Does your odds increase tremendously by millions just like this? Read for details.

Reply #29

dr san — Thu, 11 Dec 2014 16:52:03 GMT

Hello, metro, the lottery 49/6 decades we have 5 or 4 groups of 10 numbers and a group with 9 numbers, sometimes if we have 5 groups then is are drawn 6 numbers of the groups always have at least two numbers that is a pair, one could also filter by position by position vertically using the temina oes

example

12,15,25,36,45,49

1st place 12 Termination (last digit) is 2 = filter their adjacent 1.3, your mirror 7 (see table)

Then himself in position = 2, 1.3 and its adjacent mirror 7

Reply #28

LottoMetro — Thu, 11 Dec 2014 15:28:00 GMT

Well, there's a tricky line between playing numbers you think have shown up too much or too little and those that actually have. I don't really play draw games much anymore but when I do, usually I will perform a chi-square test on the numbers; basically it is a statistical test for bias/independence. There are other and more vigorous tests but this one is the simplest. You come up with an 'expected' number of appearances (basically 6*draws/49), compute the chi-test statistic, and then calculate... [ More ]

Reply #27

newtolottoagain — Thu, 11 Dec 2014 14:20:39 GMT

Regarding the cold number, thats exactly what I'm saying. But I try to stay away from that, in my ToTo lottery there's a big example of why. I noticed the number 35 keeps coming up as a bonus number but never as the main and was wondering if it's due for the main number. I checked and discovered it's been 42 draws since it last appeared (which is the longest of all number, the second due number is 19 draws ago), and not only that, since ToTo began, it has appeared the least times (80 times).... [ More ]

Reply #26

LottoMetro — Thu, 11 Dec 2014 13:16:08 GMT

if there's been 5 draws with no previous numbers, chances are high that it's going to come up in the next. It's not exactly predicting but comes very close. Better than using 2 more random pairs.

Well, if it's any consolation, there has been very loose evidence in academia that 'cold' numbers (basically similar to what you describe) are slightly better to play. This is simply because eventually every number will come out. However, any advantage (if any even exists) is really mitigated by the... [ More ]

Reply #25

newtolottoagain — Thu, 11 Dec 2014 12:57:25 GMT

I understood your explanation of why your probabilities were low and knew your equation was correct, but I realized we were talking about different stuff that's what I was trying to clear up. :D

Yeap, it's actually 1/8 for any 6 group of numbers, not just the previous draw. Actually, I have to admit I sorta blanked that out, because in reality, I don't buy 15 every week, I just do that whenever there's been quite a few draws with no previous numbers. So in a way it gave me an illusion of c... [ More ]

Reply #24

SergeM — Thu, 11 Dec 2014 12:49:26 GMT

K: 1 2

W: 3 4 5 6

KW:

1 2 3 4 5

1 2 3 4 6

1 2 3 5 6

1 2 4 5 6

Outcome:

K score: 0 1 2

W score: 0 1 2 3 4

K: 2 + W: 4 cannot

Probability win over full system, costs and odds when

1 AND 2 are correct and the rest of numbers are in the wheel in any order.

Conclusion:

Chances for a jackpot improve when all the keys are correct.

Chances for a jackpot are none when more than no key is wrong.

Do you improve your odds? Only when your keys are right, then y... [ More ]

Reply #23

LottoMetro — Thu, 11 Dec 2014 12:29:51 GMT

Ohhhhhkay, I think I understand clearly now. I was just about to post that I may have been thinking the wrong thing (I tend to confuse myself times ) because I checked ToTo and another lotto and the data didn't seem to make sense.

What I really should have been saying/thinking all along is that because your chances for any pair are 1 in 7.55 or whatever, then it doesn't matter what pair you choose for the next draw. I basically got hung up on specific pair in specific sequence, which is why p... [ More ]

Reply #22

newtolottoagain — Thu, 11 Dec 2014 11:58:18 GMT

I don't need them to show up back to back, because I'm taking the numbers, no matter what it is, from the first draw.

So I'm not looking for any specific 2 numbers to appear for the first draw, just any number.

Today it came out 12, 15, 20, 33, 45, 49.

Ok, great.

Now I take the 6 numbers and arrange them in pairs, for any combination, so it takes me 15 tickets to get all combinations of 2.

Now the next draw comes. The chance that one of my tickets (based on the previous 6... [ More ]

Reply #21

LottoMetro — Thu, 11 Dec 2014 10:57:43 GMT

1/7.55 = 13.245%. My original calculations I gave the exact number of combinations that match, which is how the lottery comes up with their odds.

You seem to contradict yourself here, if you don't need them to pop up in 2 draws, then why would we need them to pop up in one draw followed by another (1+1=2)? The 1/8 or whatever is for a specific pair to show up in any one draw, not back-to-back or multiple draws. Example: there is a 13.2% chance that 1 and 3 will show up in any drawing. See, th... [ More ]

Reply #20

newtolottoagain — Thu, 11 Dec 2014 10:46:06 GMT

I'm actually not following, isn't 1/8 (or 7.55) 12.5%?

Oh, so you're saying the actual percentage for two numbers to pop up in 2 draws... Ok, I get it now, but well, we don't need them to pop up in 2 draws, we're choosing numbers from the first draw that might pop up in the next, so we only really need to calculate for 1, which is 1/8. Isn't that so

Reply #19

LottoMetro — Thu, 11 Dec 2014 10:03:42 GMT

And I looked it up, the probabilities of matching 2 numbers with your ticket is about 1 in 8. Remember we're talking about 649 lottery.

That's the odds, but nevertheless the answer is identical.

1/8 (technically it's 1 in 7.55) = 0.132 which is the exact same I gave for matching 2 numbers.

And again, matching 2 numbers 2 drawings in a row (or 2 numbers showing up 2 drawings in a row) is simply that number squared. There is a difference between matching 2 numbers, and doing so consecutiv... [ More ]

Reply #18

newtolottoagain — Thu, 11 Dec 2014 09:58:54 GMT

There's no way my state's lottery is bias, it's one of the strictest and non corrupted country in the world. I'm not talking just about my lottery, you can go to any country's lottery history and see the same, every 10 or so draws there's a 2 number match. And I looked it up, the probabilities of matching 2 numbers with your ticket is about 1 in 8. Remember we're talking about 649 lottery.

Infact I just did a test run with a lottery generator, I get 2 numbers right frequently.... [ More ]

Reply #17

LottoMetro — Thu, 11 Dec 2014 09:31:29 GMT

If it were easy, lotteries wouldn't make money 85% of combinations in 6/49 (11,872,042) will match one or no numbers in the drawing.

The probability of any 2 given numbers showing up in a drawing is equivalent to you matching 2 numbers on your ticket.

Every drawing is independent. Thus, the probability of matching 2 numbers two drawings in a row is simply the above probability squared.

For 6/49, there are 1,851,150 possible 2-number combinations. 1,851,150/13,983,816 = 0.132378. This is... [ More ]

Reply #16

newtolottoagain — Thu, 11 Dec 2014 09:10:54 GMT

Are you sure the probability of 2/6 numbers is that low? Looking at the histories of multiple lottos, just in the first page of about 50 draws, there's an average of 10 draws that has 2 or more numbers from the last draw.

Reply #15

LottoMetro — Thu, 11 Dec 2014 01:48:04 GMT

Same if you decide to play 1-5-15-26-41-48. You're only playing 6 numbers and only covering 1 line of possibilities.

My bad, didn't type out enough numbers in my post, but you get the idea.

I think people focus too much on patterns and playing numbers without realizing that the winning ticket is an entire line and not just select numbers. You can believe all you want that last drawing's numbers affects the next, but that doesn't change the fact that there are X possibilities and X-t ways t... [ More ]

Reply #14

LottoMetro — Thu, 11 Dec 2014 01:17:49 GMT

In no way does this strategy improve your odds. More tickets give you more chances, but arbitrarily picking numbers does not. It doesn't matter what numbers you play, because there are still X ways that can lose. If you decide to play 1-2-3-4-5-6, there are still 49-6 = 43 numbers you didn't play and 13,983,815 ways to lose. Same if you decide to play 5-15-26-41-48. You're only playing 5 numbers and only covering 1 line of possibilities.

1 line = 1 combination. So buying 15 lines still leaves... [ More ]

Reply #13

dr san — Thu, 11 Dec 2014 00:21:12 GMT

And here is for a 6/49 lottery, the number of combinations for each of the possible ration of odd to even numbers:

0 odd / 6 even: 134,596

1 odd / 5 even: 1,062,600

2 odd / 4 even: 3,187,800

3 odd / 3 even: 4,655,200

4 odd / 2 even: 3,491,400

5 odd / 1 even: 1,275,120

6 odd / 0 even: 177,100

Reply #12

dr san — Thu, 11 Dec 2014 00:20:16 GMT

Reply #11

SergeM — Wed, 10 Dec 2014 23:48:26 GMT

1. 02-12-22-32-42-52 sum: 162 even: 6 odd: 0 o|e: eeeeee delta: 10-10-10-10-10

2. 02-12-22-32-42-53 sum: 163 even: 5 odd: 1 o|e: eeeeeo delta: 10-10-10-10-11

3. 02-12-22-32-43-52 sum: 163 even: 5 odd: 1 o|e: eeeeoe delta: 10-10-10-11-9

4. 02-12-22-32-43-53 sum: 164 even: 4 odd: 2 o|e: eeeeoo delta: 10-10-10-11-10

5. 02-12-22-33-42-52 sum: 163 even: 5 odd: 1 o|e: eeeoee delta: 10-10-11-9-10

6. 02-12-22-33-42-53 sum: 164 even: 4 odd: 2 o|e: eeeoeo delta: 10-10-11-9-11

Reply #10

dr san — Tue, 09 Dec 2014 23:15:00 GMT

Hello, with 64 conditions of code in 100% objective is positional in ascending order

Numbers of previous draws, where there has not sweepstakes or number,

This pattern of 64 conditions is basic, separating the last digit, you can also make the

High low in out etc. (Monel)

So in addition to the odd and even still have bottom, inside out, mirror etc ....

View monel

Reply #9

dr san — Tue, 09 Dec 2014 23:09:53 GMT

Hello, ok first you have to put the draws in ascending order to see this

Sometimes sweepstakes that has the last sweepstakes, raffles there are in or out a number

And sometimes, 1,2,3, then we have 4 conditions of previous = 0,1,2,3 sweepstakes, numbers to the next from the last. The 64 patterns pairs / odd serves to position and for odd example of a code is

Eee000 = the first three positions is odd numbers = 19,15,23. And aso other pairs

O - O - O - O -O - O - X

E - O - O - O -O

Reply #8

SergeM — Mon, 08 Dec 2014 21:22:30 GMT

With the tail wheel generator is a link to a text with how you can use it. It is not limited to your simple key wheel. You have to keep busy with it to understand its facets. When you have such a wheel done, you can do the input in Covermaster free version. Let it show the covering. So first make your wheel, second put it in covermaster and third run the covering. Fourth substract the wins from the losses in Excel. Calculate the payout ratio. See Lotterypost for that part.

This answers more t... [ More ]

Reply #7

Igamble — Mon, 08 Dec 2014 19:22:32 GMT

It certainly looks like the odds are greatly improved.In Nj for cash 5 game -5/43,it pays 15bucks for match 3 ,costs 1 dollar so if i use your system to cover previous any 2 ,odds to pick the other 4 can decrease from 1-976.000 to 1-50 ??

I need some input on this one

Reply #6

newtolottoagain — Mon, 08 Dec 2014 18:35:53 GMT

So it's true your odds are tremendously improved by just betting on all combinations of 2 previous draw numbers? + 4 more

Reply #5

RJOh — Mon, 08 Dec 2014 17:56:45 GMT

You don't have to be good at math to use an odds calculator program. The odds of matching 4 in a 6/49 game are 1:1032.

Reply #4

newtolotto — Mon, 08 Dec 2014 15:22:49 GMT

SergeM: Sorry, but I don't really get the relevance or what I'm really supposed to do with the wheel generators there.

Reply #3

newtolotto — Mon, 08 Dec 2014 15:21:43 GMT

Tialuvslotto: I was afraid of something like that. I suck at math but if I recall, the formula for the odds isn't exactly what you wrote, something about the numbers not having to be in order or something. I got my odds of 1:1000 from wikipedia (look up 649 lotto wiki), so I'm trusting that. It also makes sense to me as it seems relatively easy to get 3 numbers (I got it 5 times in the span of a month), so 4 shouldn't be in the millions. Well, we'll wait for the math gurus.

Reply #2

SergeM — Mon, 08 Dec 2014 14:31:37 GMT

Systems : for making such wheels.

Reply #1

Tialuvslotto — Mon, 08 Dec 2014 13:35:05 GMT

Well, even if your 2 repeat digits are correct you still have to pick 4 numbers out of the remaining 43. I calculated this as 1/43*1/42*1/41*1/40 which gives odds of 1 in 2.95M. Substantially less than 1:13.9M, but still high.

Any math/probability gurus out there can tell me if I'm right

Does your odds increase tremendously by millions just like this? Read for details.

newtolotto — Mon, 08 Dec 2014 11:53:33 GMT

Ok, hope you guys can follow me here. I'm no math person but I need to know if this is true.

Numbers from the previous draw frequently appear in the next draw. Usually 1, sometimes 2, and once in a few months, 3.

You need 15 tickets to make sure you cover any combination of 2 of the previously drawn 6 numbers.

Let's say the 6 numbers are 1, 2, 3, 4, 5 and 6.

So it'll be like.

1 2 A B C D

1 3 A B C D

1 4 A B C D

1 5 A B C D

1 6 A B C D

2 3 A B C D

2 4 A B C... [ More ]