The last one.

Clever answer, Jay, but we need a number. Game 1, 2, 3, 4, 5, 6 or 7.

Craig, I'm going to ask for one more hint. Can the answer be found in math/logic, or does one have to know the historical success rates of how teams fare after winning a particular game in a seven-game baseball series?

No, it's a pure math question. Consider the two teams as being exactly as good as one another, and so each has a 50/50 shot of winning each game. I forgot to mention that except in the "story".

Craig B,

Seeing as you asked, and this is thoroughly non-controversial, the answer, I think, is that it is Game 7, under which the Cubs have about 67% chance of winning the series. My instinct was that it would be any one of Games 1-4, but unless I've totally messed up the mathematics, that would increase the odds to only about 59%. But, heck Craig, you're the tax lawyer.

And how can one have a Cubs story involving probabilities with the Lord as a character, but not have Lucifer as a counterweight.

I'm no math major - but it seems to me that if you have exactly 50% chance of winning all the other 6 games and 100% chance of winning one game, then no matter what game you pick you always end up with a 65.625% chance of winning the series

Alright my wise-guy guess wasn't it. Let me try another. Without using any mathematics, I will say the choice doesn't matter so long as it is one of game 1, 2, 3 or 4. The gambler's fallacy bias tells me that one a flip of a coin, the outcome of one event has no bearing on the outcome of the next event. Given that, a win is a win, regardless of the game it falls in. My next step is to guarantee I take advantage of the gift win. The only way to do that is for it to fall in the first four games.

Now this seems to be too easy and that is why I know I am wrong. I look forward to the correct answer and hope that whoever gets it, posts their formulation.

Here are my thoughts on this one.

It's not Game 1, because that just gives you a 50% chance of winning the series (you have to win 3 of the next 6).

If it's Game 2, then you're either down 0-1 or up 1-0 already. If it's the former, then you have to win 3 of the next possible 5, a 60% burden. If you've already won Game 1, then now you need only win 2 of the next possible 5, a 40% burden. Taken together, that gives you a 50% chance of winning the Series, same as Game 1.

It's the same with Game 3. You're either down 0-2, tied 1-1 or up 2-0 going into Game 3; after winning that match, your burden is either 75% (win 3 of 4), 50% (win 2 of 4) or 25% (win just 1 more of 4). Again, this works out to 50%.

Game 4 is where it gets interesting. If you were up 3-0, then the Series is over. For the first time, you're leading (1-0) in possible Series victories. You couldn't have been swept: since you won Game 4, you can't have lost the Series 4-0. That leaves you with a burden of either 100% (you need to win 3 of 3 remaining), 67% (win 2 of 3 remaining), or 33% (win 1 of 3 remaining). This adds up, so to speak, to a 200% burden over 3 outcomes, or 66% per outcome (I know that's not mathematically correct, but it's the best I can express what I think to be the odds here). But this is still pretty good, because although your remaining possibilities require a must-win on average 67% of the time, at least you have one Series victory in the bank.

Now there's Game 5. Going into it, you were either down 3 games to 1, tied 2-2 or ahead 3-1. These are the only possibilities; if the previous sequence was 0-4 or 4-0, the Series is over. Since the odds of sweeping or getting swept in the first 4 are equal, they cancel each other out and we're not worried about them. Your burden going forward, after winning Game 5, is either 100% (win 2 of 2 remaining), 50% (win 1 of 2 remaining) or you've won the Series. So once again, you're ahead 2-1 in possible Series outcomes -- either you lost 0-4, or you won 4-0 or 4-1. The one result that is now impossible is losing 4-1; you can't have lost the Series 4-1 if you won Game 5. So you're up 2-1 in possible Series victories, which is great. But at the same time, you also face a tougher burden in the remaining possibilities: you're either down 3-2 in games or ahead 3-2 in games, meaning that you have two options: win both (100%)or win just 1 of 2 (50%). So your burden is 75%, which is worse than had you won Game 4. Your possible Series advantage is in your favour, 2-1, but you face a higher burden (75%) than if you'd won Game 4 (67%).

Now there's Game 6. If you're still playing at Game 6, you were either up 3-2 or down 3-2 in games. So either you win Game 6 and win the Series, or you win Game 6 and force a 7th game, which is a must-win. So now you're still doing well, because you lead in possible Series results 3-2 (at this point, the results of possible completed Series are 0-4, 1-4, 4-0, 4-1, 4-2; you cannot have lost the Series 4-2 because you won Game 6). But your burden is now 100%: there's only one possible game left, and you must win it, so it's a 100% burden. This is less attractive than Game 4 or Game 5.

Finally, we come to Game 7. if you're playing this game, that means everything that has come before you is equal. The six possible completed Series up till now have gone 0-4, 1-4, 2-4, 4-0, 4-1 and 4-2. You're dead even. But here you are in Game 7, and you won it. This means that your possible Series results are now 4-3; what's more, this result is guaranteed; there are no more possibilities going forward. You will come out on top, because you've guaranteed a victory in the only possibile tie-breaking game.

This is better than winning Game 4, which would have given you a one-game lead in possible Series victories but also left you with a healthy two-thirds burden of having to win the remaining games. But winning Game 7 has no strings attached; you're up by one game in possible Series victories and there are no more games left to play. Guaranteeing a Game 7 victory is the only way to guarantee that no matter what, the odds are substantially better than even that you'll win the Series.

So my choice is, Game 7.

I have to go with Game 1. (Though I suppose 2-4 would work out the same, I can't ignore the mental advantage of being ahead)

If it is guaranteed that you win game 1, then you have a 1-8 chance of winning in four games, and a 0-8 chance of losing in four.

IF the game goes to 5 games, you have a 3-14 chance of winning the series in 5, and only a 1-14 chance of losing the series in 5.

IF the game goes to 6 games you have a 6-20 chance of winning the series in 6, and a 4-20 chance of losing the series in 6.

IF the game goes to 7, you (obviously) have even odds, 5-10 of winning or losing.

Your odds worsen the deeper into the series you go. To some that would support the idea that you should get Game 7 guaranteed, but I think that is wrong. If you have no guaranteed games, with an equally matched team, your odds start even, and adjust as the series advances. To get to game seven, you have already had to "win the coin toss" and survive to game seven. By winning game 1, you have improved your odds at EVERY STAGE, except game 7. Since Game 7 only comes around every (31%?? Something like that) of the time, I believe it would behoove you to inprove your odds of winning the series at 4-6 games instead.

Jordan...you left out the probability of actually playing games 5-7. I think (but am not positive) that will affect the odds enough to eliminate game 7. Anxious to hear the results.

http://economics.about.com
It's not Game 1, because that just gives you a 50% chance of winning the series (you have to win 3 of the next 6).

I'm not going to answer Craig's riddle, but do you really think there's *no* value in winning the first game?

Mike

Mike is right. Winning Game 1 does not give you a 50% chance of winning, because you don't play all six other games if the series ends early.

Based on my math the chance of winning the series (assuming evenly matched teams) after winning game 1 is 65.6245%

Let me double check to find the chance of losing...34.3725%

There is some rounding there so I might have blown some minor nimbuers, but nothing to change the real outcome.

You want to win game 1.

To begin with, there are 128 possible outcomes (including series where teams win 5, 6 or 7 games). Choosing any of games 5, 6 or 7 does not preclude the series from ending before reaching those games, so you do not want to choose any of those games.

By choosing any of games 1, 2, 3 or 4, you narrow the possible outcomes to 64. From there, you have a 65.625% chance of winning the series (42 of 64 outcomes gives you at least 3 more wins). But by winning game 1, you lock in a 65.625% chance of winning.

By winning games 2, 3 or 4, you are not controlling the events of the previous games. So you end up betting that you will win at least a fixed amount of the previous games (1 for games 2 and 3, and 2 for game 4) in order to give the team a better chance of winning the series.

So winning game 1 gives you the best chance of winning the series without knowing the other outcomes.

Jason

It doesn't matter what game you pick.
The odds change from 50% to 65.625% chance of winning.

For example, the odds of playing a game 7 are about 31.25%
Therefore your chances of winning the series are:

(1-31.25%) * 50% (6 or less games) + 31.25% * 100% = 65.625%

It's pretty easy to make a little spread sheet to figure
this out.

Rob

From that reknowned mathematician Don Cherry:

"Game Four is the most important game in a seven game series. Whoever wins Game Four usually wins the series"

If the outcome of each game is independent of the previous games' outcomes (with the obvious exception that games 5, 6, and 7 may not be played at all), then how could it matter which of games 1-4 one picks?

Logic aside, I would pick #3, to maximize the chances of being up 2 games to 1, because this seems like good "momentum". This may be complete superstition, however.

I agree with you gid, there is no difference between games 1-4, I just picked 1 for the same reason you picked 3: momentum.

I am trying to run the numbers on what Rob Bateman posted. I think he is right, but I am trying to make sure of it...

Jordan,

Your logic excluded the forward burden should you win the series, in other words, from game 4 - 7 you should have included a 0% burden for a series win, thereby giving an average burden of 50% for games 4 - 6.

The bottom line is still the same though, winning game 7 leaves you with 0% burden while losing game 7 leaves you with...

You're missing the obvious...if the teams are exactly even, the series would go to a seventh game...so game 7 is what he should choose.

Rob Bateman is probably right. I checked his number for Game 7 (I had remembered 3-3 splits in bridge as 34%, not 31.25%). I'm guessing that he's go the numbers right for 1-6, and I made a mathematical error.

Incidentally, Rob, the odds of a Game 7 are precisely 31.25% (3-3 splits occur precisely 1/2 of the rate that 3-2 splits do, which is 5/8 or 62.5%).

Here's my friend's take. His opinion is kind of between those of Jason Robar and Rob Bateman. Bottom line: Any of the first four games, because you know they will certainly be played.
________________________________

If there are n games, with a probability p for success, then the probability that the Cubs will win EXACTLY k of these
n games is given by

P(X = k) = (n choose k) * (p^k) * (1 - p)^(n-k)

In this case, p = 0.5.

For X (the number of games the Cubs win)=0, n=4., and Probability is = 4C0 (.5^4) * (.5^0)

= .0625

For X=1, n=5, P = (5C1)-1 (-1 because we want to exclude the case where the other team wins 4 and then Cubs win 1) (.5^5) = .03215 * 4

= .125

For X=2, n=6, P = .15625

For X=3, n=7, P = .15625 (!)

So if both teams are even, there is an even likelihood that a team will win in six as that a team will win in 7. (you will also note my math here is correct, as these probabilities add up to .5.

If we fix a game (say the first) then the P(lose the next 4) = .0625; P(lose 4 out of next 5) = .125; P(lose 3 of next 6) = .15625

This should hold true if we fix any of the first four games. So, if we fix game 1, 2, 3 or 4, P (Cubs lose) = .34375. P(Cubs win)=.65625

However, P(Cubs make it take game 5) = 1-(.0625*2) or .875. The Cubs at that point will have either 2 wins and 3 losses, or 3 wins and 2 losses after game five. They will have two games left to play.
P(Cubs lose in 6 or 7, after making it to the fixed game five) = .875 [(.5)(.75) + (.5)(.25)] = .4375. P(Cubs win) = .5625

I we fix game 6, P(Cubs don't make it there) = .375.

P(Cubs make it to game 6) = .625. Cubs had to have won at least 2 games prior to game six. They either won 2 and lost 3, or won 3 and lost 2. So there is a .3125 chance of winning in 6, or a .3125 chance of being tied at 6. The chances of winning game 7 after that are .5. So P(Cubs winning in six) = .3125+(.3125*.5)=.46875.

If we fix game seven, P(Cubs not making it there) = .6875
P(Cubs making it) = .3125. If they make it to game seven, they win. So they have a .3125 chance of winning.

So, I think the best chance of winning is fixing game 1, 2, 3 or 4.

I made a couple of transcription errors. First, in the paragraph that begins "If we fix a game...," the probability given is for the Cubs losing 4 of the next 6, not 3 of the next 6.

Later, it should read "If we fix game 6..."

Mike Green:

"Incidentally, Rob, the odds of a Game 7 are precisely 31.25% (3-3 splits occur precisely 1/2 of the rate that 3-2 splits do, which is 5/8 or 62.5%). "

That was sloppy typing on my part.
I originally had about 31%, then I changed to the real number.

Rob

This has been most entertaining. A great deal of skull sweat and ingenuity has been displayed.

Mike Moffatt, who I originally enjoined from posting here because I knew he'd get it right (even though he actually got one *wrong* in the Primer thread that inspired this question) did indeed e-mail me with the correct answer. I call on Mike to demonstrate (if he would be so kind) the right answer, which is...

1, 2, 3, 4, 5, 6 or 7... they're all the same. It doesn't matter which game he picks, and the chances of the Cubs winning is 21 in 32 - 65.6%.

Originally, I also thought when I was first formulating the question that any of 1, 2, 3 or 4 would be OK, but not 5, 6, or 7. Then I realized I was going wrong.

Incidentally, Rob Bateman got it right.

Your logic excluded the forward burden...

My logic excludes many things. That's why I employ my own brand so often. :-) In the result (lousy trick question), this is why I was an English major.

To a friend of Mike D's,
Your calculations and Rob Bateman's both show that the probability of the Cubs winning if any of Games 1-4 were 65.625%. Rob Bateman and I both have the probability of the Cubs winning if Game 7 was fixed also at 65.625%. What do you have the Cubs probability if Game 7 was fixed? Your post seems to say 31.25%, but I'm not sure that's what you mean.

Just to take it away from pure Math for a second...

Are they starting at home or on the road? (What is their Winning% at home for the last three years?)

Do they have a dominant starting 3 in the rotation? (a la Cubs 2003)

If Wood is starting game one (on full rest), I go Wood/Prior/Zambrano/WIN!/Wood/Prior

I think you're looking for the Math answer though...

(To quote talking Barbie: "Math is tough!")

The Cubs answer would be game #3 - the one that Prior and Woods wouldn't be pithching.

http://economics.about.com
Thanks Craig. The Primer thing was embarrassing because I posted despite being pretty sure my answer was wrong (that's why my post there sounded somewhat hesitant). Problem was my logic was right, but I didn't look up Pascal's Triangle so I got the math wrong. I won't make that mistake again.

The easiest way to calculate the odds is to use a combination of Brut Force and Excel. There's two things you need to calculate.

1. The probability that you'll get to game 5, game 6, game 7.
2. The probability that *given* you've gotten there, what impact will winning the game have.

I'll explain how you do 1 first, and post how you do number 2 later today (unless someone wants to beat me to it)

1. The probability that you'll get to games 1 through 4 is 100%.

Take game 5. In the first 4 games the following 5 things can happen:

Cubs win 4, lose 0
Cubs win 3, lose 1
Cubs win 2, lose 2
Cubs win 1, lose 3
Cubs win 0, lose 4

You can calculate how many different ways this can happen by using Pascal's Triangle. Look for the row that has 5 digits. That's the row labeled "1 4 6 1 4". That gives us the following:

Cubs win 4, lose 0 (1 way)
Cubs win 3, lose 1 (4 ways)
Cubs win 2, lose 2 (6 ways)
Cubs win 1, lose 3 (4 ways)
Cubs win 0, lose 4 (1 way)

1+4+6+4+1 = 16, so we have 16 possible scenarios. Since each scenario is equi-probable, when can calculate the odds of being in each one by dividing the number of "ways" by 16.

Cubs win 4, lose 0 (prob = 1/16)
Cubs win 3, lose 1 (prob = 4/16)
Cubs win 2, lose 2 (prob = 6/16)
Cubs win 1, lose 3 (prob = 4/16)
Cubs win 0, lose 4 (prob = 1/16)

Note that only in the middle 3 situations we'll get to a game 5. So the probability of getting to a game 5 is: 4/16 + 6/16 + 4/16 = 14/16. Note that you can get the same figure by starting at 1 and subtracting the probabilities that you do not get to game 5.

----

GAME 6

There's a ton of different ways you can calculate the probabilities for game 6. The easiest way is to assume that game 5 is played no matter what and use Pascal's triangle again. Then you'd have:

Cubs win 5, lose 0 (prob = 1/32)
Cubs win 4, lose 1 (prob = 5/32)
Cubs win 3, lose 2 (prob =10/32)
Cubs win 2, lose 3 (prob =10/32)
Cubs win 1, lose 4 (prob = 5/32)
Cubs win 0, lose 5 (prob = 1/32)

The Cubs only play game 6 if they won 3 or 2 of the first 5 games. So the odds they get to game 6 is 10/32 + 10/32 = 20/32.

The more elegant way of doing it is to take the Game 5 probabilities and re-calculate the probabilities assuming that only the game 5 is only played if necessary, and each time won Game 5 with a probability of 1/2. That leads to the following:

Cubs win 4, lose 0 (prob = 1/16)
Cubs win 3, lose 1 (prob = 4/16) * (prob = 1/2) Cubs win game 5
Cubs win 3, lose 1 (prob = 4/16) * (prob = 1/2) Cubs lose game 5
Cubs win 2, lose 2 (prob = 6/16) * (prob = 1/2) Cubs win game 5
Cubs win 2, lose 2 (prob = 6/16) * (prob = 1/2) Cubs lose game 5
Cubs win 1, lose 3 (prob = 4/16) * (prob = 1/2) Cubs win game 5
Cubs win 1, lose 3 (prob = 4/16) * (prob = 1/2) Cubs lose game 5
Cubs win 0, lose 4 (prob = 1/16)

Re-arrange and calculate, and this yields:

Cubs win 4, lose 0 (prob = 2/32)
Cubs win 4, lose 1 (prob = 4/32)
Cubs win 3, lose 2 (prob =10/32)
Cubs win 2, lose 3 (prob =10/32)
Cubs win 1, lose 4 (prob = 4/32)
Cubs win 0, lose 4 (prob = 2/32)

So the probability that they get to game 5 is 10/32 + 10/32 = 20/32, or as the same as before. So I hope I've convinced you that the Pascal's Triangle method is easier to use for games 6 and 7.

---
You end up with the following probabilities that the games are played:

Game 1 = 100%
Game 2 = 100%
Game 3 = 100%
Game 4 = 100%
Game 5 = 87.5%
Game 6 = 62.5%
Game 7 = 31.25%

Wow.. that's really hard to explain. Anybody have any questions? :)

Cheers,

Mike

I didn't look up Pascal's Triangle so I got the math wrong

Heh. I am such a stats geek that I actually have it memorized out to eight iterations. From there, I always "grow my own" (i.e. further the triangle by adding the two diagonal numbers above a space to get the new number.)

For those not in the know... Pascal's Triangle.

I am amending my comments that Mike posted for me. I rethought my calculations. I forgot that, for instance, if the Cubs do not make it to game 7, half the time they will have WON in 6 games. Factoring in that in each of my scenarios, I agree that, mathematically, it should not matter which game is chosen to be fixed.

Many regrets.

Thanks, Craig and Mike M for the explanation and the Pascal Triangle reference,

Given that the probability is the same, there is probably a very elegant way of demonstrating this without using the brute force method. If my brain was 25 years younger, I might have had a chance at it.

http://economics.about.com
Given that the probability is the same, there is probably a very elegant way of demonstrating this without using the brute force method.

There is actually a really fast way of getting the answer. I wasn't 100% sure of the logic, so I used the brute force method to verify.

If you assume that all 7 games are played, using Pascal's triangle you can show that the odds of winning 3 (and losing 4) of 7 is 35/128. Now draw one of these at random and replace it with a win. The odds you'll get a "loss" is 4/7. So the joint prob of winning 3 and drawing a loss is 35/128 * 4/7 = 5/32 = 15.625%

These are now "wins" you wouldn't have gotten before. Since before you had a 50% chance of winning the series, now you have a 65.625%. The end result doesn't matter from which game we drew from, so the overall winning percentage is independent of the game we chose to win.

Cheers,

Mike

Where's my acknowledgement for the first correct answer? ;)

Thanks Mike M. On the streetcar home, the same answer came to me, but in a different format. Indulge me the conclusion to Craig's story:

Joey wondered which game to choose, but he just couldn't figure it out. He begged and pleaded with the Lord, and finally the Lord answered his prayer. Three angels came to Joey's assistance and whispered to him: "it doesn't matter which game you choose, the Cubs' chances are 65.625%".

Joey looked baffled and asked the Lord "how come the result is the same no matter which game I choose?" The Lord replied: "are you a failed math and philosophy student?" Joey bowed his head in shame and nodded yes. The Lord continued: "we don't answer those kinds of questions; Lucifer has all the failed math and philosophy students, and he'll tell you. But I must warn you if you ask his advice, you must bear a burden".

Visions of eternal damnation in front of him, Joey asked the Lord what the burden was. "A cheap advertising ploy" came the answer. Joey called out "well, bring on the devil then".

Lucifer arrived with an assistant in tow and let loose a hearty "Here in Hell, we make things easy". "So, you want to know how come the odds of the Cubs winning don't change no matter which game is chosen..my assistant will explain it to you". Lucifer's assistant explained it to Joey this way:

"In Hell, we play our baseball series a little differently. In a best of seven series, we play all seven games, even if one team wins the first four games, all seven games must be played. It doesn't change the outcome, but the boss particularly enjoys watching the losing team play out a series when it has no chance in winning. Sometimes, when one team complains too much, he tells the other team that he'll fix one game, and believe me when the boss knows a thing or two about fixing games.

Anyways, in our series, it's obvious that no matter which game is fixed, the chances of one team winning will be the same. It is no different in the other place. That's true whether it's a 3 game series, a 5 game series or a 129 game series."

The explanation given, the assistant told Joey that it was time for the cheap advertising ploy. The assistant disappeared and Lucifer returned. He put his arm around Joey's shoulder.
"Joey, my boy, how would you like to see some games in our park. We've got plenty of ballplayers, owners, managers, general managers, but we don't have enough fans, just a few math and philosophy students. Our franks are great, you know, really well grilled".

With that Lucifer was off, saying: "I hear the other guy's messing with my pal Ahh-nold, so I've got a date with Wesley Clark and Howard Dean"

You guys have forgotten about the homefield advantage!

http://economics.about.com
You guys have forgotten about the homefield advantage!

It's funny. For some reason the Cubs in this story (which are different than the Cubs in real life) play in a stadium with a short right field fence, but unfortunately have a lot of right-handed sluggers. Similarly the opponents play in a stadium with a short left-field porch but have a lot of lefty sluggers. Because of that, each team is more suited to their opponents ballpark which completely negates the homefield advantage. Pretty weird, eh?

Mike

Missed it, Geoff, sorry!

ALL HAIL GEOFF. HE HAD THE FIRST CORRECT ANSWER.

Yah Mike,

Except that that isn't stated in the original scenario. Therefore, I can only conclude that you've made it up out of whole cloth.

http://economics.about.com
Except that that isn't stated in the original scenario. Therefore, I can only conclude that you've made it up out of whole cloth.

Jeez, I was being sarcastic.

Actually we've fully accounted for home field advantage, seeing as we don't know who is going to have that advantage for each game.

Take an extreme case where the home team wins 90% of the time. If you think each team as likely as the other to be the home team in game 1, then each team has a 50% chance of winning. Since Bartman doesn't know who the home team will be in Game 1, it's a reasonable assumption to make that each team is equally as likely to be the home team.

Obviously if Bartman can say, "I want to pick a road game" and God lets him, that's a different kettle of fish.

Mike

Because of that, each team is more suited to their opponents ballpark which completely negates the homefield advantage.

Heh, heh. Good work, Moffatt.

If God had them play a team EXACTLY as good as them, that means they'd be mirror images of each other and therefore the first game would never end. Therefore you'd pick game one and let Bud Selig cancel the other six games due to inevitable ties (he's not about to argue with the odds God makes). Cubs win the series 1-0 and advance to the WS with a rested pitching staff.

I guess "someone" made some kind of arrangement in game four in this last world series, Fat Wells was punished as the most obvious target and here you go series for the Marlins against all odds...
you guys gave the answer too soon...
an interesting way to go from here is God answering " OK you got an answer from your math guy, now go back and choose which game Dusty wants..."

Lurker,

No. Even two teams of clones would be influenced by random chance. The random elements ensure that each game has a finite end.

http://economics.about.com
No. Even two teams of clones would be influenced by random chance. The random elements ensure that each game has a finite end.

There's no such thing. The outcome would be part of GOD'S PLAN and randomness is just a shorthand for things we cannot predict because we are not HIM and we do not know the details of HIS plan.

Seriously, I don't think there's a need for a theological debate in this forum.

Mike

This is from the opening scenario:

"... And I'll make sure that they get play a team exactly as good as they are."

The two teams are exactly as good, but they are not necessarily identical. This statement should be interpreted as meaning that the two teams have equal skill, such that over an infinite number of games they would each win 50% of the time, but without any further qualifications in regard to their composition.

Suppose Team A has two great starters and two mediocre ones, and Team B has 4 starters with very little difference between best and worst (the recent 2001 World Series comes to mind as an example of this). If Team A is the Cubs, then it makes sense to ask for a guaranteed win in games 3, 4 or 7 because the #3 and #4 starters are more likely to start those games.

The correct answer is: indeterminate.

Robert, please see the fourth comment, where I clarified the meaning of the problem. Sorry for the confusion.

I assumed a .56 to .44 home field advantage for each team. I constructed a chart based on the probability of each team winning from any given series score. Then I calculated the chances of the series reaching that series score. Using the information in the charts I derived, I considered two scenarios: 1) No divine intervention; 2) Divine intervention in one of the games.

For each of the 7 games, the chance a team that starts the series at home (designated "home" team) has of winning the series was 51.89% (and the other team - designated "road" team for the entire series - had a 48.11% chance of winning the series, of course).

For Games 1,2,6 and 7, a guaranted win for the "home" team led to a 65.76% chance of winning the series; in games 3,4 and 5, a guaranted "home" win led to 69.44% chance of wining the series. But a guaranteed win for the "road" team in games 1,2,6 and 7 led to a 65.76% chance of winning the series; and a guaranteed win for the "road" team in games 3,4, and 5 led to a 61.90% chance of winning the series.

The NET EFFECT of the intervention is to raise the "home" team's expected series wins by .1387 in games 1,2,6 and 7, and .1755 in games 3,4 and 5. For the "road" team, the net effect (for an intervention in their favour) is a gain of .1765 in games 1,2,6 and 7, and a gain of .1379 in games 3,4 and 5.

Assuming the chances of the Cubs being the home team/road team are 50/50, the expected benefit of divine intervention on the Cubs behalf is .1576 in games 1,2,6 and 7 and .1567 in games 3,4 and 5. Unless I have made a mistake in my calculations, games 1,2, 6 and 7 have more swing than the middle games.

Later today, I will assume an unrealistic HFA of .90 to .10 to see if the HFA actually has an effect or else that I've made some kind of error.

http://economics.about.com
In the case where each game is 50/50, the improvement is .15625.

You've got it such that improvement is higher for *both* the road team and the home team than it is in the "equal" case. This seems counterintuitive, so I suspect there's a problem with the math somewhere.

It shouldn't make a difference, because of the reasoning given in message 34. Since the ex-ante probabilities of winning any game is 50%, the logic shouldn't change. In fact, if the ex-ante probabilities are any number between 0 and 1, so long as they're equal the logic will hold but the value of the improvement will be different.

If the ex-ante probabilities are different, you'll always do the best by picking the game that you're least likely to win.

Mike

Mike,

I checked the math. Given an HFA of 90/10, the weighted net gain (assuming 50/50 chance of home or road) is .2956 in games 1,2,6 and 7 and only .1225 in games 3,4 and 5 in terms of expected winning percentage.

The reason is that the team that starts at home is already heavily favoured because of the huge HFA - with a 73.65% chance of winning the series without divine intervention. Giving them a guaranteed win at home has very little effect (it bumps them up to 79.56%). Giving them a win on the road makes them a near certain victor (95.70%), but that's not a huge gain in win expectancy. Conversely, the underdog is favoured to win their home games so giving them a win in games 3,4, or 5 only bumps them to 28.8% from 26.35% win expectancy.

The big deal is when the pre-series underdog wins a game on the raod (games 1,2,6 or 7). This effectively steals HFA from the pre-series favourites and raises the road team's series expected win pct from 26.35% to 79.56%. This swamps the gain made by the favourites winning a game on the road in games 3,4 or 5.

Craig,

I missed that comment first time around. Assuming some sort of HFA makes the problem more interesting, though.

http://economics.about.com
Interesting stuff. It's much easier to see the effect win the probs are .1 and .9.

I'll go over your figures today or tomorrow, but you're probably right as nothing jumps out at me as being erroneous.

Mike

http://economics.about.com
Ahh.. I see where the logic of #34 breaks down.

Suppose that the home team *always* won. If each team is equally as likely to be the home team, the ex-ante probability of winning any particular game is .5.

Now suppose we can choose which game want to win. Obviously this will only change the outcome if we win three games.

However, if we *know* we won exactly three games, then our ex-ante probabilities are no longer .50. We can know with certainty that we games 3,4,5 and lost games 1,2,6,7. So by choosing one of 1,2,6,7 we can gurantee ourselves victory.

This only works because we know that the home team always plays 1,2,6,7 and the road team plays 3,4,5. If the order of the games were random, we gain no information by knowing that we won only 3 games. Since it's not, we do gain some information from knowing that we won exactly 3 games. And if we didn't win exactly 3 games, our choice is irrelvant.

Interesting stuff Robert.

Mike

BTW, if we move the series to a neutral site, then no matter which team is favoured (as long as the odds of winning each game are the same) the choice of game doesn't matter. I.E. it doesn't matter if Team A has a 55%, 60%, 65%, 80% etc. chance of winning each game - as long as that chance is the same in EVERY game - the Cubs can ask for any of the games and end up gaining the same WRT a series win.

In the HFA scenario, the chance of Team A winning a given game is NOT constant (i.e. varies according to whether they are at home or on the road).