Talk about TireIron starting out at PF just because of familiarity with
that position but later as he gets used to NBA game, "playing out on
the floor more than inside ultimately" according to Skiles.

Which is interesting because that puts him in the
Noc/Denger/Grif/Khryapa group of small forwards at some point. Kinda
crowded.

Which I think means Noc is going to get a fair amount of time at PF this year.
Guess he'd be out there with either "Body" or "PJ" playing C.

There's been a lot of talk about taking time to get all the new players
on the same page with leftovers and each other. And the two rookies
getting up to speed. And the extreme schedule out of the gate. All of
which are tough problems.

But ya know, I think the bigger problem is actually Skiles figuring out
a rotation because he has so very many possible lineups to run out
there. Pax was right that all 12 "could be" in the rotation.

Mathematically, how many combinations of five can be made out of these
12? And maybe a few on the inactives too. Gotta throw out some because
"Body Ben" is not going to be playing point with "Jets Ben" playing C.

In article <2006100420063843658-otherwise3@maccom>,
BoneDry <otherwise3@mac.com> wrote:



You just had to ask that... some guys in this group live for questions
like that...

In alt.sports.basketball.nba.chicago-bulls on Thu, 05 Oct 2006 02:14:37 GMT
Granville Waiters' Ghost wrote:


12 choose 5 = 12! / (5! * 7!) = 792.

There's a way to do this with Pascal's Triangle, but computers have made a
lazy fuck out of me.

In real life, there's more possible combinations, since most teams go
through more than 15 players in a season. Also in real life, teams use about
400 different lineup combinations, maybe 100 combinations who play more than
5 minutes together.


Why, I never!

Fistpout Trebuchet wrote:

I consider myself a casual fan of mathematics more than an actual
participant. But I love any equation with exclamation points.

12! 5! 7! FIGHT! GO!

In article <p5s8i2d8aojplak4gp8lbn2i2u3uvrpr4m@4ax.com>,
Fistpout Trebuchet <humehnity@example.com> wrote:




Puvyrna, V xarj lbh pbhyqa'g yrnir hf.

Granville Waiters' Ghost wrote:

Hey, I resemble that!

Depends on what the meaning of the 'combination' is. (Say that in your
best Bill Clinton.)

Call a 'group' a collection of 5 Bulls players. Call a lineup a group
of 5 Bulls players that have been assigned to the positions PG, SG, SF,
PF, and C. There are many ways to assign a group of 5 Bulls players to
a lineup. The exact number is 5!=120. Of course, some of these lineups
would be insane (e.g., with Ben Wallace playing PG and Duhon C).

If you consider a 'combination' a particular group of 5 players (that
Skiles assigns the most appropriate lineup), then the answer is
12!/(5!7!)=792.

If you consider a 'combination' a particular lineup of 5 players, then
the answer is 12!/7!=95040.

Mathematically, the distinction is between 'combinations' and
'permutations'. For more information:
http://mathforum.org/dr.math/faq/faq.comb.perm.html

Marc Heiden wrote:

At my alma mater, at football games, the band would stand at 3:14 and
yell "pi!".

At 3:18, they would stand and yell "e!". (Think about that for a
moment.)

In fact, my freshman year, to encourage more students to attend
football games, they announced that everyone would receive a free 3.5"
floppy disk, which at the time was worth about $1. It made Sports
Illustrated, and not in a good way. Alas, I no longer have my disk.

On 2006-10-04 21:12:40 -0600, "sv0f" <sashankvarma@yahoo.com> said:


Can I rephrase the question?

With 5 different given positions as commonly labeled used for all lineups,
and a pool of 15 players available,
10 of which can be used for two given positions only,
5 of which can be used for one given position only,
how many unique lineups could Skiles produce?

In alt.sports.basketball.nba.chicago-bulls on Wed, 4 Oct 2006 22:12:04 -0600
BoneDry wrote:


Noooooooooo!


Okay, this has now officially crossed the line between complicated but
interesting to complicated but mindfucking.

Your description left out a pertinent parameter: the distribution of primary
and secondary positions among the 15 players. I'm going to make it easy and
assume that positions are distributed evenly:

Player PrimaryPos SecondaryPos
1 1 -
2 1 2
3 1 2
4 1 -
5 2 1
6 2 3
7 3 -
8 3 2
9 3 4
10 4 5
11 4 3
12 5 -
13 5 -
14 5 -
15 5 4

This gives us an equal number of players who can play positions 1 through 5:
five different players can play each position.

Ignoring your conditions for the moment, the max number of lineups possible
from a 15 player roster is 15 choose 5 = 15!/(5! * 10!) = 3003. That is our
ceiling.

Let's choose a lineup: Player 1 at the 1. For the 2, we have our choice of
five players, let's take Player 2. At the 3, we have five players, let's
take player 6. At the 4, we have our choice of 5 players, let's take Player
11. For the 5, we have our choice of 5 players -- I'll take Player 15.

You may have noticed that for all of these positions we had our choice of 5
players. The number of possible combinations is therefore 5^5 = 3125, which
is more than the number of possibilities calculated earlier (3003). Why is
that? Because in some cases, we'll have a choice of only four players: say
we take Player 2 for the PG spot -- we only have a choice of 4 player for
the 2 now.

None of this gets us closer to answering your question. What it does is set
an upper limit of 3003 possibilities, regardless of my assumptions about the
distribution of positions among players. To calculate the actual number of
possibilities, which will be something less than 3003, you'd need some kind
of Markov chain or something -- or maybe a simulation would help you
eliminate the impossibilities. In either case, it's beyond my abilities. My
guess is that if you use my assumptions, you'll get an answer between 2800
and 3000. If you assume a different distribution of positions -- say, you're
overloaded at the guard and thin up front -- the number of possibilities
declines in proportion to the imbalance of your lineup. If your roster
contains only one player who can play center, your possibilities max out at
14 choose 4 = 1001.

Fistpout Trebuchet wrote:

I remember Hinrich and Piatowski playing power forward over the last
two years, but did anyone ludicrous ever make it to center? My
recollection is that Skiles kept the center spot relatively
conventional (stay golden, Dali-boy) even when surrounded by four point
guards.