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Yes, this logic is correct as far as I can tell. I came up with the exact same thing using the geom(p)=1/p and p=1/n. Since he always has n keys at his disposal, this being the average number he needs to try would be logical. | Yes, this logic is correct as far as I can tell. I came up with the exact same thing using the geom(p)=1/p and p=1/n. Since he always has n keys at his disposal, this being the average number he needs to try would be logical. | ||
Evan Clinton | Evan Clinton | ||
Revision as of 14:09, 23 September 2008
So I have in my notes that E[x] of a Geometric RV = 1/p. and if p=1/n (from the hint provided by Hamad) then does that mean the expected value is n? that does not sound right to me.
Actually after thinking about it a little longer that does make sense because the prof could choose the same wrong key many times, so it would make sense that the expected value is actually the number of keys. But please let me know if there is something wrong with my logic.
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Yes, this logic is correct as far as I can tell. I came up with the exact same thing using the geom(p)=1/p and p=1/n. Since he always has n keys at his disposal, this being the average number he needs to try would be logical. Evan Clinton
