I love the office pool (even though I don't work in an office and the school I work at doesn't do a pool). There's the classic pool where you get a certain number of points for each correct pick and the points awarded increase by some amount. There are lots of alternatives. Some involve giving greater weights to upsets. One of my favorites is for you to pick five players as your tournament team. You get as many points as they score in the tournament (which means that you want to pick players who can score and who play on teams that will do well and therefore play more games). Heck, you could also run a tournament where you try to pick the losers making the first round the most important round (one pool I was a part of for a number of years gave second to last place 10% of the winnings).
Anyway, speaking of mathematics there are some great mathematical questions here: a classic What Can You Do With This?
- How many people would need to be in your office to guarantee a perfect bracket (a nice counting problem)?
- How many total games are played (a nice counting problem that has a very clever solution involving no counting)?
- Using the Vegas odds of each team winning the tournament, what's the most likely bracket outcome? What's the probability of this being the actual outcome?
- And finally...how many hours of basketball am I going to be watching over the next month?
For the first question, and an unspecified number of complete (2^n) rounds, I am getting 2^(n(n+1)/2) - not what I expected. The partial round is a minor complication.
ReplyDeleteMy own favorite odd variant, but no one will put it in a pool - big loser. Not 2nd place. Not the worst bracket. But selecting the team that loses in the first round to a team that loses in the second to the team that loses in the third... If wins followed seedings, 9 seeds would be top candidates. But it rarely works that way.
Jonathan
JD: Do you mean 2^n rounds or 2^n teams with n rounds? I think you have some "off by 1" issues going on. If n=1, you have 2^1 rounds. 2 rounds mean there are 4 teams. There are 8 different brackets you could make with 8 teams. Your expression gives us just 2.
ReplyDeleteI *love* your variant. Seems hard to pick. I've thought about it a bit and I think that this means:
a)You'd have to pick the correct winner of the entire tourney
b)there is exactly one "big loser" path for any tournament. no more. no less.
c)picking a 9 seed wouldn't be the best choice is the higher seed won every game because the 9 seed would lose to the 8 seed which would lose to the 1 seed, which would be favored to win the next game.
d)If you expected higher seeds to win every game, you'd pick an 11 seed (which would lose to a 6 seed, the 6 seed would then lose to a 3 seed, the 3 seed would lose to a 2 seed, and the 2 seed would lose to a 1 seed.
Related to Avery's comment: this is why 11 seeds do so well. The "538" guy had a great piece on this, backed up with data... and suddenly an 11 seed runs into the Final Four, again.
ReplyDeletehttp://fivethirtyeight.blogs.nytimes.com/2011/03/15/when-15th-is-better-than-8th-the-math-shows-the-bracket-is-backward/
I always look for the Big Loser at the end of a bracket tournament. Been doing it for a long time and so glad I'm not the only sicko.
This year's Big Loser: #4 seed Louisville.
ReplyDelete