Audition Online is a music game where players compete to gain the most number of points in a given song. In each song, there are a certain number of bars - in each bar, a player has to execute a certain sequence of keystrokes and press spacebar on the 4th beat. A judgment is made based on how accurately the spacebar is pressed - from best to worst, "perfect", "great", "cool", "bad", and "miss". When a player does 2 or more "perfects" consecutively, he will get bonus points for doing so. Within a song, the largest number of consecutive "perfects" a player gets is called his "chain".
A song has \(n\) number of bars. Assume that a given player has a constant probability of getting a "perfect", \(p\), on each bar. The chain for a particular song (largest number of consecutive "perfects") is denoted as \(c\).
What is the expected value of \(c\) in terms of \(n\) and \(p\)?
Add-on: What is the probability of getting a particular \(c\) in terms of \(n\) and \(p\)? (In essence, what is the probability distribution of \(c\)?)
A valid solution must minimally satisfy the following cases:
We know that given \(p=1\) (i.e. a completely perfect player), \(c=n\).
We also know that given \(p=0\) (i.e. a completely horrible player), \(c=0\).
A song with \(n=1\) basically has only one bar, hence \(c=p\).
A song with \(n=2\) has two bars and the expected value of \(c\) can be trivially calculated to be \(2 p\).
Details and assumptions
- A gameplay of P-G-P-P-G-G-G-P-G-P-P-P-P-P-G-G-P-G counts as having a chain of 5 (P is "perfect", G is "great").
- For simplicity, "greats", "cools", "bads" and "misses" are treated as the same thing. In reality, a "miss" on one bar will forfeit the next bar, hence \(n\) may not be treated as a constant.