Different Patterns

Investigating the availability of information as well as the effectiveness to find - or to learn - them we have until now handled every possible pattern $g \in$$\cal{G}$ as having equal probability $1/2^l$. If one looks to specific patterns like '00' or '11' in a set of binary strings with length $l=2$ then each of these patterns has the probability $1/(2^2) = 1/4$ or according to the probability theory with the axioms of Kolmogorov (cited in [52]:132f) the events $'00' \vee '11'$ have the probability $E2 = 1/4 + 1/4 = 1/2$ while the events $E1 = '10' \vee '01' \vee '00' \vee '11'$ - abbreviated as '**' - have the probability of $1/4 + 1/4 + 1/4 + 1/4 = 1$ The lower probability of E2 follows directly from the fact that $E2 \subset E1$. An example is given below for $n=4, N=256$. The decimal value corresponds to the strings $0 := '00', 1 := '01', etc.$

-->l=2,n=4,N=256,show=0,[DIST2,STD, MEAN, FREQ,STD1, MEAN1, FREQ1,FX,
POPX]=frequCheck(l,n,N,show)

Number of Events n * N = 1024

    DEC   OCC     FREQ         FREQ [%] IDEAL
    -----------------------------------------
    0.    257.    0.2509766    100.39063  
    1.    257.    0.2509766    100.39063  
    2.    253.    0.2470703    98.828125  
    3.    257.    0.2509766    100.39063

If one looks to the event of compound strings like '00' or '01' as a process starting with one element, then concatenating another one, etc, then we have the situation that every time we have the decision between '0' or '1' with a probability of $1/2$ each. If it does not matter which element follows then we have always $1 * 1 = 1$. If it does matter which elements should be concatenated then we have $1/2 * 1/2 = 1/4$. Thus starting with '1' or '0' yielding '00' or '01' or '10' or '11' - abbreviated as '**' - has the probability of '1'. To produce from '1' the event '11' or from '0' the event '00' has the probability $1/2 * 1/2 = 1/4$ in every case.

If we would continue with such a production we would go from '**' to '***' with $1 * 1 * 1 = 1$ or from '00' to '000' with $1/2 * 1/2 * 1/2 = 1/8$ or from '11' to '111' with $1/2 * 1/2 * 1/2 = 1/8$

Thus the general formula for the probability of a certain string can be given as $p(e)_{1} \times p(e)_{2} \cdots p(e)_{l}$ where $p()_{i}$ is the probability of the event within the space of possible events.

A string like $11*0*$ has then the probability $1/2 * 1/2 * 1 * 1/2 *1 = 1/8$ (decimal 0.125), because the string $11*0*$ is represented by $11101 = 26$ or $11100 = 28$ or $11001 = 25$ or $11000 = 24$. Thus we have $1/32 + 1/32 + 1/32 + 1/32 = 1/8 = 0.125$.

-->l=5,n=32,N=64,show=0,[DIST2,STD, MEAN, FREQ,STD1, MEAN1, FREQ1,FX,
POPX]=frequCheck(l,n,N,show)
   
Number of Events n * N = 2048

    0.     59.    0.0288086    92.1875   
    1.     47.    0.0229492    73.4375   
    2.     63.    0.0307617    98.4375   
    3.     50.    0.0244141    78.125    
    4.     64.    0.03125      100.      
    5.     76.    0.0371094    118.75    
    6.     64.    0.03125      100.      
    7.     64.    0.03125      100.      
    8.     59.    0.0288086    92.1875   
    9.     72.    0.0351563    112.5     
    10.    56.    0.0273438    87.5      
    11.    64.    0.03125      100.      
    12.    69.    0.0336914    107.8125  
    13.    74.    0.0361328    115.625   
    14.    63.    0.0307617    98.4375   
    15.    68.    0.0332031    106.25    
    16.    62.    0.0302734    96.875    
    17.    74.    0.0361328    115.625   
    18.    70.    0.0341797    109.375   
    19.    76.    0.0371094    118.75    
    20.    61.    0.0297852    95.3125   
    21.    54.    0.0263672    84.375    
    22.    76.    0.0371094    118.75    
    23.    61.    0.0297852    95.3125   
    24.    51.    0.0249023    79.6875   
    25.    66.    0.0322266    103.125   
    26.    64.    0.03125      100.      
    27.    63.    0.0307617    98.4375   
    28.    66.    0.0322266    103.125   
    29.    51.    0.0249023    79.6875   
    30.    69.    0.0336914    107.8125  
    31.    72.    0.0351563    112.5

Therefore we can compute the probability of certain subsets $\cal{G}$$^{i}$ $\subseteq \cal{G}$. If we assume that a goal set $\cal{G}$$^{*}$ is a true subset of $\cal{G}$ then the probability to find this set is a special probability.