At this posting, the Giants are up 2 - 0 in the World Series. Reports say that historically, the team that is up 2 - 0 has gone on to win the World Series 79% of the time.

Suppose that the games remaining to be played were decided by a 50 - 50 coin toss. If this were the case, what would be the probability that the Giants will win two of them before the Tigers win four of them? IOW, in this case, what would be the odds that the Giants would win the World Series? Please show your thinking.

Sadly, sports do not follow any of the mechanics of random probabilities. The 79% statistic is purely empirical, and if you were to decide the outcome of the next two games via a coin instead of actual play, it would completely change the psychological and emotional momentum of the contest.

If you want to define a sport in which teams of differing skill levels have a fixed probability to win any given game, then state your problem as such. Real sports, however, don't work that way.

Originally posted by forkedknight
Sadly, sports do not follow any of the mechanics of random probabilities. The 79% statistic is purely empirical, and if you were to decide the outcome of the next two games via a coin instead of actual play, it would completely change the psychological and emotional momentum of the contest.

If you want to define a sport in which teams of differing s ...[text shortened]... in any given game, then state your problem as such. Real sports, however, don't work that way.

I think it was a fair enough question, although to clarify how many games are left? Is it best out of 7 (ie first to 4)?

Originally posted by JS357
At this posting, the Giants are up 2 - 0 in the World Series. Reports say that historically, the team that is up 2 - 0 has gone on to win the World Series 79% of the time.

Suppose that the games remaining to be played were decided by a 50 - 50 coin toss. If this were the case, what would be the probability that the Giants will win two of them before the Tig ...[text shortened]... e, what would be the odds that the Giants would win the World Series? Please show your thinking.

Originally posted by forkedknight
Sadly, sports do not follow any of the mechanics of random probabilities. The 79% statistic is purely empirical, and if you were to decide the outcome of the next two games via a coin instead of actual play, it would completely change the psychological and emotional momentum of the contest.

If you want to define a sport in which teams of differing s ...[text shortened]... in any given game, then state your problem as such. Real sports, however, don't work that way.

Originally posted by deriver69
I think it was a fair enough question, although to clarify how many games are left? Is it best out of 7 (ie first to 4)?

Originally posted by SwissGambit
The possible outcomes:
Giants win in 4, 5, 6 or 7 games
Tigers win in 6 or 7 games

The probability of the Giants winning the Series is 2/3.

Hmm, this is why I hate probability problems. Do I need to consider that WLW and WWL transpositions are different possibilities?

There is one way the Giants can win in 4.
2 ways to win in 5.
3 ways to wi ...[text shortened]... l possibilities in which the Giants win 10 of them.

Revised probability is 10/16 = [b]5/8[/b]

Reveal Hidden Content

Originally posted by SwissGambit
The possible outcomes:
Giants win in 4, 5, 6 or 7 games
Tigers win in 6 or 7 games

The probability of the Giants winning the Series is 2/3.

Hmm, this is why I hate probability problems. Do I need to consider that WLW and WWL transpositions are different possibilities?

There is one way the Giants can win in 4.
2 ways to win in 5.
3 ways to wi ...[text shortened]... l possibilities in which the Giants win 10 of them.

Revised probability is 10/16 = [b]5/8[/b]

Reveal Hidden Content

I would go for B(5,0.5) distribution and look for P(X>=4) as 4 wins and a loss or 5 wins would do it.

1-5C4*(.5^5)+5C5*(.5^5) = 1-6*.5^5

Originally posted by forkedknight
You are correct about the number of ways to win in 4,5,6, or 7 games, but not about the probabilities of each.

There is a 25% chance of any given ordering of two games
12.5% for 3 games
6.25% for 4 games
3.125% for 5 games

Therefore
[hidden]25% + 2 * 12.5% + 3 * 6.25% + 4 * 3.125% = 81.25% chance of winning for the Giants[/hidden]

Originally posted by forkedknight
You are correct about the number of ways to win in 4,5,6, or 7 games, but not about the probabilities of each.

There is a 25% chance of any given ordering of two games
12.5% for 3 games
6.25% for 4 games
3.125% for 5 games

Therefore
[hidden]25% + 2 * 12.5% + 3 * 6.25% + 4 * 3.125% = 81.25% chance of winning for the Giants[/hidden]

Originally posted by tomtom232
You said the 79% is purely empirical yet your answer is quite close. We'll see after another couple hundred years of baseball.