Bonus puzzle -- World Series

Hint

Alex Y

I better give out a hint fast, since it looks like the real thing is about over (top of the 7th in game 6 as I type this).

No computations are needed to answer this; general reasoning will suffice.

Re: Bonus puzzle -- World Series

Larry Barrett

I would rank these A, E, D, C, B; A is the fairest with probability .522 for the team that plays 4 at home. B, the NBA plan, is the least fair, with a probability .610 for the team that plays 4 at home.

The prob that I have are:

A .522

E .527

D .529

C .580

B .611

  I don't think so

Alex Y

Your calculations do not conform to my general reasoning solution. Why don't you post details on a couple of your calculations (D and E might be easiest) and I will see if they poke a hole in my reasoning, or if I disagree with the calculation.

Anybody else with a general reasoning approach to this?

Or can any of you computer gurus run a little monte carlo simulation?

Re: Hint

david weaver

Reasoning would make it so that the fairest of the scenarios would be the one that has the longest expected play....as in, the fairer the scenario, the longer team 1 would be expected to play (team 2 is playing if team 1 is, so it needs to be done only for team 1).

The fairest scenario is probably going to be the one that averages a probability closest to 0.5^n, (after each game, the probability average is as close to 0.5^n).

Because of that, on the surface just by reasoning, I would want the series that alternates back and forth the most evenly.

I would choose B without calculation because of that. I'm following the rules by not doing calculations for now.

Re: Hint

Alex Y

David,

You bring up some good alternative criteria, but they do not equate with fairness to the two teams.

The logic leap I was looking for is to imagine that the full seven-game series is played out. A team will win that seven-game series if and only if it is the first one to reach four games. In real life, they would not play the remaining games (to empty stands and risking injury), but the equivalence still stands. And if they play four games at one field and three games at the other, Team A's chances of winning 4,5,6,or7 games are identical no matter the order of their four home and three away games.

I found this equivalence hard to accept at first. But for doubting Thomases, I have a spreadsheet that Larry prepared calculating the odds. I modified the sheet to allow different home game orders. If you'd like to see that to convince yourself that the odds are the same independent of order, just drop me an email and I will try to hunt up that file.

Your post brings up some other criteria that likely are NOT indifferent to order:

* Maximize length of series (I guess that this would require the favored team to get the home field for games 5,6,and 7)

*Fairness to the fans of each city (alternating each game or HHAAHAH?)

*Others?

Re: Hint

david weaver

I resisted the urge to run combinatorics exercises and figure out what the actual answer was.

One thing I remember doing in college was that every time a roommate and I would work something out that was probability related, his brother would actually code a simulation and run it to see if it worked out in practice.

Or at least simulated pratice.

If I was good at coding anything, this one would be pretty easy to code in basic and run a million times or so to get all kinds of things.

Thinking back on fairness now, you're right, I was too narrow in thinking I should just stretch it out to 7 games, it's perfectly fair if both teams have an equal chance of winning the series in 4, too.

I'm not as smart as I once was!