@Tushar Bindal
No need of long calculations :)
here is a shortcut, actually in O(1) time :)
for calculating chances of any two entities to collide in given different
species is just take underoot of it.
here underoot of 365 is approx 19.....he he enjoy the solution.
For more details just go through cryptographic mathematics :)
On Thu, Jul 7, 2011 at 11:47 PM, Tushar Bindal <tushicom...@gmail.com>wrote:
> Sory once again for that incomplete answer.
> The complete one is here.
>
> probability that i win standing at second position: 1/365
> probability that i win standing at third position : 364/365*2/365 =
> 1/365)*(628/365)
> probability that i win standing at fourth position : 364/365*363/365*3/365
> probability that i win standing at 4th position :
> 364/365*363/365*362/365*4/365
>
> probability that i win standing at (n+1)th position:
> (365-1)*(365-2)*(365-3)*(365-
> 4)*(365-5).....*(365-(n-2))*(365-(n-1))*(n)*(1/365)^n
>
> there is a pattern in the probabilities
> let probability of winning standing at nth position be x
> probability of winning standing at (n+1)th position = x * {(365 - n
> +1)*(n)} / {365*(n-1)}
>
> maximum probability is at nth position if at (n+1)th position,
> {(365 - n +1)*(n)} / {365*(n-1)} <= 1
>
> This is true for n>=20
>
> For n=19,
> {(365 - n +1)*(n)} / {365*(n-1)} > 1
>
> So max probability is when *n=19*
> i.e., n+1 = 20, which is my position.
>
> So standing at 20th position gives me maximum chance of winning
>
>
> Just hope I haven't got anything wrong here.
>
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SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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