Probability of 0 bit in ASCII text files

In summary: What is "odds"?I meant the ratio. We can't know what the probability actually was, but we can calculate the ratio of how likely it was to be one thing compared with another. The answer may be written as a fraction, decimal, or percentage.In summary, the probability of drawing a 0 in the second time, given that the first drawing was also a 0, is calculated using Bayes formula as 2/9. This is based on the assumption that the file consists of randomly distributed bits with an equal probability of 0 and 1, as well as a probability of 1/8 for the Most Significant Bit (MSB) being 0.
  • #36
Cylab said:
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?
 
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  • #37
haruspex said:
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.
 
  • #38
Cylab said:
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 ...?
 
  • #39
haruspex said:
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 ...?

haruspex said:
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.
 
  • #40
Cylab said:
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).
 
  • #41
haruspex said:
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] =?
 
  • #42
Cylab said:
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?
 
  • #43
haruspex said:
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?
 
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  • #44
Cylab said:
Is (N1)Prob[0] = (N2)Prob[0] right,
Of course. The bits don't know how many others were chosen.
 
  • #45
haruspex said:
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]
...
 
  • #46
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.)
 
  • #47
haruspex said:
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.
 
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  • #48
Cylab said:
1st. (N1 case) : {(9X/16)C2 * (7X/16)C5 } / xC7.
I have no idea what that notation means.
 
  • #49
haruspex said:
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.
 
  • #50
Cylab said:
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.
 
  • #51
haruspex said:
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
 
  • #52
Cylab said:
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?
 
  • #53
haruspex said:
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.
 
  • #54
Cylab said:
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.
 
  • #55
haruspex said:
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.
 
  • #56
Cylab said:
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.
 
  • #57
haruspex said:
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
 
<h2>1. What is the probability of a 0 bit occurring in an ASCII text file?</h2><p>The probability of a 0 bit occurring in an ASCII text file depends on the specific file and its contents. However, in general, ASCII text files are made up of 8-bit characters, so the probability of a 0 bit occurring is approximately 1/8 or 12.5%.</p><h2>2. How does the probability of a 0 bit change in different types of ASCII text files?</h2><p>The probability of a 0 bit occurring in an ASCII text file can vary depending on the type of file and its contents. For example, a file with mostly lowercase letters will have a higher probability of 0 bits compared to a file with mostly numbers or special characters.</p><h2>3. Is the probability of a 0 bit in ASCII text files affected by file size?</h2><p>The probability of a 0 bit in ASCII text files is not directly affected by file size. However, larger files may contain a higher number of 0 bits simply due to the increased number of characters in the file.</p><h2>4. Can the probability of a 0 bit be calculated for a specific ASCII text file?</h2><p>Yes, the probability of a 0 bit can be calculated for a specific ASCII text file by counting the total number of bits in the file and then determining the number of 0 bits. The probability can then be calculated as the number of 0 bits divided by the total number of bits.</p><h2>5. How does the probability of a 0 bit in ASCII text files relate to data compression?</h2><p>The probability of a 0 bit in ASCII text files is important in data compression because it can help determine the most efficient way to compress the file. If a file has a high probability of 0 bits, it may be more easily compressed using certain algorithms compared to a file with a lower probability of 0 bits.</p>

1. What is the probability of a 0 bit occurring in an ASCII text file?

The probability of a 0 bit occurring in an ASCII text file depends on the specific file and its contents. However, in general, ASCII text files are made up of 8-bit characters, so the probability of a 0 bit occurring is approximately 1/8 or 12.5%.

2. How does the probability of a 0 bit change in different types of ASCII text files?

The probability of a 0 bit occurring in an ASCII text file can vary depending on the type of file and its contents. For example, a file with mostly lowercase letters will have a higher probability of 0 bits compared to a file with mostly numbers or special characters.

3. Is the probability of a 0 bit in ASCII text files affected by file size?

The probability of a 0 bit in ASCII text files is not directly affected by file size. However, larger files may contain a higher number of 0 bits simply due to the increased number of characters in the file.

4. Can the probability of a 0 bit be calculated for a specific ASCII text file?

Yes, the probability of a 0 bit can be calculated for a specific ASCII text file by counting the total number of bits in the file and then determining the number of 0 bits. The probability can then be calculated as the number of 0 bits divided by the total number of bits.

5. How does the probability of a 0 bit in ASCII text files relate to data compression?

The probability of a 0 bit in ASCII text files is important in data compression because it can help determine the most efficient way to compress the file. If a file has a high probability of 0 bits, it may be more easily compressed using certain algorithms compared to a file with a lower probability of 0 bits.

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