Pure Chance with no Inheritance

In the case of 'pure chance without inheritance' the function $\gamma$ has to be organized as a sequence of substitutions where there is no connection between two succeeding generations $\cal{G}$$_{i}$, $\cal{G}$$_{i+1}$. Every generation is generated by random selection from the set of all possible states $Ps(G)$ given as the set of all binary strings with length $l$, written as $Ps(G) = 2^l$ or abbreviated $K = 2^l$ .

The probability to select a certain string $s \in Ps(G)$ is with classical probability assuming equal conditions for every possible value $1/K$. According to Bernouli's Ars conjectandi (1713) ([27]:IV.4, cited in [104]:29), holds the law, that increasing the number $N$ of events towards infinity will produce enough events for each kind of possible value. This means every possible value will approach the limit $N^{\infty} \times (1/K)$ (with $N^{\infty}$ going towards infinity).

With regard to the fundamental question whether this procedure can 'in principle' find all possible values $\cal{G}$ $^{*} \subseteq \cal{G}$ one can say that random selection can select any possible value out of $\cal{G}$. The open question is the minimal size for a set $\cal{G}$ $^{\emptyset}$ in the beginning of such a process that all important values can be included simultaneously.

Within the fact that in principle all possible values can be found there remains the question of the quality of the intermediate sets $\cal{G}$$_{i}$. Measuring the quality of each intermediate set $\cal{G}$$_{i}$ one can define the velocity of approaching the solution set $\cal{G}$ $_{i} \subset \cal{G}$$^{*}$by measuring the number of necessary selections.

To answer these questions some empirical results will be given first. We will start with the following examples ($n$ := number of individuals in the set$\cal{G}$ $^{\emptyset}$; $N$ := number of selections with n elements each): (i) n=1,N=32, (ii) n=32, N=32, (iii) n=1, N=32x32.

In these examples the function 'freqCheckRep(l,n,N,K)' computes for a population with $n$ many members of binary strings with length $l$ and $N$-many events with $K$-many repetitions the frequency of each individual converted into the percentage of the ideal values. From each run with $N$ events the mean values $MEAN0$ is computed as well as the half of the mean difference between the maximum and the meinimum of these frequencies.

This is illustrated in the example below for a population with $n=1$ member and $N=32$ events. The 1st row shows the decimal value of a binary string between [0,31]. The 2nd row shows the frequency of a value during $N=32$ events. The 3rd row shows the percentage of these frequencies compared to the ideal values according to Bernoulli's formula. Thus in this one run with $N=32$ events we get a mean values $MEAN0$ of $150$ and the half distance $DIST2$ of $150$.

-->L=5, n=1, N=32, [DIST2,STD, MEAN, FREQ,FX,
POPX]=frequCheck(l,n,N) 

    0.     1.    100.  
    1.     2.    200.  
    2.     1.    100.  
    3.     0.    0.    
    4.     0.    0.    
    5.     0.    0.    
    6.     0.    0.    
    7.     2.    200.  
    8.     1.    100.  
    9.     2.    200.  
    10.    1.    100.  
    11.    0.    0.    
    12.    3.    300.  
    13.    3.    300.  
    14.    2.    200.  
    15.    2.    200.  
    16.    0.    0.    
    17.    0.    0.    
    18.    0.    0.    
    19.    1.    100.  
    20.    1.    100.  
    21.    1.    100.  
    22.    0.    0.    
    23.    0.    0.    
    24.    0.    0.    
    25.    0.    0.    
    26.    3.    300.  
    27.    0.    0.    
    28.    2.    200.  
    29.    0.    0.    
    30.    1.    100.  
    31.    3.    300.  
 FREQ  =
 
    300.    4.   
    200.    6.   
    100.    8.   
    0.      14. 
 
 MEAN  =
 
    150.  

 STD  =
 
    107.76318  

 DIST2  =
 
    150.

If one repeats such a run $K=10$-many times then one gets the following results:

-->l=5, n=1, N=32, K=10,[MEANX, DIST2X, DIST20,
MEAN0]=freqCheckRep(l,n,N,K)

 MEAN0  =
 
    167.5  

 DIST20  =
 
    200.  

 DIST2X  =
 
    150.  
    150.  
    150.  
    150.  
    200.  
    150.  
    200.  
    200.  
    150.  
    200.  

 MEANX  =
 
    150.  
    150.  
    150.  
    150.  
    175.  
    150.  
    200.  
    200.  
    150.  
    200.

This example shows that the values in case of $n=1$ are vary 'unstable' covering a range with deviations of 150 - 200% above and below the 'mean'.

Increasing the numbers either with regard to the events $N=32^2$ or $n=32$ shows some improvements.

-->l=5, n=32, N=32, K=10,[MEANX, DIST2X, DIST20,
MEAN0]=freqCheckRep(l,n,N,K)

 
 MEAN0  =
 
    100.65639  

 DIST20  =
 
    32.8125  

 DIST2X  =
 
    37.5     
    40.625   
    31.25    
    37.5     
    35.9375  
    29.6875  
    39.0625  
    29.6875  
    46.875   
    32.8125 
 
 MEANX  =
 
    100.86806  
    103.32031  
    98.125     
    104.91071  
    101.75781  
    97.65625   
    98.784722  
    97.851563  
    101.15132  
    102.13816 

-->l=5, n=1, N=32*32, K=10,[MEANX, DIST2X, DIST20,
MEAN0]=freqCheckRep(l,n,N,K)

 MEAN0  =
 
    101.38233  

 DIST20  =
 
    32.8125  

 DIST2X  =
 
    28.125   
    31.25    
    35.9375  
    28.125   
    34.375   
    39.0625  
    34.375   
    34.375   
    31.25    
    32.8125 
 
 MEANX  =
 
    102.91667  
    101.38889  
    102.43056  
    103.95833  
    100.78125  
    99.826389  
    101.5625   
    102.02206  
    99.804688  
    99.131944

While the mean values are in these examples close to the 'ideal' of 100% the deviation with about 30% above and below the mean values is still vary high. The interesting question is how 'fast' the deviations will approach zero or some $\epsilon > 0$. For this we have another empirical experiment.