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# Do some number combinations have better odds?

Topic closed. 5280 replies. Last post 5 years ago by rdgrnr.

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 Posted: July 7, 2012, 6:24 am - IP Logged

Boney & mc both get 4+0.

Great work everyone.

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Can RJ (or anyone) please tell me how many groups of 28 numbers there are in the MM game, and which group (or groups) have the highest winning percentages??

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 Posted: July 7, 2012, 11:31 pm - IP Logged

Mathematically there should only be 38.4 distinct groups (of 28).........

but I cant find a way to create the groups.

mid-Ohio
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 Posted: July 8, 2012, 12:50 am - IP Logged

Mathematically there should only be 38.4 distinct groups (of 28).........

but I cant find a way to create the groups.

MegaMillions has 56 numbers. If you group those numbers in groups of 28 starting with 1-28, 2-29, 3-30 and etc...., you would have 28 groups when you reach 29-56. If you go back and change the last number in the first group of 1-28 to 29, 30, 31 and etc...to 56, you would have an addition 27 groups which makes 55 distinct groups of 28 which is a lot more than 38.4. Doing that to just the 28 groups you started out with will give you more than 400 groups of 28.  You are just getting started and you haven't included the groups of 28 odd or even numbers yet.

According to a Keno calculator I saw on the web, there are more than 7 trillions distant groups of 28 numbers in 56 numbers. I don't know if that's right or not but I suspect it's closer to being right than 38.4.

* you don't need to buy more tickets, just buy a winning one *

Kentucky
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 Posted: July 8, 2012, 12:51 am - IP Logged

Can RJ (or anyone) please tell me how many groups of 28 numbers there are in the MM game, and which group (or groups) have the highest winning percentages??

Considering we're using 56 numbers, the number of 28 number groups is astronomical. It starts with almost 4 million groups of five numbers and each group uses 51 numbers to create a six number group. There are 7,575,968,400 nine number groups and then multiple that by 47 to get the number of ten number groups and that times 46 to get the number eleven number groups and so on. Continue to multiple until you get to 29.

The number is the same as the number of combos in a 28/56 lotto game. Another way is to multiply 56x55x54x53x52x51x50x49x48x47x46x45x44x43x42x41x40x39x38x37x36x35x34x33x32x31x30x29.

Don't know any shortcuts but when you find that total, divide it by 30,488,834,461,171,386,050,150,400,000 (30 Octillion?) to get the total number of 28 number groups.

By the way, all five PB numbers and the bonus number were odd tonight.

Kentucky
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 Posted: July 8, 2012, 1:02 am - IP Logged

MegaMillions has 56 numbers. If you group those numbers in groups of 28 starting with 1-28, 2-29, 3-30 and etc...., you would have 28 groups when you reach 29-56. If you go back and change the last number in the first group of 1-28 to 29, 30, 31 and etc...to 56, you would have an addition 27 groups which makes 55 distinct groups of 28 which is a lot more than 38.4. Doing that to just the 28 groups you started out with will give you more than 400 groups of 28.  You are just getting started and you haven't included the groups of 28 odd or even numbers yet.

According to a Keno calculator I saw on the web, there are more than 7 trillions distant groups of 28 numbers in 56 numbers. I don't know if that's right or not but I suspect it's closer to being right than 38.4.

It's similar to the math question, "if you save \$1 on the first day of the month, \$2 on the second day, \$4 on the third, \$8 on the fourth day, doubling each day, how much will you have by the end of the month"?

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 Posted: July 8, 2012, 1:03 am - IP Logged

Considering we're using 56 numbers, the number of 28 number groups is astronomical. It starts with almost 4 million groups of five numbers and each group uses 51 numbers to create a six number group. There are 7,575,968,400 nine number groups and then multiple that by 47 to get the number of ten number groups and that times 46 to get the number eleven number groups and so on. Continue to multiple until you get to 29.

The number is the same as the number of combos in a 28/56 lotto game. Another way is to multiply 56x55x54x53x52x51x50x49x48x47x46x45x44x43x42x41x40x39x38x37x36x35x34x33x32x31x30x29.

Don't know any shortcuts but when you find that total, divide it by 30,488,834,461,171,386,050,150,400,000 (30 Octillion?) to get the total number of 28 number groups.

By the way, all five PB numbers and the bonus number were odd tonight.

Thanks Stack, I just cant bring myself to be active in predicting PB as the cost is double to actually play. I do play 2 PB lines and 4 MM lines on every draw and pay the same for each.

mid-Ohio
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 Posted: July 8, 2012, 1:06 am - IP Logged

Considering we're using 56 numbers, the number of 28 number groups is astronomical. It starts with almost 4 million groups of five numbers and each group uses 51 numbers to create a six number group. There are 7,575,968,400 nine number groups and then multiple that by 47 to get the number of ten number groups and that times 46 to get the number eleven number groups and so on. Continue to multiple until you get to 29.

The number is the same as the number of combos in a 28/56 lotto game. Another way is to multiply 56x55x54x53x52x51x50x49x48x47x46x45x44x43x42x41x40x39x38x37x36x35x34x33x32x31x30x29.

Don't know any shortcuts but when you find that total, divide it by 30,488,834,461,171,386,050,150,400,000 (30 Octillion?) to get the total number of 28 number groups.

By the way, all five PB numbers and the bonus number were odd tonight.

I suspect the Keno calculation is correct when you plug in 56 numbers to draw from, 28 numbers drawn and 28 played when it comes out 7 trillions or the next value 3 places to the left.  It takes about 20 seconds to check the last 734 MM drawings with one group of 28 numbers using a program I wrote using GWBasic and complied with a QB compiler, I wouldn't even attempt to do it with 7 trillion groups.

* you don't need to buy more tickets, just buy a winning one *

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 Posted: July 8, 2012, 1:11 am - IP Logged

MegaMillions has 56 numbers. If you group those numbers in groups of 28 starting with 1-28, 2-29, 3-30 and etc...., you would have 28 groups when you reach 29-56. If you go back and change the last number in the first group of 1-28 to 29, 30, 31 and etc...to 56, you would have an addition 27 groups which makes 55 distinct groups of 28 which is a lot more than 38.4. Doing that to just the 28 groups you started out with will give you more than 400 groups of 28.  You are just getting started and you haven't included the groups of 28 odd or even numbers yet.

According to a Keno calculator I saw on the web, there are more than 7 trillions distant groups of 28 numbers in 56 numbers. I don't know if that's right or not but I suspect it's closer to being right than 38.4.

LOL. Thanks RJ, I saw the same Keno numbers and it just confused me. I'm not trying to get all 28 numbers right obviously.

If 28 numbers covers 98,820 of the 3.8 million total combinations there should be 38.4 groups in there somewhere that cover all the possible combinations.... There has to be...   lol. lol.

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 Posted: July 8, 2012, 1:18 am - IP Logged

It's similar to the math question, "if you save \$1 on the first day of the month, \$2 on the second day, \$4 on the third, \$8 on the fourth day, doubling each day, how much will you have by the end of the month"?

I know that one Stack. I dream of hitting the craps table with \$20. and walking away with \$20,000. after 10 passes. Too bad things almost never turn out the way we expect them to.

mid-Ohio
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 Posted: July 8, 2012, 1:27 am - IP Logged

LOL. Thanks RJ, I saw the same Keno numbers and it just confused me. I'm not trying to get all 28 numbers right obviously.

If 28 numbers covers 98,820 of the 3.8 million total combinations there should be 38.4 groups in there somewhere that cover all the possible combinations.... There has to be...   lol. lol.

If 28 numbers covers 98,820 of the 3.8 million total combinations there should be 38.4 groups in there somewhere that cover all the possible combinations.... There has to be...

That's like saying 5 numbers must have 1.25 combinations of fours  in it since 5/4=1.25 when we know there are 5 combinations of fours in five numbers.

* you don't need to buy more tickets, just buy a winning one *

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 Posted: July 8, 2012, 1:53 am - IP Logged

If 28 numbers covers 98,820 of the 3.8 million total combinations there should be 38.4 groups in there somewhere that cover all the possible combinations.... There has to be...

That's like saying 5 numbers must have 1.25 combinations of fours  in it since 5/4=1.25 when we know there are 5 combinations of fours in five numbers.

No, because 3.8 million is a distinct and limited number of lines. And that number divided by 38.4 = 98,820.

So there may not be a way to go 1-28.... 2-29.... and so on, But a full 56 number wheel could be divided into 38.4 parts easily enough.

And there would be 38.4 groups.

United States
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 Posted: July 8, 2012, 2:13 am - IP Logged

So if that's the case..... (and it is)

Why cant there be.... this 28 numbers and that 28 numbers.......

until all the 3.8 million lines are used up????

I will have to sleep on this..... I'm still confused. lol. lol. lol.

Whiskey Island
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 Posted: July 8, 2012, 2:13 am - IP Logged

It's not the number combinations that has better odds . The System/Method analyzing the the right Data from Previous draws is the only way to WIN . But , you must use the right Data Regression Analysis to generate the number string it could be the last 5 draws or 500 draws you will need .

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 Posted: July 8, 2012, 3:03 am - IP Logged

It's not the number combinations that has better odds . The System/Method analyzing the the right Data from Previous draws is the only way to WIN . But , you must use the right Data Regression Analysis to generate the number string it could be the last 5 draws or 500 draws you will need .

D R A does have valid uses and improves chance as you have shown CW4...but its more about the right skip mix... which is always in flux...thats is what really improves the odds...and allows consistent winning...

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