>>12060819

It's not about "subsets", because order matters. For [1,500,500] and [500,1,500] the number of plates seen is 3 but for [500,500,1] it's 2.

So you're dealing with conditional probabilities: the probability of a win after n plates is the probability that a sequence of n plates includes a 1000-sum pair, given that the initial n-1 plate subsequence doesn't.

IOW, the nth plate must make a 1000-sum pair with at least one plate from the initial n-1 plates, but no two plates within the initial n-1 plates make a pair. If the initial n-1 plates are all distinct, the probability of the nth plate forming a pair with one of them increases but so does the probability that a pair already exists. If the initial n-1 plates contain many duplicates, it's less likely that they contain a pair but also less likely that the nth plate will form a pair.

The main symmetry is that you don't particularly care exactly which numbers are present in the initial n-1 plates (other than 000, which will never form a pair with another plate), only the probability that n-1 plates will contain k distinct, non-zero numbers.

From the probability of n plates having k distinct numbers, you can determine the probability of n-1 plates containing no pairs and the probability of the nth plate forming a pair. For the latter: if there are k distinct numbers, there's a k/1000 probability that the nth plate will be a duplicate, a (k-1)/1000 probability that it will be distinct and non-zero, and 1/1000 that it will be zero. If it's distinct and non-zero, the probability of it forming a pair is k/999 if none of the k are zero and (k-1)/999 if one of them is zero. There's a k/1000 probability that one of the k is zero.