Consider a bag with N red balls. At each step, you select two balls from the bag and if they are the same colour, you repaint one of them a new unique colour. If they are different, then do nothing. Either way, place the two balls back in the bag and begin a new step. Given long enough, eventually all the balls in the bag will be of different colours, on average how many steps does it take?

Its clear that this is some sort of reverse of the original problem, however, it is not clear that this is actually the "best" reverse. One could also give the newly coloured balls "random colours from the list of N colours" instead of new unique ones, but it turns out this problem is much more awful.

Anyway, I'm going to just solve this reverse problem on the spot, it seems like the thing to do. Let F(k) be the average number of steps it takes to get from a state with k red balls to a state with one red ball. Note that knowing the number of red balls totally specifies a state, since all non-red balls are of unique colours. F(1)=0 trivially, and we seek F(N), the state with N red balls. The probability that we move from a state with k red balls to a state with k-1 red balls (I will call this P(k)) is the chance that we draw exactly two red balls from the bag, that is:

P(k)=k/N*(k-1)/(N-1)

Of course, if something happens with probability p, then on average we need 1/p tries to have it happen once (I proved this back in the post where I gave the solution to the prisoner problem). Naturally then, we have:

F(k)-F(k-1)=(N-1)N/(k-1)k

and since F(1)=0, we can just express F(m) as F(m)-F(m-1)+F(m-1)-F(m-2)+...+F(2)-F(1), to give

F(m)=Σ N(N-1)/k(k-1)

summing k from 2 to m.

Therefore, for F(N) we have:

N(N-1)Σ 1/k(k-1)

=N(N-1)Σ [1/(k-1)-1/k]

summed k from 2 to N. This series telescopes, the terms look like 1/1-1/2+1/2-1/3+1/3-1/4+...-1/(N-1)+1/(N-1)-1/N. The final result is then that:

F(N)=N(N-1)[1-1/N]

=N(N-1)[(N-1)/N]

=(N-1)^2

Weird.

Alright, so this (much easier) problem has the same answer as the "forward" ball problem. I guess in some sense it is the correct reversal of the initial problem, since it has the same answer. I would love to find a way to show that these two problems have the same answer (that does not rely on using the fact that they both give (N-1)^2), so that way we can solve the harder one just by doing the easier one. Anyway, thats all I got here, if you can find a way to show that these problems are equivalent, feel free to give it in the comments.

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