Tuesday, December 01, 2009

Math would help you win The Amazing Race

SPOILER ALERT:  In The Amazing Race this week, the only team in which the players are always kind to each other -- Harlem Globetrotters "Flight Time" and "Big Easy" -- got eliminated because Big Easy didn't know math.  Here's why you should pay attention in math class, kids! 

Technically he did not need to know math--he needed to figure out how to unscramble the letters N, A, R, F, Z in a "Kafkaesque" challenge involving a bureaucratic nightmare in Prague (dozens of ringing phones, repeated filling out of forms, etc.)  (So, absent math,  you could have figured it out if you'd recognized that "Franz" is a word, but I could see not noticing that because of the whole it-might-be-in-a-different-language thing.)  The more annoying brother in the Evil Gay Brothers team told Big Easy they'd work together, but when Annoying One figured it out, he would only tell Big that the word started with an F. 

What a douchebag!  But, OK, a word that starts with an F, and has four other unique letters.  Combinatorics, my favorite branch of math from my olden days as a math major, teaches us if there are 4 options for the second letter, then there are 3 for the third letter, 2 for the fourth letter, and only 1 for the last letter, the number of possible words is 4 x 3 x 2 x 1 = 24.  Twenty-four words!  Even if it took 2 minutes to fill out the rest of the Kafkaesque bureaucratic form, it would only take 48 minutes to go through all the possibilities.  Instead, Big Easy got totally flustered and the Globe Trotters took the 4-hour penalty rather than complete the challenge.  The HUMANITY!  The LACK OF MATH SKILLS!

Now The Amazing Race is left with the aforementioned Evil Gay Brothers, the Whiny Miss America and her Excessively Patient Husband (accompanied by their unofficial sidekick, Their Interracial Relationship), and the Two Boring Blond Ones.  I guess I am rooting for the boring blondes out of a lack of options.  So sad!  Globe Trotters, I hated to see you go. 


B^2 said...

A combinatorics problem on television that wasn't on the Big Bang Theory!

Good point! With no hints they have 5! = 5*4*3*2*1 = 120 permutations

Two minutes/form = 240 minutes

(240 minutes) / (60 minutes/hour) = 4 hours.

They could probably cut down on the time it takes to fill out each form and avoid getting lost using a derangement. A derangement is a permutation where each object is in a different position

N, A, R, F, Z


A, N, F, R, Z

are derangements of each other the !n represents the number of derangements in an n element set.

!n = [n!/e] the brackets, [], mean round the number to the nearest integer. n! is the factorial and e is about 2.7

!4 = [4!/e] = [24/2.7] = 9

Now we fix one point (actually they were given F as the fixed point in the 5 element set.)

For partial derangements with k fixed points:

P(n,k) = C(n,k)[(n-k)!/e]

we can choose any k of the n points to be fixed and derange the remaining n-k points. C(n,k) is just the binomial coefficient C(n,k) = n!/[k!(n-k)!]


P(4,1) = C(4,1)[(4-1)!/e]

P(4,1) = 4[3!/e] = 8

P(4,2) = 6

P(4,3) = 0 because if only 1 point is not fixed then it cannot be deranged so there are zero derangements.

P(4,4) = 1 (our starting arrangement)

The sum of the derangement and all the partial derangements is equivalent to the total number of permutations.

9+8+6+0+1 = 24.

This is overkill for a 4 element set. For a 5 element set or higher it provides a general method (they should split the work too) and you can check your answer for each fixed point to ensure you don't miss any permutations.

They shouldn't feel bad though because I'm terrible at basketball.

Amy B. said...

Ahhh! Math majors! I watched it feeling dismayed that Big Easy had probably never read The Metamorphosis. Then again, Franz Kafka's granddaughter was my very first creative writing teacher (fun fact!). Anyway, farewell Globetrotters. The finale will be exponentially (math!) more annoying without you.