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Old 2007-06-03, 06:48   #13
RedGolpe
 
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Aug 2006
Monza, Italy

4916 Posts
Default Math explained

Quote:
Originally Posted by paulunderwood View Post
I don't see this. Please explain why you use this assumption.
The chance of a random number n to be prime is 1/log(n) and for a random odd number, it of course becomes 2/log(n). See http://primes.utm.edu/howmany.shtml#3 for more information. I just don't see why Cullen numbers should behave any differently.

Quote:
If my chance of throwing a "six" is 1/6 then by throwing twice my chance does not become 1/3, but rather 1-(5/6)^2. That is the chance of being unsuccessful is 5/6 at the first throw and at the second throw it is (5/6)^2, meaning my chance of success at the second throw is 1-(5/6)^2 which is 11/36.

Am I missing something?
All you wrote is correct. Still, if you throw two dices, the expected number of 6's is 1/3. You computed the chance of having at least one 6, which is a bit different. Anyway, for numbers much smaller than 1, these values are very close: indeed, 11/36 differs from 1/3 only by less than 10%. Things change when the chance gets closer to 1: for example, the chance of having at least a 6 with 12 throws is 1-(5/6)^12>88.7%, and the expected number of 6's is 2.

Last fiddled with by RedGolpe on 2007-06-03 at 06:49
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