MHB Find the probability of being same

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The discussion revolves around calculating the probability that two files, A and B, of 8192 bytes (65536 bits) are identical. Initially, the probability is determined to be 1/2^65536, assuming any bit sequence is possible. However, the problem becomes more complex when it is specified that both files contain 16384 bits filled with 1s and the rest with 0s, suggesting that the probability of them being the same would increase. The challenge lies in recalculating the probability given this new constraint, which complicates the initial approach. The consensus is that the question is not straightforward and requires a more nuanced analysis.
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This question is somewhat related to computer but deals with a big lot of probability.I have been given two files (A and B) of size 8192 bytes and is asked to find the probability of that file B is same as file A.I know that for two files to be same, they should have same bit sequence i.e same sequence of 1s and 0s.
As the files are of 8192 bytes so they have (8192*8) i.e 65536 bits.
Now our sample space (S) includes all the probable bit sequences of file A.
Using the formulae variation^length, I got that n(S)=2 ^65536
[Since , variations are 1 and 0 i.e 2 variations and length =65536 bits]
Now , Since file B can have only one sequence of bits
\therefore n(E)=1 [let E denote the set of favourable outcomes to the event that file B is same as file A]
\therefore probability of the matching of the two files = n(E)/ n(S) =1/2^65536 = 2^-65536

Upto this it is clear to me but the real problem starts in the next part :
In the next part of problem it is given that file A as well as file B have 16384 bits filled with 1s and the rest bits filled with 0s.
Now certainly the above found probability would increase as in the above part we have considered that both file A and B can have as many 1s and 0s required to fill the complete sequence of bits.
Now I am not able to figure out what the probability of both the files being same would be!
Plz help me soon...
 
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Diganta28 said:
This question is somewhat related to computer but deals with a big lot of probability.I have been given two files (A and B) of size 8192 bytes and is asked to find the probability of that file B is same as file A.I know that for two files to be same, they should have same bit sequence i.e same sequence of 1s and 0s.
As the files are of 8192 bytes so they have (8192*8) i.e 65536 bits.
Now our sample space (S) includes all the probable bit sequences of file A.
Using the formulae variation^length, I got that n(S)=2 ^65536
[Since , variations are 1 and 0 i.e 2 variations and length =65536 bits]
Now , Since file B can have only one sequence of bits
\therefore n(E)=1 [let E denote the set of favourable outcomes to the event that file B is same as file A]
\therefore probability of the matching of the two files = n(E)/ n(S) =1/2^65536 = 2^-65536

Upto this it is clear to me but the real problem starts in the next part :
In the next part of problem it is given that file A as well as file B have 16384 bits filled with 1s and the rest bits filled with 0s.
Now certainly the above found probability would increase as in the above part we have considered that both file A and B can have as many 1s and 0s required to fill the complete sequence of bits.
Now I am not able to figure out what the probability of both the files being same would be!
Plz help me soon...

Your approach to the problem would be fair if any combination of 8192 bytes had the same probability, and this clearly is not true ... I think your question isn't easy to answer ...

Kind regards

$\chi$ $\sigma$
 
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