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We follow the intersection of column n = 6 with row k = 4 and find the maximal value of cij to be 1. This means we will choose the lines of the system such that any two of them will not contain more than 1 common number. Here is a simple example of a 11-line system in which this condition is fulfilled: V1: 1 2 3 4 5 6 V2: 1 7 8 9 10 11 V3: 12 7 13 14 15 16 V4: 17 18 13 19 20 21 V5: 22 23 24 19 25 26 V6: 27 28 29 30 25 31 V7:3233343536 1 V8:3738394041 1 V9:3742434445 2 V : 46 47 48 49 45 3 10 V : 12 18 24 30 36 4 11 Any two lines of the previous system contain no more than 1 common number; thus the exclusiveness condition is satisfied, and thus the system’s winning probability, at least in the third category, grows linearly with the number of lines: 11 PV4 =11P(V4 ) =11⋅0.000987 =1.0857% i1 i=1 We want to build a playing system for the 5/55 lottery which will ensure the linearity of the winning probability for at least the second winning category (minimum 4 winning numbers). What is the maximal number of common numbers any two lines of the system may contain? How about for the third winning category (minimum 3 winning numbers)? 33PDF Image | THE MATHEMATICS OF LOTTERY Odds, Combinations, Systems
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