leagues to change this. In the lopsided 2/3-probability