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加拿大28分析“不对称”原则你掌握了吗

2021年04月01日

  在加拿大28指标分析中,“偏斜”表示指标的中点在很长一段时间内不会在指标的统计表中对称。分类号1的大数或中号或十进制数都是独立的指标。在3D等级或等级3统计信息表中,您可以看到对称性。例如,有时分类号1的十进制数会出现在统计信息表中。状态是非常常规的:遗漏,遗漏——遗漏,遗漏,如果发生这种情况,则判断它们将是下一个遗漏间隔中的遗漏,错误遗漏,形成对称的遗漏,遗漏,遗漏——遗漏,遗漏——遗漏和遗漏。

  In indicator analysis, "skew" means that the midpoint of the indicator will not be symmetrical in the statistical table of the indicator for a long time. The large number, medium number or decimal number of classification number 1 are independent indicators. You can see symmetry in the 3D level or level 3 statistics table. For example, sometimes the decimal number of classification number 1 appears in the statistics table. The state is very conventional: omission, omission-omission, omission. If this happens, it will be judged that they will be the omission in the next omission interval, wrong omission,

  该分析是否有理论依据?答案是不。实际上,这种判断只是一个巧合。根据历史统计,加拿大28“不对称”现象的发生次数比“对称”现象的发生次数高得多。例如,通过排序统计表的统计信息可以看出,如果有序数的大量索引丢失了1个周期,则该大数会被正确丢弃,然后在丢失2个周期后,大数是正确的并且消失了,然后连续3个周期后,大数丢失了。该帐户仍然正确。因此,如果跳过连续的4个周期,大量数字能否正确显示?

  遗漏1期∶遗漏—正确中出;

  遗漏2期∶遗漏—遗漏—正确中出;

  遗漏3期∶遗漏—遗漏—遗漏—正确中出;

  遗漏4期∶遗漏—遗漏—遗漏—遗漏—正确中出?

  我们认为在错过4个周期后很难正确退出。为什么?因为此时省略周期1,省略周期2和省略周期3是对称增加的现象。根据加拿大28指标的“非对称”原理,省略了4个点,中间刚好退出的可能性非常小。

  同样,如果每次校正的状态完全相同,则这是对称偏移现象,也可以完全消除。例如:纠正纠正纠正——纠正纠正纠正——纠正,纠正,纠正,纠正,纠正,纠正,纠正,纠正,纠正,纠正,连续3次,然后在遗漏后不太可能再次纠正3次时期。因此,尽管在实际的抽奖中可能出现对称性发展,但是这种状态小于不对称性的可能性。我们不能仅仅因为指标在某个阶段对称地发展而在所有期望中追求对称。

  一般而言,加拿大28规则图案出现的可能性比不规则图案出现的可能性要低得多。彩票中奖号码分布不均是最大的法则!因此,一旦出现三个以上对称发展,就必须第四次坚决排除这种情况。

  symmetrical omission, omission, omission-omission, omission-omission and omission. Is there a theoretical basis for this analysis? The answer is no. In fact, this judgment is just a coincidence. According to historical statistics, the frequency of "asymmetry" is much higher than that of "symmetry". For example, it can be seen from the statistical information of the sorting statistics table that if a large number of indexes of an ordered number are lost for one cycle, the large number will be discarded correctly, then after two cycles are lost, the large number is correct and disappears, and then after three consecutive cycles, the large number is lost. The account is still correct. Therefore, if four consecutive cycles are skipped, can a large number of numbers be displayed correctly? Missing Phase 1: Missing-Right Out; Missing 2 issues: missing-missing-correct; Missing 3 periods: missing-missing-missing-correct; Missing 4 periods: missing-missing-missing-missing-missing-correct? We think it is difficult to quit correctly after missing 4 cycles. Why? Because cycle 1 is omitted at this time, cycle 2 and cycle 3 are symmetrically increasing phenomena. According to the "asymmetry" principle of indicators, four points are omitted, and the possibility of just exiting in the middle is very small. Similarly, if the state of each correction is exactly the same, it is a symmetrical offset phenomenon, and it can be completely eliminated. For example, correct, correct, correct, correct, correct, correct, correct, correct, correct, correct, correct, correct, correct, for three consecutive times, and then it is unlikely to be corrected three times after omission. Therefore, although symmetry may develop in the actual lottery, this state is less than the possibility of asymmetry. We can't pursue symmetry in all expectations just because indicators develop symmetrically at a certain stage. Generally speaking, the probability of regular patterns is much lower than that of irregular patterns. Uneven distribution of lottery winning numbers is the biggest rule! Therefore, once there are more than three symmetrical developments, this situation must be resolutely ruled out for the fourth time.


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