Список проголосовавших за комментарий "Number of consecutive matches is calculated by formula N = log(1-C)/log(1-P),<br/> where N is step, P is probability, C is reliability of probability.<br/> <br/> Substituting C and P: N = log(1-0.99)/log(1-0.5) = 6.7 = natural value 7,<br/> that means that 7th step of distribution should include<br/> about 1% of half data, due to counting repetitions and 0 and 1, in amount of 100%.<br/> <br/> Distribution step number:<br/> at C = P = 0.5; N = 1 = log0.5/log0.5 = log(1-1/2)/log(1-1/2) = 1<br/> at C = 0.25; P = 0.5; N = 2 = log0.75/log0.5 = log(1-1/4)/log(1-1/2) = 2, etc.<br/> <br/> Multiplication of constant probabilities c+p^n = 1<br/> personifies reliability of probability and creates a formula<br/> N = log(1-c)/log(1-p)<br/> c — probability of winning guaranteed<br/> p — probability of winning event.<br/> <br/> for example: with a probability of 99% for a probability of 48.65%<br/> number of mismatches in a row n = log(1-0,99)/log(1-0,4865) = 7<br/> and that means about 50% probability is easy to guess 7 times in a row.<br/> <br/> it is simpler to calculate by formula N=7+(5*(1/p-2))<br/> for example p = 0.1 N = 47 is normal and p = 0.78 N = 4 is normal<br/> and same formulas are valid for probabilities above 50%.<br/> <br/> <a class="imgpreview" href="/uploads/images/05/70/34/2020/03/30/617731.png"><img src="/uploads/images/05/70/34/2020/03/30/fd2512.png"/></a><br/> <br/> what I was required to prove<br/> what we wanted to prove<br/> <br/> <a class="imgpreview" href="/uploads/images/05/70/34/2020/03/30/40bd4a.gif"><img src="/uploads/images/05/70/34/2020/03/30/bd2c2c.gif"/></a><br/> <br/> "