N Bottles of Beer

This week’s Riddler Classic is puzzle about the world’s most annoying song.

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?

My solution:
[Show Solution]

2 thoughts on “N Bottles of Beer”

    1. I thought about this at the time and came to the same realization; that the original problem specification was ambiguous but no matter how you interpreted it, the limiting value of $E_1/n$ would be unchanged.

      My interpretation was that the song ends once we finish singing the last verse, no matter whether we forgot it was the last verse or not.

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