Australia Member #126495 April 11, 2012 7 Posts Offline

Posted: April 11, 2012, 7:05 am - IP Logged

I know that in a 6/45 lottery, there are 8,145,060

Lets say you split the 45 numbers into two groups, with group A having the numbers 1-30 and group B 31-45. You make a prediction that 5 of the 6 numbers will be drawn from group A and 1 of the 6 will be drawn from group B. If your prediction is correct, your probability of winning has certainly increased, but has the size of combinations changed? If so, what is the formula to calculate this change.

Would this be any different if the numbers in group A and group B overlapped.

cleveland ohio United States Member #65897 October 9, 2008 275 Posts Offline

Posted: April 11, 2012, 9:48 am - IP Logged

not sure if this is what your looking for but you would take the amount of chances and multiply them together.

In your example for your first choice you have 30 numbers to choice from

your next choice would have 29 (you've already selected 1) etc

So you would have 30*29*28*27*26 next you need to factor out all the doubles because each number is unique so you have 6 choices on where to place the first number then 5 choices on the 2nd number ect all the way to 1.

You then divide the first set by the 2nd set to get your total combinations.

30*29*28*27*26/5*4*3*2*1 = 142506

you have a 1 in 15 chance at the last number so multiply them together 142506*15= 2137590

Australia Member #126495 April 11, 2012 7 Posts Offline

Posted: April 11, 2012, 10:32 am - IP Logged

If your maths is correct. Does that mean that by me making a correct prediction, I have reduced my odds from 8mil to 2mil, while still playing all 45 balls.

cleveland ohio United States Member #65897 October 9, 2008 275 Posts Offline

Posted: April 13, 2012, 12:35 am - IP Logged

Well before someone comes in here and says that the odds do not change what ever numbers you play are, Yes given the peramiters you have set your odds have lowered by a considerable amount.

bgonÃ§alves Brasil Member #92564 June 9, 2010 2122 Posts Offline

Posted: April 13, 2012, 9:10 am - IP Logged

Quote: Originally posted by nickbrownsfan on April 13, 2012

Well before someone comes in here and says that the odds do not change what ever numbers you play are, Yes given the peramiters you have set your odds have lowered by a considerable amount.

Hello, falling, the bestdivision isin 4 parts, because, whenbillions ofsimulationsof the groupswill havezero oronenumberand onewill haveat leasttwo(there may bemore butat leasttwonumbersmalways100%of the sweepstakes, you cancheckok,ieyou can seethe drive (anumber)stop andtrios ofcad4 groupsand combinesuchahotgrouptrio,will combine withanother grouppair ofcold, andanother numberfrom anothercorner.

Australia Member #126495 April 11, 2012 7 Posts Offline

Posted: April 19, 2012, 4:44 am - IP Logged

How do you calculate the odds if for the 3 balls, you select from numbers 1-45, and for the last 3 you select from 1-15? You don't know whether the first 3 numbers you select contains a number from 1-15.

mid-Ohio United States Member #9 March 24, 2001 19816 Posts Offline

Posted: April 19, 2012, 5:09 am - IP Logged

Quote: Originally posted by fallingfirst on April 11, 2012

I know that in a 6/45 lottery, there are 8,145,060

Lets say you split the 45 numbers into two groups, with group A having the numbers 1-30 and group B 31-45. You make a prediction that 5 of the 6 numbers will be drawn from group A and 1 of the 6 will be drawn from group B. If your prediction is correct, your probability of winning has certainly increased, but has the size of combinations changed? If so, what is the formula to calculate this change.

Would this be any different if the numbers in group A and group B overlapped.

You would calculate the odds like you do for PM and MM which use two groups of numbers.

5 of 30 = 142,506 x 1 of 15 =15 or 15 x 142,506 = 2,137,590

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

mid-Ohio United States Member #9 March 24, 2001 19816 Posts Offline

Posted: April 19, 2012, 5:13 am - IP Logged

Quote: Originally posted by fallingfirst on April 19, 2012

How do you calculate the odds if for the 3 balls, you select from numbers 1-45, and for the last 3 you select from 1-15? You don't know whether the first 3 numbers you select contains a number from 1-15.

Are you planing to use the numbers 1-15 twice since both groups cover 1-15?

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

mid-Ohio United States Member #9 March 24, 2001 19816 Posts Offline

Posted: April 19, 2012, 11:36 am - IP Logged

Quote: Originally posted by fallingfirst on April 19, 2012

Yes. For my first 3 numbers, I am just picking randomly from 1 to 45. For my last 3 numbers, I am picking randomly from 1 to 15.

Sounds like you're back where you started picking 6 numbers from a pool of 45. Choosing at least three of them from the lower 15 doesn't change the way the number of possible combinations are calculated.

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

Australia Member #126495 April 11, 2012 7 Posts Offline

Posted: April 19, 2012, 12:28 pm - IP Logged

Quote: Originally posted by RJOh on April 19, 2012

Sounds like you're back where you started picking 6 numbers from a pool of 45. Choosing at least three of them from the lower 15 doesn't change the way the number of possible combinations are calculated.

I'm pretty sure it does. To illustrate. In a 2 of 6 game, i get 15 possible combinations. If I were to choose one from 1-6 and the other from 1-3, I effectively eliminated the combinations, 4/5, 4/6 & 5/6, bring the number of possible combinations down to 12. I'm trying to figure out the formula to calculate this.

mid-Ohio United States Member #9 March 24, 2001 19816 Posts Offline

Posted: April 19, 2012, 6:54 pm - IP Logged

Quote: Originally posted by fallingfirst on April 19, 2012

I'm pretty sure it does. To illustrate. In a 2 of 6 game, i get 15 possible combinations. If I were to choose one from 1-6 and the other from 1-3, I effectively eliminated the combinations, 4/5, 4/6 & 5/6, bring the number of possible combinations down to 12. I'm trying to figure out the formula to calculate this.

When you figure out the formula, post it and we will learn something new. Good luck.

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

cleveland ohio United States Member #65897 October 9, 2008 275 Posts Offline

Posted: April 22, 2012, 1:47 pm - IP Logged

Quote: Originally posted by fallingfirst on April 19, 2012

How do you calculate the odds if for the 3 balls, you select from numbers 1-45, and for the last 3 you select from 1-15? You don't know whether the first 3 numbers you select contains a number from 1-15.

There is 2 ways to look at this the first way 3 of 45 14190 * 3 of 15 455 with leaves 6456450 total combinations but Im not sure that is totally correct because you can still have 2 overlapping numbers. That would be more for looking for 3 mega balls or something being able to pick the number 1 in each set sooo... I am going to say you need to pick the 3 numbers from 1-15 first the combinations there stay the same.

15*14*13/3*2*1= 455 Now for the 2nd set you have already placed 3 numbers in the 1-15 area which leaves you picking 3 of 42 (total possible numbers left to pick from)