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Am Thu, 18 Dec 2008 02:55:41 -0500 schrieb Florian Pesth:
> In this case the infinite sum is
> diverging and you can't calculate an average.
Ah... forget that, there is also an infinite number of solutions of the
same length. Even though it is an infinite sum, one can calculate it.
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>> Do a search for Coupon Collector's Problem. For equal random drawings
>> it's roughly nHn that is n * Sum k=1-n of 1/k multiplied by n
>
>
> Brain death! - n * Sum k=1-n of 1/k
Thanks Phil, when you know what it's called it's easy to find.
So, my answer is that the expected number of drawings to get all balls is
given by:
E[T] = n*(1/1 + 1/2 + 1/3 + ... 1/n)
Cool neat little formula :-)
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And lo On Thu, 18 Dec 2008 08:26:56 -0000, scott <sco### [at] scott com> did
spake thusly:
>>> Do a search for Coupon Collector's Problem. For equal random drawings
>>> it's roughly nHn that is n * Sum k=1-n of 1/k multiplied by n
>>
>>
>> Brain death! - n * Sum k=1-n of 1/k
>
> Thanks Phil, when you know what it's called it's easy to find.
Indeed, one of the major difficulties in research is trying to guess what
someone else might have named their solution ;-)
> So, my answer is that the expected number of drawings to get all balls
> is given by:
>
> E[T] = n*(1/1 + 1/2 + 1/3 + ... 1/n)
>
> Cool neat little formula :-)
Yup, of course as you recognise it's only an estimate the lower bound
being n and the upper bound being infinity; there's no reason why you
can't be pulling out the same number over and over again although you
might suspect a fiddle if it yields a high mean absolute error.
--
Phil Cook
--
I once tried to be apathetic, but I just couldn't be bothered
http://flipc.blogspot.com
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>> E[T] = n*(1/1 + 1/2 + 1/3 + ... 1/n)
>>
>> Cool neat little formula :-)
>
> Yup, of course as you recognise it's only an estimate the lower bound
> being n and the upper bound being infinity; there's no reason why you
> can't be pulling out the same number over and over again although you
> might suspect a fiddle if it yields a high mean absolute error.
On an only slightly related topic, I did manage to find this one
http://en.wikipedia.org/wiki/Benford%27s_law
without knowing what it was called beforehand.
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On Thu, 18 Dec 2008 08:38:40 +0100, "scott" <sco### [at] scottcom> wrote:
>> Strange, I've never wondered about it o_O
>
>Hehe, the exact problem I was applying it to was that I needed to save all
>the possible images generated by some website, I knew there was a certain
>number of possibilities and that it was random that each one appeared, so I
>just wondered how many times I would need to try before I got all of them.
>
>I didn't need to know exactly, eg if N=100, am I expecting 150 tries, 200
>tries, 500 tries? But knowing the exact formula would be cool as it seemed
>an interesting problem without an obvious result.
>
OK, a real world problem. I don't think that I've come across the need to work
out probabilities in RL.
>Another situation I can think of is if you need to check every item in a
>physical population, but you can only select ones at random (eg tagging fish
>in a lake or something).
>
There are more things in Heaven and Earth than in my philosophy, Scott. :)
--
Regards
Stephen
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Stephen wrote:
> OK, a real world problem. I don't think that I've come across the need to work
> out probabilities in RL.
*Clearly* you're not a betting man. ;-)
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On Thu, 18 Dec 2008 12:02:19 +0000, Invisible <voi### [at] devnull> wrote:
>Stephen wrote:
>
>> OK, a real world problem. I don't think that I've come across the need to work
>> out probabilities in RL.
>
>*Clearly* you're not a betting man. ;-)
Not with anything as trivial as money. ;)
--
Regards
Stephen
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>> *Clearly* you're not a betting man. ;-)
>
> Not with anything as trivial as money. ;)
Oh I see... You wager with beer, eh?
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On Thu, 18 Dec 2008 12:25:37 +0000, Invisible <voi### [at] devnull> wrote:
>>> *Clearly* you're not a betting man. ;-)
>>
>> Not with anything as trivial as money. ;)
>
>Oh I see... You wager with beer, eh?
Generally with mars bars.
--
Regards
Stephen
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>> Oh I see... You wager with beer, eh?
>
> Generally with mars bars.
Eww... beer and mars bars?? o_O
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