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The solution is to buy a total of 14 tickets, comprised of two sets of seven. One set of seven is every line of a Fano plane with the numbers 1-7, the other with 8-14, i.e.: 1-2-5, 1-3-6, 1-4-7, 2-3-7, 2-4-6, 3-4-5, 5-6-7, 8-9-12, 8-10-13, 8-11-14, 9-10-14, 9-11-13, 10-11-12, 12-13-14 Because at least two of the winning numbers must be either high (8-14) or low (1-7), and every high and low pair is represented by exactly one ticket, you would be guaranteed at least two correct numbers on one ticket with these 14 purchases. 21/26 of the time you will have one ticket with two numbers matched. If all three winning numbers are either high or low you would either have one ticket with all three numbers (1/26 chance of this occurring), or three different tickets that each matched two (4/26 chance). EXTERNAL LINK
I think that this theory is wrong, for it is impossible to crack a lottery system even if it seems so obvious. |
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