Relative Resonance (9/9 – 9/22)
Alice and Bob are tuning their guitar. However, they only have relative pitch, so they only care if their pitches are relatively in tune.
Given a tuple of intended notes the strings are supposed to be tuned at, and a tuple of notes they are currently tuned at, we say that this tuple is relatively in tune if the range (max – min) of the offsets is at most \(\frac{1}{2}\).
For example, if the intended notes of 3 strings are \((1, 2, 3)\) and they are tuned as \((0.8, 2.3, 3.5)\), the offsets are \((-0.2, 0.3, 0.5)\) so the range is equal to \(0.5 – (-0.2) > 0.5\). So, \((0.8, 2.3, 2.5)\) is not relatively in tune. Whereas, the strings that are tuned as \((1.8, 3.3, 4.1)\) is relatively in tune since the offsets are \((0.8, 1.3, 1.1)\) which has range \(1.3 – 0.8 \leq 0.5\).
(For the sake of the problem, assume notes are linear.)
Now, Alice and Bob are tuning the guitar (with 6 strings) with the following strategy.
Alice will check if string 1 and string 2 are relatively in tune, then check if string 1 and string 3 are relatively in tune, all the way up to string 1 and string 6.
Bob will check if string 1 and string 2 are relatively in tune, then check if string 2 and string 3 are relatively in tune, all the way up to string 5 and string 6.
Suppose Alice completed her check with her guitar and Bob with his.
Given that every 6-tuple of notes being likely to occur,
1. Find the probability that all of Alice’s strings are relatively in tune.
2. Find the probability that all of Bob’s strings are relatively in tune.
Solution
First we note that the intended tune doesn’t matter and only the offsets between each string. Let these be \(a_1, a_2, \dots, a_6\). Now, we can rephrase the question.
Alice: If \(-0.5 \leq a_1, a_2, \dots, a_5 \leq 0.5\), then what is the probability that \(\max(a_i – a_j) \leq 0.5\)?
Bob: If \(|a_i – a_{i+1}| \leq 0.5\), what is the probability that \(\max(a_i – a_j) \leq 0.5\)?
Now, consider the following problem.
We pick 6 random diameters \(A_1, A_2, A_3, A_4, A_5, A_6\) from a circle of perimeter \(1\). Then, we select 6 points \(a_1, a_2, a_3, a_4, a_5, a_6\) by selecting a point in random from a diameter. We want to find the probability that these 6 points lie on a 180 degree arc. This has probability 12 (there are 12 consecutive runs) divided by 64 (\(2^6\)) which is \(\frac{3}{16}\)
For Alice, this is an equivalent problem since from \(a_1\), the rest of the points are randomly chosen between -0.5 (left of \(a_1\)) and 0.5 (right of \(a_1\)).
For Bob, this is also an equivalent question since from \(a_i\), selecting \(a_{i+1}\) is picking a point randomly between -0.5 (left of \(a_i\)) and 0.5 (right of \(a_i\)). The reason this argument holds is because in the case that in the degenerate cases where the points loop around the circle, \((\max(a_i – a_j) \geq 1)\), we already violated our condition which is \((\max(a_i – a_j) \leq 0.5\) the turn before.
Therefore, the probability is \(\frac{3}{16}\) for both.