Probability of 0 bit in ASCII text files

[haruspex]

the analysis concentrates on MSB only (say, N=7); so the formula maybe like;
P(MSB)=P(MSB |H0)P(H0) + P(MSB |H1)P(H1)
(Let Hi be the event that there are i MSB bits in N, for i = 0, 1, 2, 3….. )
where P(MSB |H0) stands for conditional probability of MSB bit in N given it is H0 which equals 0 (no MSB) and P(MSB |H1) stands for conditional probability of MSB bit in the N given it is H1 which equals 1/7 (one MSB);

That's all true, but I'm not sure where it gets you. What are you trying to solve here?
Are we done with the N≤8 case?

 

[Cylab]

That's all true, but I'm not sure where it gets you. What are you trying to solve here?
Are we done with the N≤8 case?

How about investigation of N > 8 case?
Just can`t quiet figure out how N=8A+B works.

 

[haruspex]

How about investigation of N > 8 case?
Just can`t quiet figure out how N=8A+B works.

N=8A+B, B<8, means there are A whole bytes and B odd bits, so either A or A+1 MSBs included.
See what gaps you can fill in here:
Prob that this includes A+1 MSB's = ...?; if A+1 MSBs, prob that all N bits are 0 is ...?
Prob that this includes A MSB's = ...?; if A MSBs, prob that all N bits are 0 is ...?
Adding this up, prob that all N bits are 0 is ...?

 

[Cylab]

N=8A+B, B<8, means there are A whole bytes and B odd bits, so either A or A+1 MSBs included.
See what gaps you can fill in here:
Prob that this includes A+1 MSB's = ...?; if A+1 MSBs, prob that all N bits are 0 is ...?
Prob that this includes A MSB's = ...?; if A MSBs, prob that all N bits are 0 is ...?
Adding this up, prob that all N bits are 0 is ...?

if A+1 MSBs, prob that all N bits are 0 is ...?

That is not easy for me Sir.
Analysis: Since the N bits are drawn consecutively from ASCII, there is only 1 character (out of 2^256), which is are all 0. So only 1 MSB. Thus, the Prob =1/2^256.
Others seem to follow the conception , or I misunderstood your point?

Question:
1) Pr[0] in ASCII (assume each character appears with same ratio) equals = 1/8+1/2=5/8. Is it OK?
2) Successive 7 bits are drawn at random from ASCII bits (e.g. no bias of character distribution), what is Pr[0] in the 7 bits?
Successive 4 bits are drawn (same condition with above), what is Pr[0] in the 4 bits?
So, say, Successive N bits are drawn (same condition) , what is Pr[0] in the N bits?
Analysis:
Do you think it is same case? remember you explained that N/8*2^-N + (1-N/8)*2^-(N+1) . Does the formula apply to the case of 2).

Shed some lights on please.

 

[haruspex]

Analysis: Since the N bits are drawn consecutively from ASCII, there is only 1 character (out of 2^256), which is are all 0. So only 1 MSB. Thus, the Prob =1/2^256.

The N bits might start in the middle of one byte, span several whole bytes, and finish part way through the last. For each whole byte, prob of all zeroes is 1/128 (since MSB always zero).
If the N bits include A MSBs then how many non-MSBs do they include?
What is the prob that the A MSBs are all 0?
What is the prob that the non-MSBs are all 0?
So what is the prob that all N bits are 0?

Question:
1) Pr[0] in ASCII (assume each character appears with same ratio) equals = 1/8+1/2=5/8. Is it OK?

No. 1/8 that bit is MSB, so 7/8 that it is non-MSB. P[0] = 1/8 + 7/8*1/2 = 9/16. We went through that much earlier in the thread.

2) Successive 7 bits are drawn at random from ASCII bits (e.g. no bias of character distribution), what is Pr[0] in the 7 bits?
Successive 4 bits are drawn (same condition with above), what is Pr[0] in the 4 bits?
So, say, Successive N bits are drawn (same condition) , what is Pr[0] in the N bits?
Analysis:
Do you think it is same case? remember you explained that N/8*2^-N + (1-N/8)*2^-(N+1) . Does the formula apply to the case of 2).

I believe you are misquoting the formula. In post #27 I explained that the prob of N consecutive 0 bits (N≤8) is 2^-N(1+N/8).

 

[Cylab]

I believe you are misquoting the formula. In post #27 I explained that the prob of N consecutive 0 bits (N≤8) is 2^-N(1+N/8).

That is right and it was good explanation.
Now, say, two N consecutive bits are taken (or two groups) (_N1=7, and _N2=4).
So the prob is different in following cases in comparing of same amount of 0`s?
1st case: (in _N1=7) Pr[0], Pr[00],..,Pr[0000] =?
2nd case: (in _N2=4) Pr[0], Pr[00],..,Pr[0000] =?

 

[haruspex]

1st case: (in _N1=7) Pr[0], Pr[00],..,Pr[0000] =?
2nd case: (in _N2=4) Pr[0], Pr[00],..,Pr[0000] =?

I don't understand your question. What does Pr[0] mean in the context of N₁=7? Is it the probability that the next bit is zero given the preceding 7 were?

 

[Cylab]

I don't understand your question. What does Pr[0] mean in the context of N₁=7? Is it the probability that the next bit is zero given the preceding 7 were?

Sorry!
I meant the prob of 0 within _N1 =7 consecutively drawn from ASCII .
In other words, say, now we have a group of bits consists of many _N1, each of which is consecutive 7 bits drawn from ASCII. what is Prob[0], Prob[00] in the group respectively?

2nd case. another group of bits with same condition, where _N2=4. What is Prob[0], Prob[00] within the second group respectively?

Is (_N1)Prob[0] = (_N2)Prob[0] right, or should it be unequal?

 

[haruspex]

Is (_N1)Prob[0] = (_N2)Prob[0] right,

Of course. The bits don't know how many others were chosen.

 

[Cylab]

Of course. The bits don't know how many others were chosen.

So you are saying following are correct?

(_N1=7)Prob[0] = (_N2=4)Prob[0]
(_N1=7)Prob[00] = (_N2=4)Prob[00]
(_N1=7)Prob[000] = (_N2=4)Prob[000]
...

 

[haruspex]

To be completely clear:
If you choose N consecutive bits, the probability that the first R of those bits are all zero (R <= N), depends only on R. It cannot depend on N.
Further, if you choose N consecutive bits, then choose R consecutive bits from those N, the probability that the first R of those bits are all zero depends only on R.
(This seems so obvious that I worry that I have not understood the question.)

 

[Cylab]

To be completely clear:
If you choose N consecutive bits, the probability that the first R of those bits are all zero (R <= N), depends only on R. It cannot depend on N.
Further, if you choose N consecutive bits, then choose R consecutive bits from those N, the probability that the first R of those bits are all zero depends only on R.
(This seems so obvious that I worry that I have not understood the question.)

case: N₁=7 & N₂=4 . Assume N is taken from X bits, which is ASCII.
take R=2 bits from N₁ & N₂ respectively, what is prob that they are two 0 bits.

1st. (N₁ case) : {_(9X/16)C₂ * _(7X/16)C₅ } / _xC₇.
1st. (N₂ case) : {_(9X/16)C₂ * _(7X/16)C₂ } / _xC₄.

Seems it depends on N too.

 

[haruspex]

1st. (N₁ case) : {_(9X/16)C₂ * _(7X/16)C₅ } / _xC₇.

I have no idea what that notation means.

 

[Cylab]

I have no idea what that notation means.

Hypergeometric Distribution.

X: number of ASCII bits , from which N is taken.
Pr[0] = 9/16.

_xC₇ : The number of combinations of x , taken 7 at a time.

 

[haruspex]

Hypergeometric Distribution.

X: number of ASCII bits , from which N is taken.
Pr[0] = 9/16.

_xC₇ : The number of combinations of x , taken 7 at a time.

So how do I read _(9X/16)C₂? If I plug in X=7, that gives _(63/16)C₂, which is meaningless.

 

[Cylab]

So how do I read _(9X/16)C₂? If I plug in X=7, that gives _(63/16)C₂, which is meaningless.

C: combinations
Pr[0] = 9/16.
X: number of ASCII bits , from which N is taken.
case: N1=7 & N2=4 . Assume N is taken from X bits, which is ASCII.
Other definitions should be clear

 

[haruspex]

C: combinations
Pr[0] = 9/16.
X: number of ASCII bits , from which N is taken.
case: N1=7 & N2=4 . Assume N is taken from X bits, which is ASCII.
Other definitions should be clear

You wrote _(9X/16)C₂, and you have still offered no reasonable explanation for that notation. Did you mean (9/16)_XC₂?

 

[Cylab]

You wrote _(9X/16)C₂, and you have still offered no reasonable explanation for that notation. Did you mean (9/16)_XC₂?

X: The number of bits in ASCII.
9X/16: The number of 0 bits in the X that are classified as successes.
7 or 4: The number(s) of bits taken consecutively from X.
2: The number of 2 zeros in the 7 or 4 that are classified as successes.
_(9X/16)C₂ : The number of combinations of 9X/16, taken two 0 bits at a time.

 

[haruspex]

X: The number of bits in ASCII.
9X/16: The number of 0 bits in the X that are classified as successes.
7 or 4: The number(s) of bits taken consecutively from X.
2: The number of 2 zeros in the 7 or 4 that are classified as successes.
_(9X/16)C₂ : The number of combinations of 9X/16, taken two 0 bits at a time.

Now that you have explained that, thankyou, I can see where it is wrong.
For one thing, that analysis treats all bits as independently 0 or 1, regardless of their proximity to each other. Bits multiples of 8 positions apart will be positively correlated, and at other distances negatively correlated.
More significantly, let's look at what these represent:
1st. (N1 case) : {(9X/16)C2 * (7X/16)C5 } / xC7.
1st. (N2 case) : {(9X/16)C2 * (7X/16)C2 } / xC4.
The first is the probability of picking 7 bits that are exactly two 0 bits and 5 1 bits; the second is the prob of picking 4 bits that are exactly 2 and 2. No wonder they're different! In the problem I thought we were discussing, P[00] doesn't care what the remaining 2 or 5 bits are.

 

[Cylab]

Now that you have explained that, thankyou, I can see where it is wrong.
For one thing, that analysis treats all bits as independently 0 or 1, regardless of their proximity to each other. Bits multiples of 8 positions apart will be positively correlated, and at other distances negatively correlated.
More significantly, let's look at what these represent:
1st. (N1 case) : {(9X/16)C2 * (7X/16)C5 } / xC7.
1st. (N2 case) : {(9X/16)C2 * (7X/16)C2 } / xC4.
The first is the probability of picking 7 bits that are exactly two 0 bits and 5 1 bits; the second is the prob of picking 4 bits that are exactly 2 and 2. No wonder they're different! In the problem I thought we were discussing, P[00] doesn't care what the remaining 2 or 5 bits are.

You are right!
P[00] doesn't care what the remaining 2 or 5 bits are.
So does the calculation in the following two cases, which are the prob of P[00] taken from N1 and N2 respectively regardless of the contents of the N1 & N2.
1st. (N1 case) : {(9X/16)C2 * (7X/16)C5 } / xC7.
1st. (N2 case) : {(9X/16)C2 * (7X/16)C2 } / xC4.

 

[haruspex]

So does the calculation in the following two cases, which are the prob of P[00] taken from N1 and N2 respectively regardless of the contents of the N1 & N2.
1st. (N1 case) : {(9X/16)C2 * (7X/16)C5 } / xC7.
1st. (N2 case) : {(9X/16)C2 * (7X/16)C2 } / xC4.

Once again, I'm not at all sure what you are saying. Are you insisting that the above formulae are correct for P[00]? I have just explained to you why they are not.

 

[Cylab]

Once again, I'm not at all sure what you are saying. Are you insisting that the above formulae are correct for P[00]? I have just explained to you why they are not.

Just focusing your points.

Following link may help you clarify your analysis mentioned so far.
http://en.wikipedia.org/wiki/Hypergeometric_distribution