Summing the log of the count is equivalent to multiplying them, I was using Long and not BigInt, It is conceivable there was a simple numeric overflow. On Monday, April 29, 2013 4:37:09 AM UTC+3, Nate Bauernfeind wrote: > A correct answer I read through was very similar and I tried rewriting it in > a way that was more intuitive to me. > > I tried to build a probability of the choices by multiplying the probability > of the chance you saw k for each k and the probability it was that > combination of numbers (taking into account duplicate arrangements basically > n! / n2! / n3! / n4! ... etc > > That doesn't pass for some reason, but intuitively I would expect more > probable choices to have higher probability of happening. > > Then I tried simply summing them thinking maybe the value is too small to > compare. But still wrong answer. > > The solution I looked at summed the log of the count. It works but I can't > explain why. Because I would expect this just to be another way to do the > same thing we both tried.
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