Puzzle o' the Day 406!

[captaino]

Sometimes I present a puzzle because I want someone to come up with a more elegant answer than mine. So it is with this puzzle, where my solution is a tangle of algebra that mysteriously resolves into a simple expression. I hope you can do better

Comments

[wantonhalo]

I got confused about this for a while, but working on it with a friend helped!

Let n = the number of sides on our die, while i = the last roll we got. Define l_(n,i) be the expected number of rolls it will take to get a pair of consecutive rolls with one lower than the other, as a function of this pair of numbers. Problem b.i. asks for l_(n,1).

Then l_(n,i) = 1 + sum(k=i to n)[(1/n)*l_(n,k)]. In particular, l_(n,n) = 1 + l_(n,n)/n, so l_(n,n) = n/(n-1). If i < n, we can simplify our recurrence to:

l_(n,i) = l_(n,i+1) + l_(n,i)/n => l_(n,i) = l_(n,i+1) * n/(n-1)

Which collapses to l_(n,i) = (n/(n-1)) ^ (n-i+1). So the quantity we're looking for is (n/(n-1)) ^ n. Thus:

a. (6/5)^6 or almost 3.
b.i. (n/(n-1))^n
b.ii. It goes to e.

[captaino]

This is a lovely solution! Much more elegant than mine.