Which Friend is the friendliest?

Slate coded Friends transcripts into a database and produced an interactive display of which of the friends shared scenes together.

According to Slate's results, each friend had about equal screen time. But I wanted to ask which friend appeared in the most different combinations, a.k.a., which friend was the "friendliest"?

To start with, Joey lost 1 point for being the friend who appeared in the most scenes alone. Ross also lost a point for being the friend most commonly excluded (that is, the Chandler-Joey-Monica-Phoebe-Rachel fivesome was the most common fivesome).

Next, I looked at the dyads. Given that Joey was in a dyad on screen, Chandler was his most common companion (so Joey gives one vote to Chandler). These are not always symmetric: given that Chandler was on screen, Monica was his most common companion (so Chandler gives one vote to Monica). There was only one tie: Phoebe was equally likely to pick Monica or Rachel.

Things get a bit more complicated with triads. I started with a given dyad. Given that Chandler and Joey were on screen in a triad, Ross was the most common third member, which makes sense. Given that Monica and Phoebe were on screen in a triad, Rachel was the most common third member.

Quads were more complicated still. Given that Monica, Phoebe, and Rachel were on screen in a quad, Ross was the most common fourth member. Likewise for the quints.

And the final results are:

How to square this with the fact that each friend had approximately equal screen time? The reason is that some friends tended to dominate a few groupings: Ross was Rachel's first pick for a dyad by a wide margin; Ross was Joey and Chandler's pick for a triad by a wide margin; and Ross was the most common guy picked to join Monica, Phoebe, and Rachel. He accumulated his screen time in just a few other groupings, but none as often as these.

Other friends spent time in a much more distributed way. Monica and Chandler appeared in all sorts of different groupings frequently. And in my book, that's a reasonable enough way to measure "friendliness."

Standard debate tournament calculations

I thought it might be useful to post a list of standard probability/counting methods for debate tournaments. I list them in order of rapidity of growth -- the last listed expands most quickly, but they all increase without bound. For all of these, there a n teams at the tournament, and n is an even number.

First: once the teams have been separated into aff and neg pools, how many possible matches are there to consider?


Second: how many possible matches are there at a tournament, period? If you don't consider sides, then there are


possible matches. If you did consider sides (A aff vs. B is counted as different than B aff vs. A), then are twice as many possible matches.

Third: how many ways are there to separate the teams into aff and neg pools?


Fourth: once the teams have been separated into aff and neg pools, how many complete (perfect) pairings are possible?


This is the fastest growing number of all.

Pascal's triangle: a neat proof

There is an interesting question about how the terms in Pascal's triangle grow. All the terms in a row obviously grow (except the 1s at the extreme left- and right-hand sides of the triangle), but the rows' totals obviously grow, too. Do any of the terms in a row converge, as a percentage of the total of the row?

The most interesting example is the middle term in a row, e.g.:

1
1 1
1 2 1

the 1 and 2 are middle terms for their respective rows. The middle term is always the largest term and grows quickly. So, which grows faster: the row total or its middle term, or is there a convergence? A few examples illustrate what happens:

The 0th row -- 100%
The 2nd row -- 50%

The numerator is the formula for the middle term (see the binomial distribution); the denominator is the row's total. The largest row that my TI-84 calculator could handle before overflowing is the 332nd. As the examples show, the denominator is "winning" -- the middle term as a percentage of the total is steadily decreasing. Pretty amazing considering just how big the middle term is; for the 332nd row, the numerator is 3.8E98!

Can we write a proof of this? Certainly. The first step is to re-write the numerator. Given that the combination is really a simplification of


so, our specific numerator can be re-written as:


To re-write the limit for our whole expression, the middle term divided by the row total, would look like this:

Now the clever step. Let's re-organize that n factorial in the numerator and expand the terms in the denominator:

All the n factorial terms are there; they just been re-organized, with the even terms on one side and the odd terms on the other. It should make sense that the terms in the numerator "sit" perfectly in the "seats" provided by the denominator: there are n terms in the numerator, and there are two groups of n/2 terms in the denominator.

The proof is basically done. One simple re-write makes the final conclusion clear:

Each fraction is now on its own. The first set of fractions are all 1s. Every other fraction must be less than 1. An infinite multiplication of fractions less than or equal to 1 must converge to 0. So, in an infinite Pascal's triangle, the middle term, although it increases without bound, is 0% of the row's total.

---

The above paragraph is incorrect. A lot more work must be done to show that the product of the remaining fractions


goes to zero. I like explanations that just make intuitive sense. So, the first term is 0.5. I noticed that the product of the next four terms (terms two to five) is slightly less than 0.5. A little cancelation helps make why clear:


Thus:


And this continues. The product of the next sixteen terms (terms six to twenty-one) is also less than 0.5. Here are the terms:


A litte cancelation starts the process:


and reorganizing the denominators makes the result clear:


Each pair of terms is slightly less than one. Each pair of terms can be written in the form:


or


The product of the next 64 terms is also less than 0.5. (The number of terms is based on powers of 4.) So, the whole product can be rewritten as 0.5^infinity... that definitely goes to zero.