Of all the notions of advanced mathematics, one of the easiest to grasp is the Pigeonhole Principle:
If n pigeons are placed into m pigeonholes with and n is a number greater than m, then at least one pigeonhole must contain more than one pigeon.
This principle is the equivalent to the ‘musical chairs’ of math, where the pigeons are the children and the pigeonholes are the chairs. When there are more kids than chairs, one of those chairs is going to have multiple kids aiming to sit in it. (Belated warning: MATH ALERT.)
One can use the Pigeonhole Principle to prove all sorts of clever and/or useless things. For instance, it has to be the case that two people from Milwaukee have the same number of hairs on their head. A safe estimate for the maximum number of hair follicles on a human head is 500,000, and according to the 2010 U.S. Census, the City of Milwaukee has a population of 594,833. Since there are more people than allowable follicle counts, it’s a lock that two of them have the same number of hairs.
Or take the case of a theoretical 2013-14 BMO Harris Bradley Center sellout. In a packed house of 18,717 fans, at least two of them would be guaranteed to have the same initials. Why? Because there are 26 letters in the alphabet, and as a result, there are 26 x 26 x 26 = 17,576 possible three-letter sets of initials. Since there would be more people than allowable letter combinations, at least two would be guaranteed to be the same.
(The same cannot be said for a crowd of 11,016. That tally from Monday’s loss to Boston marked the lowest home attendance figure at the BMOHBC since the team moved over from the MECCA in 1988. While there may have been a matched monogram — and more than likely there was a match — the numbers alone don’t guarantee it.)
With just a handful of days remaining before the NBA’s trade deadline, the Bucks have something of a pigeonhole problem looming.
The NBA’s maximum roster size is 15.
1) Their own first-round pick: The Bucks are 5.0 games behind Philadelphia (and 6 behind in the win column). If they can stave off the hard-tanking 76ers for last place in the NBA’s cellar, then they’ll have a 25.0% chance at the overall #1 pick in the draft. Milwaukee will also have 21.5% and 17.8% chances at the #2 and #3 slots respectively. Their most likely spot would still be the #4 pick — with a 35.7% likelihood.
Much has already been said, and much more will be said in the future, about the best draft opportunity the Bucks have had in years, so let’s move onto their other picks.
2) Their own second-round pick: Assuming they finish in last place, the Bucks should get a functional piece here. From 2007 to 2012, the #31 picks were Carl Landry, Nikola Pekovic, Jeff (Pendergraph) Ayres, Tibor Pleiss, Bojan Bogdanovich, and Jeffery Taylor. While Pleiss and Bogdanovich haven’t played in the NBA yet, the other four have worked their way into NBA rotations (and Bogdanovich probably could too if he eschewed Europe for the NBA).
3) The Lakers’ second-round pick: At the moment, this pick would be the #36 overall selection. But the sinking Lakers, who started out the season with a 7-7 record, have gone 11-27 since — and racked up an injury list nearly as long as Milwaukee’s (Steve Nash, Kobe Bryant, Jordan Farmar, Pau Gasol, Nick Young, Xavier Henry, Jodie Meeks).
The Lakers have ripe incentive to tank, too, since they still have their 2014 first-round pick and have already traded their 2015 first-rounder to Phoenix. There’s a strong chance this pick gets a little sweeter by April.
4) The Raptors’ second-round pick: The Bucks acquired this protected pick when they traded Luc Mbah a Moute to the Kings. In return, the Bucks get the less favorable of the Kings’ second-round picks: Sacramento’s (which is protected from #56-#60), and the Raptors’ pick (which is protected from #31 to #36).
With the Kings sitting in the bottom of the Western Conference standings and the Raptors pushing for the #3 seed in the East (!), the Bucks will almost assuredly get Toronto’s pick. As of today, that pick would be the #49 overall pick.
5)-16) The 12 current players under contract for 2014-15: It is likely that the Bucks only will have three contracts that come off the books next season: Caron Butler, Ekpe Udoh, and Luke Ridnour. Unrestricted free agents-to-be Butler and Ridnour almost assuredly won’t return. The same can be said of Ekpe Udoh, who will be a restricted free agent. But that still leaves 12 contracts on the books, if you include the non-guaranteed deal of Khris Middleton, who has been one of the team’s few bright spots this season.
So the Bucks have 16 “pigeons” headed for 15 pigeonholes. If they make it to training camp October with 16 pigeons, they won’t be able to keep them all for the start of the regular season.
There are myriad ways for general manager John Hammond to deal with this issue. He could make a 2-for-1 trade at the deadline. He could trade away a player now for the rights to an already drafted player (who isn’t actually NBA bound) in a salary dump. He could use a second-rounder to secure the rights to an overseas player who won’t make it to the NBA for another year or two. He could make a 2-for-1 trade later on down the line, or trade away a pick on Draft Day for a pick in the future. A second round pick could fail to make the team. The possibilities are nearly endless.
There are too many ways to list that deal with the extra body count. However, the impending roster jam seems to make it just that much more likely that Hammond makes a move in the eight days leading up to the NBA’s trade deadline.