Recent content by Cylab

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

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) ...

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...

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

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.

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.

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] ...

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...

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]...

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...

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

OK! But 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...

Thanks. the explanations are really helpful. So in other words, 「We can assume that there are seven bits (or N<8) and that there is at most one MSB bit (which means either one MSB or zero MSB) in it. Thus we can compute the probability of having zero MSB, plus the probability of having one...

You are right. Thanks. Please let me confirm about the "1/2". 1st. Analysis: The number of "1/2 " at the right part (of +) is 2^-(N-1), and left is 2^-N. right? but Where the " number of 1/2 " comes from? Is that because of the N that is number of successive zero bits taken from ASCII? 2nd. If...

1st: the Prob of having 3 successive 0 bits(000) is ; 3/8*1/2 + 5/8*1/2*1/2 = 11/32. Since you computed of having 2 successive 0 bits is 10/32. So my calculation of (000) seems wrong? 2nd : if it follows to the rule of (2^-N)(1+N/8), Then 5 successive 0 bits(00000) would be: 2^5(1+5/8)= 13/512...