Trying to quantify probability. My null hypothesis: this is randomly occurring and not significant.
There are two alpha-bet sequences. Both of them includes number a number of elements
1st alphabet sequence[separated by a comma] ['XX','XX','XX'...] (1) ;
2nd alphabet sequence ['XX','XX','XX'...] (2)
Both have totally, 19 elements. Both have 83 characters.
Both are associated with two different character alphabet sequence (randomly selected from a-z): CCCC . The first sequence, ASDS,occurs in (1) 27, and in (2) 27 times.
The second sequence of randomly generated alphabet, AJKA, occurs in (1) 26 times, (2) 26 times.
Four similar characters sequence, in a row, occurs in both (1) and (2). MSDA. The number of character before the four characters, 66 in (1) and 66 in (2) (or seventeen, from the end, 83-66 = 17)
Notice the amount of repeated occurence is two digits, is p= 1/2^2 for single repeated instance?
I am interpreting this experiment as "The probability of getting exactly k successes in n independent trials"
n!/k!(n-k)! x p^k (1-p)^n-k
Assuming p is defined by the character range of (26 possibilities the alphabet character range) the probability would be very small. Instead I am interpreting as p= (1/2)^2.
We have 5 outcomes, with 5 successive trials; Thus n!/k!(n-k)! = 1 and
p^k (1-p)^n-k = 0.25^5*(1-0.25)^0 =0.000732.
The probability of this occuring randomly is 0.07 % is this significant? Or is there some other way to quantify it?
There are two alpha-bet sequences. Both of them includes number a number of elements
1st alphabet sequence[separated by a comma] ['XX','XX','XX'...] (1) ;
2nd alphabet sequence ['XX','XX','XX'...] (2)
Both have totally, 19 elements. Both have 83 characters.
Both are associated with two different character alphabet sequence (randomly selected from a-z): CCCC . The first sequence, ASDS,occurs in (1) 27, and in (2) 27 times.
The second sequence of randomly generated alphabet, AJKA, occurs in (1) 26 times, (2) 26 times.
Four similar characters sequence, in a row, occurs in both (1) and (2). MSDA. The number of character before the four characters, 66 in (1) and 66 in (2) (or seventeen, from the end, 83-66 = 17)
Notice the amount of repeated occurence is two digits, is p= 1/2^2 for single repeated instance?
I am interpreting this experiment as "The probability of getting exactly k successes in n independent trials"
n!/k!(n-k)! x p^k (1-p)^n-k
Assuming p is defined by the character range of (26 possibilities the alphabet character range) the probability would be very small. Instead I am interpreting as p= (1/2)^2.
We have 5 outcomes, with 5 successive trials; Thus n!/k!(n-k)! = 1 and
p^k (1-p)^n-k = 0.25^5*(1-0.25)^0 =0.000732.
The probability of this occuring randomly is 0.07 % is this significant? Or is there some other way to quantify it?
