the best way to eliminate lottery numbers 100/20?
- Thread starter jack
- Start date
Frank
Member
If I understand this lottery correctly, to choose a winner, they draw 20 balls numbered 1 to 100. This means the odds against a single bet winning are :- 535,983,370,403,809,682,970 to 1 ! That's 535 Quintillion to 1, or 535 times 10 to the power of 20, to 1.
if you are allowed to choose 50 balls in your bet, then you can cover 47,129,212,243,960 lines of 20 balls or 47 trillion combinations of 20 balls. That's 4.7 times 10 to the power of 13 lines. So that brings the odds down to 5.3/4.7 times 10 to the power of 7, to 1 or approximately 11 million to 1.
How expensive is it to bet 50 numbers, bearing in mind it's still 11 million to 1 if you do ?
Id be inclined to eliminate the whole lottery and try something easier, if it's expensive.
hello, Frank, congratulations grateful to enjoy the post, but this right, are 11 million,
The awards are set = = 15,16,17,18,19,20
are 6 tracks awards
frank this lottery'm noticing this pattern,
of polindromos, eg 23,32, or 89, 98
never is drawn pair them together or is drawn 23 or 32
never 98. 89, and so with the 45 possible pairs of polindronos, we have 10 pairs gemeas 00-99,
as conbinar without the pair ?? care google translate bad
The awards are set = = 15,16,17,18,19,20
are 6 tracks awards
frank this lottery'm noticing this pattern,
of polindromos, eg 23,32, or 89, 98
never is drawn pair them together or is drawn 23 or 32
never 98. 89, and so with the 45 possible pairs of polindronos, we have 10 pairs gemeas 00-99,
as conbinar without the pair ?? care google translate bad
Frank, this example is an official draw of the managing
the average is 0 to 3 double polindromos
using the standard?
example if I play 45,26,78,01,15,59 ..... can not play
54 62 78 10 51 95 ....
but when fixed = 45,26,78,01,15,59., and are drawn 26,51,01, 45,78,15,59 the digits are not randomly chosen not throw
54 87 51 59, but the draw is drawn 54 87
= conclusion must set the correct groups of numbers? This is the question!!
the average is 0 to 3 double polindromos
using the standard?
example if I play 45,26,78,01,15,59 ..... can not play
54 62 78 10 51 95 ....
but when fixed = 45,26,78,01,15,59., and are drawn 26,51,01, 45,78,15,59 the digits are not randomly chosen not throw
54 87 51 59, but the draw is drawn 54 87
= conclusion must set the correct groups of numbers? This is the question!!
frank, below are the last 23 official sweepstakes
the question how to use the pair of polindromos correctly?
23 draws official =
01 04 10 13 17 40 43 45 50 56 59 60 63 67 70 76 78 84 89 92
02 04 17 26 49 50 55 59 62 65 67 68 69 70 76 81 84 86 89 95
01 05 08 11 13 14 19 36 39 45 54 65 71 74 77 83 90 91 92 98
05 10 18 24 27 33 34 42 44 47 48 58 71 79 80 81 86 93 96 98
03 08 09 14 17 28 29 30 31 32 39 52 60 63 69 77 82 88 94 00
03 07 10 14 15 16 19 22 34 37 39 42 49 51 53 60 76 82 84 98
12 13 15 40 47 52 53 54 60 62 68 69 71 75 78 80 81 85 91 00
06 10 19 21 30 33 35 52 53 54 64 67 73 74 82 90 91 92 95 98
07 17 22 24 32 37 39 40 49 52 54 60 63 68 84 85 88 92 93 97
09 12 18 21 34 35 38 42 43 48 50 54 59 71 79 87 88 90 92 96
09 16 36 39 42 51 53 57 61 62 67 68 71 77 78 79 83 85 89 98
03 05 07 10 14 26 29 34 42 43 46 49 50 66 69 71 73 76 79 82
02 05 06 16 25 28 31 33 35 40 54 57 64 70 77 78 82 88 93 99
03 10 15 28 32 36 41 43 47 52 53 55 63 66 67 68 72 77 86 90
16 20 36 39 45 47 49 52 54 56 57 62 70 74 78 81 89 93 95 98
02 04 16 19 26 28 29 30 45 59 61 66 68 72 76 77 85 86 96 99
01 08 15 18 19 21 22 34 41 50 52 53 55 61 67 68 72 78 88 92
03 05 08 18 20 27 29 41 45 50 60 61 62 69 71 78 87 88 94 96
06 07 09 11 19 21 25 38 39 42 50 51 59 61 62 67 68 96 98 99
01 09 17 22 31 33 37 43 47 55 62 66 67 68 70 77 83 92 95 99
07 09 10 15 19 29 31 36 37 44 56 58 67 72 81 88 92 93 95 00
09 12 14 17 19 20 22 23 33 34 39 42 68 71 72 76 82 87 96 97
02 04 05 07 13 23 24 28 30 35 43 44 52 54 58 65 77 90 94 95
the question how to use the pair of polindromos correctly?
23 draws official =
01 04 10 13 17 40 43 45 50 56 59 60 63 67 70 76 78 84 89 92
02 04 17 26 49 50 55 59 62 65 67 68 69 70 76 81 84 86 89 95
01 05 08 11 13 14 19 36 39 45 54 65 71 74 77 83 90 91 92 98
05 10 18 24 27 33 34 42 44 47 48 58 71 79 80 81 86 93 96 98
03 08 09 14 17 28 29 30 31 32 39 52 60 63 69 77 82 88 94 00
03 07 10 14 15 16 19 22 34 37 39 42 49 51 53 60 76 82 84 98
12 13 15 40 47 52 53 54 60 62 68 69 71 75 78 80 81 85 91 00
06 10 19 21 30 33 35 52 53 54 64 67 73 74 82 90 91 92 95 98
07 17 22 24 32 37 39 40 49 52 54 60 63 68 84 85 88 92 93 97
09 12 18 21 34 35 38 42 43 48 50 54 59 71 79 87 88 90 92 96
09 16 36 39 42 51 53 57 61 62 67 68 71 77 78 79 83 85 89 98
03 05 07 10 14 26 29 34 42 43 46 49 50 66 69 71 73 76 79 82
02 05 06 16 25 28 31 33 35 40 54 57 64 70 77 78 82 88 93 99
03 10 15 28 32 36 41 43 47 52 53 55 63 66 67 68 72 77 86 90
16 20 36 39 45 47 49 52 54 56 57 62 70 74 78 81 89 93 95 98
02 04 16 19 26 28 29 30 45 59 61 66 68 72 76 77 85 86 96 99
01 08 15 18 19 21 22 34 41 50 52 53 55 61 67 68 72 78 88 92
03 05 08 18 20 27 29 41 45 50 60 61 62 69 71 78 87 88 94 96
06 07 09 11 19 21 25 38 39 42 50 51 59 61 62 67 68 96 98 99
01 09 17 22 31 33 37 43 47 55 62 66 67 68 70 77 83 92 95 99
07 09 10 15 19 29 31 36 37 44 56 58 67 72 81 88 92 93 95 00
09 12 14 17 19 20 22 23 33 34 39 42 68 71 72 76 82 87 96 97
02 04 05 07 13 23 24 28 30 35 43 44 52 54 58 65 77 90 94 95
Frank
Member
Firstly, if it costs more than $10 to bet 50 numbers, I would not play the lottery. The cost to odds ratio would be too great. If you believe that there will be no palindromes in a row of 20 , then choose whatever numbers you think are due whilst avoiding palindromes. Simple method.
There must be billions of combinations which satisfy the rule, so I cannot help you any more than that. You could do worse than using random.org to generate some integer sets, then check them for palindromes , rejecting numbers as required.
Good luck!
There must be billions of combinations which satisfy the rule, so I cannot help you any more than that. You could do worse than using random.org to generate some integer sets, then check them for palindromes , rejecting numbers as required.
Good luck!
Frank
Member
Jack, I think you are asking the wrong person. I have no interest in deltas, I don't see the point of them. They do not reduce the odds against winning. When you have 20 numbers to pick there are a lot more deltas to get correct in both value and position, at astronomical odds, making them even less useful.
Apart from that , I really don't fancy having my PC cycle through 535,983,370,403,809,682,970 combinations of 20 numbers doing the calculations, as I'd like to use it for other things for the rest of the year, I doubt it would survive the ordeal. Sorry.
Apart from that , I really don't fancy having my PC cycle through 535,983,370,403,809,682,970 combinations of 20 numbers doing the calculations, as I'd like to use it for other things for the rest of the year, I doubt it would survive the ordeal. Sorry.