Trying to calculate the number of combinations with criteria

Reply #14

nickbrownsfan — Sun, 22 Apr 2012 17:47:54 GMT

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 t... [ More ]

Reply #13

RJOh — Thu, 19 Apr 2012 22:54:45 GMT

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

Reply #12

fallingfirst — Thu, 19 Apr 2012 16:28:41 GMT

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.

Reply #11

RJOh — Thu, 19 Apr 2012 15:36:07 GMT

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.

Reply #10

fallingfirst — Thu, 19 Apr 2012 09:53:09 GMT

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.

Reply #9

RJOh — Thu, 19 Apr 2012 09:13:26 GMT

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

Reply #8

RJOh — Thu, 19 Apr 2012 09:09:34 GMT

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

Reply #7

fallingfirst — Thu, 19 Apr 2012 08:44:45 GMT

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.

Reply #6

RL-RANDOMLOGIC — Wed, 18 Apr 2012 20:35:34 GMT

ff

Conditional odds. Setting these types of perimeters is a great way to play. Even if your

assumptions are incorrect because your tickets still carry the same odds as a QP. If on

the other hand you are correct then you have a much better chance of winning something.

Removing one ball from a 5-59 matrix reduces the pool by 424,270 sets. If you can remove

one digit from a drawing successfully, say digit 7 for instance, you reduce the pool by 2,036,701

sets for the above matrix a... [ More ]

Reply #5

fallingfirst — Sat, 14 Apr 2012 02:51:16 GMT

I don't understand what you're trying to say

Reply #4

dr san — Fri, 13 Apr 2012 13:10:17 GMT

Hello, falling, the best division is in 4 parts, because, when billions of simulations of the groups will have zero or one number and one will have at least two (there may be more but at least two numbers m always 100% of the sweepstakes, you can check ok, ie you can see the drive (a number) stop and trios of cad 4 groups and combine such a hot group trio, will combine with another group pair of cold, and another number from another corner.

Reply #3

nickbrownsfan — Fri, 13 Apr 2012 04:35:50 GMT

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.

Reply #2

fallingfirst — Wed, 11 Apr 2012 14:32:15 GMT

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.

Reply #1

nickbrownsfan — Wed, 11 Apr 2012 13:48:11 GMT

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 fi... [ More ]

Trying to calculate the number of combinations with criteria

fallingfirst — Wed, 11 Apr 2012 11:05:56 GMT

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 number... [ More ]