Number of consecutive matches is calculated by formula N = log(1-C)/log(1-P),
where N is step, P is probability, C is reliability of probability.
Substituting C and P: N = log(1-0.99)/log(1-0.5) = 6.7 = natural value 7,
that means that 7th step of distribution should include
about 1% of half data, due to counting repetitions and 0 and 1, in amount of 100%.
Distribution step number:
at C = P = 0.5; N = 1 = log0.5/log0.5 = log(1-1/2)/log(1-1/2) = 1
at C = 0.25; P = 0.5; N = 2 = log0.75/log0.5 = log(1-1/4)/log(1-1/2) = 2, etc.
Multiplication of constant probabilities c+p^n = 1
personifies reliability of probability and creates a formula
N = log(1-c)/log(1-p)
c - probability of winning guaranteed
p - probability of winning event.
for example: with a probability of 99% for a probability of 48.65%
number of mismatches in a row n = log(1-0,99)/log(1-0,4865) = 7
and that means about 50% probability is easy to guess 7 times in a row.
it is simpler to calculate by formula N=7+(5*(1/p-2))
for example p = 0.1 N = 47 is normal and p = 0.78 N = 4 is normal
and same formulas are valid for probabilities above 50%.
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what we wanted to prove
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Number of consecutive matches is calculated by formula N = log(1-C)/log(1-P),
where N is step, P is probability, C is reliability of probability.
Substituting C and P: N = log(1-0.99)/log(1-0.5) = 6.7 = natural value 7,
that means that 7th step of distribution should include
about 1% of half data, due to counting repetitions and 0 and 1, in amount of 100%.
Distribution step number:
at C = P = 0.5; N = 1 = log0.5/log0.5 = log(1-1/2)/log(1-1/2) = 1
at C = 0.25; P = 0.5; N = 2 = log0.75/log0.5 = log(1-1/4)/log(1-1/2) = 2, etc.
Multiplication of constant probabilities c+p^n = 1
personifies reliability of probability and creates a formula
N = log(1-c)/log(1-p)
c - probability of winning guaranteed
p - probability of winning event.
for example: with a probability of 99% for a probability of 48.65%
number of mismatches in a row n = log(1-0,99)/log(1-0,4865) = 7
and that means about 50% probability is easy to guess 7 times in a row.
it is simpler to calculate by formula N=7+(5*(1/p-2))
for example p = 0.1 N = 47 is normal and p = 0.78 N = 4 is normal
and same formulas are valid for probabilities above 50%.
what I was required to prove
what we wanted to prove