Examples A, B, and C can be easily incorporated into a binomial model to assess how improbable it would be for the results to occur purely by chance.
(A). The Sun’s (sidereal) rotation period = 24.66225 Earth days.
TWICE The CUBE of this number =
24.662253 x 2 = 30,000.47
(B). The Sun’s synodic rotation period (as viewed form Earth) = 26.44803 Earth days. TWICE The CUBE of this number =
26.448033 x 2 = 37,000.7
(C). The Sun’s synodic rotation period, as viewed from Mars = 24.99931 Mars sidereal days.
TWICE The CUBE of this number =
24.999313 x 2 = 31,000.59
It is fairly obvious that this cannot happen THREE times in a row from exactly the same operation each time. In fact, the statistical odds against just these above three results alone happening purely by chance are ONE CHANCE IN TWENTY TWO THOUSAND!
To see how these odds are calculated, see below:-
Examples A, B, and C (above) can be easily incorporated into a binomial model to assess how improbable it would be for the results to occur purely by chance.
In a binomial model there are 4 variables. These are:-
p = the probability of a successful outcome in one single specific trial.
q = the probability of an unsuccessful outcome in one single specific trial.
n = the number of trials
r = the number of trials that are successful.
To calculate p, we ask the following question:-
Taking the three numbers 30,000.47 and 37,000.7 and 31,000.59 which of these three numbers are FURTHEST from a perfect multiple of A THOUSAND?
The answer is 37,000.7 This number deviates from a perfect multiple of A THOUSAND by 0.7
p = the probability that any single specific randomly generated number will be either 0.7 greater than or 0.7 less than a perfect multiple of A THOUSAND. In that case, p = (0.7 x 2) ÷ 1000 = 0.0014
q = the probability of an unsuccessful outcome in one single trial = (1 minus p) = 0.9986
r = the number of trials that are successful. There are 3 successful trials, and therefore r = 3
Now we have to decide on the value for n, where n = the number of trials.
We will place certain restrictions on the “trial set”. We will only involve The Four Inner Solar System planets. We will only multiply the result by 1 or by 2. We will only raise numbers (periods) to the second or the third power.
The Sun’s sidereal rotation period can be expressed in Earth solar days, or in Earth sidereal days. (2 options).
The Sun’s synodic rotation period, as viewed from Mercury, can be expressed in Earth solar days, or in Earth sidereal days. (2 options).
The Sun’s synodic rotation period, as viewed from Venus, can be expressed in Earth solar days, or in Earth sidereal days. (2 options).
The Sun’s synodic rotation period, as viewed from Earth, can be expressed in Earth solar days, or in Earth sidereal days. (2 options).
The Sun’s synodic rotation period, as viewed from Mars, can be expressed in Earth solar days, or in Earth sidereal days, or in Mars solar days, or in Mars sidereal days. (4 options). (Note:- This is because Earth and Mars are fast rotating planets with very short days, but Mercury and Venus are slow rotating, with very long days. In that case, we cannot express any of these periods in Mercury days, or in Venus days.)
So far, we have 12 options.
All of the above options are periods which can be either squared or cubed.
So far, we have 12 x 2 = 24 options.
All of the above options are numbers which can be multiplied either by 1 or by 2.
We now have 24 x 2 = 48 options.
In that case, n = 48.
Now we can incorporate these values into a binomial model.
p(r ≥ 3) = 48C3 x 0.998645 x 0.00143 = 0.00004456
Or statistical odds against chance occurrence of 1 chance in (1 ÷ 0.00004456) = 1 chance in 22,441
ie:- ONE CHANCE IN TWENTY TWO THOUSAND!!
48C3 means the number of possible combinations of 48 items, taken three at a time. This is used to “dilute” the odds. If you did not include this “dilution factor”, you would have odds of one chance in 364 million!
The term 0.998645 also “dilutes” the odds further still. This is the way that a binomial model works.