What is the frequency of winning a 7 game series losing the first game?

I’m thinking hockey right now, but in any sport, there must be some general stats indicating the chances of winning a series if you win/lose the first game. I know that home and away are confounding variables. And some sports configure games 3 through 7 differently. But is it possible to say that In general, when a team wins the first game of a scheduled 7 game series, their chances of winning the series is X % ?

If the probability of winning each game is 50%, and they are independent, then the chance of winning a 7 game series after losing the first game is 22/64 = 34.375%.

Of course the team that wins game one is more likely the better team and so probably has a win percentage > 50%.

So that would say that winning the first game leaves you with a 65% chance of winning the series in seven games. As you say, sometimes the team that wins the first game is the better team. They may also win the series in fewer than seven. p.s. how do you derive 22/64? And, mainly, I wonder if there are stats that reflect the actual results of scheduled seven game series in all sports up to now?

Historically, using MLB, NBA and NHL best-of-7 playoff series, the team that wins the first game wins the series 71.6% of the time. If the first game is won at home, that percentage rises to 79.3%. If it’s won on the road, the percentage falls to 55.9%. If you want to calculate the percentage of time the first game loser wins the series, subtract those percentages from 100.

For hockey specifically, the percentages are 68.9% overall (first game winner wins the series), 76.7% after winning the first game on home ice, and 54.2% after winning the first game away.

These numbers and lots more (including stats for other win-loss combos, though annoyingly not ties) here: http://www.whowins.com

Exactly. Independence is a very strong assumption, and should never be your first bet.

Slight hijack, but I’d want to see a lot of data before I assumed that. An astoundingly large percentage of sports seasons, including ones where fans believe there’s a dominant team that’s clearly better than the others, pass a chi square test (i.e. they’re indistinguishable from random results if all teams had the same odds and you just flipped a coin for each game). This is particularly true in progressional sports where a (relatively speaking) very small number of players are selected from a fairly uniform pool for the same basic attributes. Folks dramatically underestimate the role of chance in sports, particularly with regard to “streaks”, “hot teams,” and the like.

As for the OP, here’s the long form (assuming 50/50 odds in each game, the last two posts were made while I was writing this):

Tthe odds that the other team will win in four games, given that they’ve already won the first, is 1/8. (1/2 * 1/2 * 1/2). That’s pretty straightforward.

The odds that they’ll win in exactly five games is more complicated. There are three ways they can do it: lose the second, lose the third, or lose the fourth. (Lose the fifth is the same as the case above, so we can ignore it). The odds of each of these is 1/16 (1/2 for each of the four games), and each is equally likely, so we’ve got a total of 3/16

So the odds of the other team winning in four or five games is 5/16 (1/8 + 3/16).

To win in exactly six games means they win the first (given) and last game, with two wins and two losses in between. The “middle games are:” WWLL, WLWL, WLLW, LWWL, LWLW, and LLWW with equal liklihood of 1/32 each, for a total of 6/32.

So the odds of the other team winning in 4-6 games is 5/16 + 6/32 = 16/32.

Finally, to win in exactly seven games they need to win the first and last game, with two wins distributed in between. Possibilities (again for the in-between games) are:
WWLLL, WLWLL, WLLWL, WLLLW, LWWLL, LWLWL, LWLLW, LLWWL, LLWLW, and LLLWW. (10 possibilities, 1/64 each, for a total of 10/64).

So total chance of the other team winning is 16/32 + 10/64 = 42/64. Subtract from 1 to get the odds you were looking for: 22/64 = 11/32.

I really love this site.

That latter number suggests that home ice can actually be a disadvantage. Strange.

I’m not sure that’s what it means. If you win the first game on the road (at least in the NHL), you’re generally the weaker team – that’s why you’re opening the series on the road, and not at home. So if you’re on the road to open, you’re the weaker team, and generally not expected to take the series. But if you win game 1, you increase your chances of winning the series to about 55%. I think that’s what the stat meant.

But you’ve reversed the other team’s (purported) home ice advantage, and still win the series less often than if you’d lost Game 1. That’s what’s strange.

The 56% figure, being over 50%, does imply that the visiting team is a somewhat better playoff team than the home team, anyway. A better record in a long regular season does not necessarily mean the team is better, btw - a veteran squad will sometimes coast through most of it, then crank up their efforts when the games start to actually matter.

I don’t understand what you’re saying. According to Garfield’s numbers, if you win the first game on the road, you win the series 56% of the time. If you lose the first game on the road, you win the series only 21% of the time (100% - 79%).

Independence and a winning percentage of > 50% are not the same thing. one team could have any winning percentage and a win or loss in any game could (or could not) be independent of other games’ results.

In addition for baseball over 162 game season a team with winning 94 games (.580 winning percentage) has a 95% confidence interval for wins of 81.4 to 106.6 so we could conclude they were indeed better than average with reasonable reliability. The best team in each league almost always has this many wins. Of course it would be much easier to conclude they were better than the worst team and much harder to conclude they were better than the other teams they’d face in the playoffs.

This is much harder in Basketball and hockey since the season is half as long (though the best teams usually do have much higher winning percentages). It is virtually impossible in football. I would suspect, however, that most sports fans think it is easiest to pick winners in football playoffs and hardest in baseball.

Which is less than if you win it at home.

If you win Game 1 on the road, you now have home ice advantage. If you win Game 1 at home, you still have home ice advantage, but you’re somehow not as likely to win the series. Why? (“Home ice advantage” either way, here, means you don’t need to win any more games on the road, while the losing team does.)

The reason is that the team that opens on the road will be the inferior team in most cases.