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We follow the intersection of column n = 5 with row k = 4 and find the maximal value of cij being 2. This means we will choose the lines of the system such that any two of them will not contain more than 2 common numbers. Here is a simple example of a 10-line system in which this condition is fulfilled: V1: 1 3 5 7 9 V2: 10 12 5 V3: 19 21 23 V4: 27 29 31 V5: 36 38 40 V6: 45 47 49 V7: 54 51 50 V8: 46 44 42 V9: 36 35 34 V : 31 30 29 10 16 18 25 26 33 35 42 44 51 53 49 48 39 37 33 32 28 27 Any two lines of the previous common numbers; thus the exclusiveness condition is satisfied, and thus the system’s winning probability, at least at the third category, grows linearly with the number of lines: 10 PV4 =10P(V4 ) =10⋅0.0000722 = 0.0722% i1 i=1 system contain no more than 2 For the very next lower winning category (minimum 3 winning numbers), we follow the intersection of column n = 5 with row k = 3 and find the maximal value of cij being 0. This means that any two variants should contain no common numbers, and the system must be built accordingly. (In the calculations of both previous examples, we used numerical probability results from the section titled Cumulated winning probabilities.) 34PDF Image | THE MATHEMATICS OF LOTTERY Odds, Combinations, Systems
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