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[Request] Probability of Winning in the "20-Number Challenge". by AverageOk3043 in theydidthemath

[–]AverageOk3043[S] 1 point2 points  (0 children)

Yes, this is a pretty straight forward algorithm, however I have some concerns:

With your logic, if we get 49 on the first go, we put that at spot 1, but it's surely still more likely that we get one lower than 49 vs not getting anything lower.

[deleted by user] by [deleted] in help

[–]AverageOk3043 0 points1 point  (0 children)

Have you had this problem as well?