How many bottles of beer is it again?

You and your friends are singing the traditional song, “99 Bottles of Beer.” With each verse, you count down the number of bottles. The first verse contains the lyrics “99 bottles of beer,” the second verse contains the lyrics “98 bottles of beer,” and so on. The last verse contains the lyrics “1 bottle of beer.”

There’s just one problem. When completing any given verse, your group of friends has a tendency to forget which verse they’re on. When this happens, you finish the verse you are currently singing and then go back to the beginning of the song (with 99 bottles) on the next verse.

For each verse, suppose you have a 1 percent chance of forgetting which verse you are currently singing. On average, how many total verses will you sing in the song?

Extra credit: Instead of “99 Bottles of Beer,” suppose you and your friends are singing “$N$ Bottles of Beer,” where $N$ is some very, very large number. And suppose your collective probability of forgetting where you are in the song is $1/N$ for each verse. If it takes you an average of $K$ verses to finish the song, what value does the ratio of $K/N$ approach?

Let's define some variables. Let's say that $B_n$ is the number of bottles of beer on the wall in the $n$th verse and let's say that you start with $B_1 = N$ bottles of beer. Then the transition from $B_n$ to $B_{n+1}$ is given by $$B_{n+1} = \begin{cases} B_n - 1, &\text{with probability $1-p$}\\ N, &\text{with probability $p$}\end{cases}$$ which holds even if $B_n = 1.$ That is, the last verse is "1 bottle of beer on the wall, 1 bottle of beer, take 1 down pass it around, no bottles of beer on the wall," but somehow perhaps having ingested a few too many of the bottles of beer, your crew could still end up singing "1 bottle of beer on the wall, 1 bottle of beer, take 1 down pass it around, $N$ bottles of beer on the wall" and you'd have to make your way all the way back down. So in this setup, we need to also state that the transition with $B_{n+1} = 0$ if $B_n = 0,$ $\forall n.$ Finally, let's define the expected number of verses as $V_n = \mathbb{E}\left[ \min \{ k : B_k = 0 \} \mid B_1 = n\right],$ for any $n = 1, \dots, N.$

Now since there are only two possibilities at the end of each verse, we have \begin{align}V_n &= \mathbb{E} \left[ \min \{k : B_k = 0\} \mid B_2 = n-1, B_1 = n \right] \mathbb{P} \{ B_2 = n-1 \mid B_1 = n \}\\ &\quad\quad + \mathbb{E} \left[ \min \{k : B_k=0\} \mid B_2 = N, B_1 = n \right] \mathbb{P} \left\{ B_2 = N \mid B_1 = n \right\} \\ &= (1 + V_{N-1}) (1-p) + (1 + V_N) p\\ &= (1-p) V_{n-1} + pV_N + 1,\end{align} for all $n = 1, \dots, N,$ with $V_0 = 0.$

So in particular, we have $$V_N = (pV_N + 1) + (1-p) V_{N-1}.$$ We can use this as the base case to prove that $$V_N = (pV_N+1) \sum_{i=0}^{k-1} (1-p)^i + (1-p)^k V_{N-k}$$ for all $k = 1, \dots, N.$ In particular, assume that for some $k > 1,$ then substituting in the recurrence relationship for $V_{N-k}$ we get \begin{align*} V_N &= (pV_N+1) \sum_{i=0}^{k-1} (1-p)^i + (1-p)^k ( (1-p) V_{N-k-1} + pV_N + 1) \\ &= (pV_N+1) \sum_{i=0}^k (1-p)^i + (1-p)^{k+1} V_{N-k-1},\end{align*} that is, the formula is also true for $k+1.$ So, by induction, the formula must hold for all $k = 1, \dots, N,$ including for instance $k=N,$ so that \begin{align*}V_N &= (pV_N+1) \sum_{i=0}^{N-1} (1-p)^i + (1-p)^N V_0 = (pV_N+1) \sum_{i=0}^{N-1} (1-p)^i \\ &= (pV_N+1) \frac{1 - (1-p)^N}{p}.\end{align*}

Solving for $V_N$ we get $$V_N = \frac{1 - (1-p)^N}{p(1-p)^N}.$$ So in particular if $N = 99$ and $p = 0.01,$ then we get an average of $$\frac{1-(0.99)^{99}}{0.01 * (0.99)^{99}} \approx 170.467$$ verses. However, if we set $p = 1/N,$ then we get $$K=V_N = N\frac{1-(1-1/N)^N}{(1-1/N)^N},$$ so the fraction $$K/N = \frac{1 - (1-1/N)^N}{(1-1/N)^N} = (1-1/N)^{-N} - 1 \to e-1$$ as $N\to \infty.$