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Estimation of the probabilities

Method I

As I mentioned above, Table 1 containing the results obtained from intersections with ISRAEL strikes even with its appearance alone. It is easy to calculate the odds of all the 10 numbered states above the line drawn by the number of letters in the Torah to be safe. The number of the non-shaded states is 13; the number of the states shaded in dark is 15. This makes the probability, p1, of 10 items of the first kind to occupy the first 10 positions out of 28 as follows:

                 13       12       11     10       9         8        7         6       5         4

P1 = ——?——?——?——?——?——?——?——?——?—— = 1:45885 =

                 28       27       26      25      24      23       22       21     20      19

= 2.17936?10^-5 ≈ 0.002%.

The probabilities were calculated taking into account only the consecutive occurrences of states of one kind. The phenomenon depends also on the number of the letters in the Torah. So it must be kept in mind that the calculation of the real probability should take into account also the probability of no state of the other kind occupying the space remaining between the last of the 10 states, America (265,620), and the number of the letters in the Torah (304,805), i.e. no “risky” state falling into the frame marked out by the latter number. This should lower the probability even further. The simplest way to do that is to regard the Number of letters in the Torah as the eleventh safe state, which yields

P_{Method I} = 1:95,048, or 0.00105%.

Even with the three Baltic States, the according calculation for 10 + 1 states out of 13 + 18 = 31 states occupying the first 11 positions gives

P2 = 1:16448 = 6.0799?10^-5 ≈ 0.006%.

The Baltic States influence in the approximately same manner all calculations, so they were excluded from the further calculations.

With the states that do not appear in the Torah excluded from calculation, 12 states of each kind will remain and the probability will be

P3 = 1:29716 = 3.3652?10^-5 ≈ 0.0034%.

The same calculations carried out for the results with the Sons of Israel, Table 3, produce the following probabilities:

P1? = 1:17.3 = 0.057746 ≈ 5.8%;

The increase of probability is about 2,650 times in relation to P1 but more than 5,500 times in relation to P_{Method I}!

There is a tremendous increase in probabilities when passing from ISRAEL to SONS OF ISRAEL, the latter word being just an “elongation” of the former one!

Method II

Another approach produces even lower probabilities. This approach sets directly the Number of letters in the Torah, 304,805, as a parameter. Let the whole one-dimensional space of P-s is divided into two compartments: values less and more than 304,805, respectively. A fair consideration, in my opinion, should exclude those states that have no chance to fall in the first compartment. So, the calculation should be limited only to the states in Table 2 save Greece.[1] States having List numbers could be divided into two groups: twelve “white” and eleven “black” items. The calculation can be performed for tossing two sets of coins: what is the probability of ten out of twelve “white” coins falling heads up and the remaining two falling tails up, together with all eleven “black” coins falling tails up[2]. This probability is

                                         132

P = —— = 1:127,100 = 7.87?10^-6 = 0.00079%.

                                          2²⁴

If the calculation of the probability takes into account the condition of not more than two white coins fall tails up and all black coins fall tails up, the overall probability will be a little bit higher:

                                   132       12         1

P_{Method II} = —— + —— + —— = 1:106,185 = 9.42?10^-6 = 0.00094%.

                                    2²⁴       2²³        2²³

This is in a very good agreement with the probability calculated according to Method I.

If a similar calculation is carried out with all 28 states assigned with List numbers in Table 1, the odds will drop down to almost 1 in 1,000,000!

With the Sons of Israel, the probabilities rise respectively to ≈ 1:40 = 2.5% and ≈ 1:50 = 2.0%. These results apparently bring the intersections with SONS OF ISRAEL to the brink of a phenomenon. The overall probability P’_overall, however, calculated under the condition of not more than four white coins fall tails up and not more than four black coins fall heads up, increases up to about 1:20, or 5 %. This value is over 5,000 times higher than the corresponding probability for the intersections with ISRAEL. This value is also in good agreement with the probability (about 5.8%) obtained using Method I.

This approach, however, is mathematically reasonable only if a substantial part of the combined number of both states fall into one of the compartments. The ideal case is when the borderline divides the list into two. Otherwise it cannot be assumed that the probability of falling into any one of the compartments is ?, i.e., that the coin is “fair”. (Consider the probability of, say 27 out of 28 coins falling on tails, as, most probably, is the case with ABRAHAM, Table7). This requirement is met only in Tables 1 and 3.

All other probabilities (Tables 4 to 10) do not disclose anything extraordinary. States are spread randomly above and below the corresponding lines and do not show any sig-nificant order of preference among either kind in the tables, compared to that in Table 1.

I do not take upon myself to estimate the probability of the encoded number 933 mentioned above to be there by chance. In any case, the probability, especially if combined with the other probability, is all too obvious millionths if not tens of millionths.

[1] And USSR too. Remind you that Russia has been used in the calculations of the probabilities.

[2] The coins are considered to be similar in every way except a mark – in our case a colour – that by no means affects the ? probability for falling on each side.