[fnord_too]

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At some point you will will be 1 or 2 away from every number on the low side (i.e. before you pass or hit any number you will be exactly one or two less than it). For each of these two equally likely occurences

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I think you need to distinguish between on the one hand eventually being 1 or 2 below, and on the other hand reaching 1 below without having first reached 2 below (or vice versa).

For eventually reaching 1 or 2 below, then potentially both can happen, so I don't think that is what you mean.

On the other hand, if you mean reaching 1 below without having first reached 2 below, or vice versa, then it is not obvious, in fact I think it is false, that they are equally likely.

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I do mean the one you arive at first. Let me see if I can prove it.

For any integer k, prove that ariving at k-1 before k-2 has the same probability as ariving at k-2 before k-1.

Assume one can arrive at k-1 first more often than you can arive at k-2 first. Then, for (k+1) one arrives at (k+1)-2 more often than you are ariving at (k+1)-1, which is a contradicion. Ergo P(K-1) <= P(K-2).

Ok, getting rid of that '<' is not trivial. I need to think about this some more.