# World Series Poser

JS357
Posers and Puzzles 26 Oct '12 15:35
1. 26 Oct '12 15:35
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.
2. forkedknight
Defend the Universe
26 Oct '12 15:53
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.
3. 26 Oct '12 17:10
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.
Thank you for your comment.
4. 26 Oct '12 18:17
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)?
5. SwissGambit
Caninus Interruptus
26 Oct '12 18:482 edits
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.
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 win in 6.
4 ways to win in 7.

There is one way the Tigers can win in 6.
5 ways to win in 7.
----
16 total possibilities in which the Giants win 10 of them.

Revised probability is 10/16 = 5/8
6. SwissGambit
Caninus Interruptus
26 Oct '12 22:02
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.
Don't get hung up on the sports thing - he's really just asking for the probable result of some coin-flipping sequence.
7. 27 Oct '12 01:36
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)?
Yes, that was what "what would be the probability that the Giants will win two of them before the Tigers win four of them?" was meant to say.
8. 27 Oct '12 15:35
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]
I don't think that's right, but I don't want to say too much more right now. I will propose my answer for your consideration after a little while. I'll just say, I think
the World Series system where fewer than 7 games might be played, should be ignored, and the focus should be on how many ways 5 coins can be flipped such that at least 4 of them are tails. This translates to focusing on the Tigers, not the Giants.

Do not use Reply&Quote if you reply to this post, or it will display the hidden text.
9. forkedknight
Defend the Universe
27 Oct '12 18:522 edits
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]
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
25% + 2 * 12.5% + 3 * 6.25% + 4 * 3.125% = 81.25% chance of winning for the Giants
10. 27 Oct '12 19:253 edits
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
11. 28 Oct '12 03:32
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]
I have the same as you, and so, am not hiding this reply. There is an interesting implication, below.

Using the Detroit POV, these are the outcomes that win the series for Detroit, with the probabilities:

WWWW 0.5^4 = 0.0625*
LWWWW 0.5^5 = 0.03125
WLWWW 0.5^5 = 0.03125
WWLWW 0.5^5 = 0.03125
WWWLW 0.5^5 = 0.03125

Their total is 0.1875.

(*ignoring MLB rules, 0.0625 could also be arrived at by p(WWWWL) + p(WWWWW).)

One minus this is 0.8125 = 81.25% for mi Gigantes. And they just won again.

Interestingly, the reported historical record shows 0.79 (79ðŸ˜µ of 2-0 winners go on to win the series. No big difference from 81.25.% Does this mean that WS opponents are basically evenly matched, and some series just happen probabilistically to go to 2-0 at first? IOW the 2-0 winners are not really superior?
12. 05 Nov '12 22:50
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]
You said the 79% is purely empirical yet your answer is quite close. We'll see after another couple hundred years of baseball.
13. forkedknight
Defend the Universe
05 Nov '12 23:221 edit
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.
Wouldn't you expect a team that wins the first two games of the series to be better, on average, than the other team?

The way the math works out, the team that wins the first two games is less than 50/50 to win any of the future games...