Money Talks

Let's explore the mathematics of Looney Tunes!

To begin with, please watch the following video:

In particular, watch the scene at time index 2m34s in which the mafia boss divides the spoils of his criminal syndicate with his second in command.

His exact words are: "One for you, and one for me. Two for you, and one, two for me. Three for you, and one, two, three for me..."

The mathematical question is: at iteration n of such an apportionment algorithm, what percentage of the overall cache is owned by the mafia boss?

In order to solve this question, we must first understand how much the mafia boss owns at iteration n. This is answered by the fact that 1+2+3+4+..., otherwise represented as Σ(i=0→n){i}, is equal to (n*(n+1))/2.

With this information, we can next investigate how much the second in command has at iteration n. This is much easier to answer, as he has 1+1+1+1+..., or Σ(i=0→n){1}, which is equal to n.

So, now we know enough to represent the mafia boss's overall share of the cache. If M is the mafia boss's pot, and S is the second in command's pot, the mafia boss's share is simply M/(M+S).

Next we must represent the big, messy equation. M=(n*(n+1))/2, and S=n. So M/(M+S)=((n*(n+1))/2)/((n*(n+1))/2 + n)

Finally, we must simplify this equation, so that it is more straight-forward. The requisite steps are below:

^{((n*(n+1))/2)}/_{((n*(n+1))/2 + n)}

= ^{((n*(n+1))/2)}/_{((n*(n+1))/2 + 2*n/2)}

= ^{((n*(n+1))/2)}/_{((n*(n+1)+2*n)/2)}

= ^(n*(n+1))/_{(n*(n+1)+2*n)}

= ^(n*(n+1))/_{(n*((n+1)+2))}

= ⁽ⁿ⁺¹⁾/_((n+1)+2)

= ⁽ⁿ⁺¹⁾/_(n+3)

And that is the solution! At iteration n, the mafia boss owns (n+1)/(n+3) percentage of the overall cache. So at iteration 50, the mafia boss owns 51/53=96.23% of the cache.

Thank you for bearing with me as I explained the mathematics of Looney Tunes, and stay tuned for more posts. That's all, Folks!