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fallingfirst

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April 11, 2012
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Apr 11, 2012, 7:05 am

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.
nickbrownsfan

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Apr 11, 2012, 9:48 am

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
fallingfirst

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Apr 11, 2012, 10:32 am

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.
nickbrownsfan

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Apr 13, 2012, 12:35 am

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.
dr san

bgonÃ§alves
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Apr 13, 2012, 9:10 am

Quote: Originally posted by nickbrownsfan on Apr 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 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.
fallingfirst

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Apr 13, 2012, 10:51 pm

I don't understand what you're trying to say
RL-RANDOMLOGIC

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Apr 18, 2012, 4:35 pm

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 and any set you play still has as good of odds as any QP regardless

of what you choose to remove. A pick-5 QP removes all the numbers but 5 and many swear by

them, go figure!! If I remove 20 numbers from the pool and play the rest I am considered an

idiot by others who spout QP's as being superior somehow. This appears to me to be stupidity

gone to seed. Don't let the odds slow you down, play on.

Welcome to LP

RL

....
fallingfirst

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Apr 19, 2012, 4:44 am

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.
RJOh

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Apr 19, 2012, 5:09 am

Quote: Originally posted by fallingfirst on Apr 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 every combination, just the winning ones *
RJOh

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Apr 19, 2012, 5:13 am

Quote: Originally posted by fallingfirst on Apr 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 every combination, just the winning ones *
fallingfirst

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Apr 19, 2012, 5:53 am

Quote: Originally posted by RJOh on Apr 19, 2012
Are you planing to use the numbers 1-15 twice since both groups cover 1-15?

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.
RJOh

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Apr 19, 2012, 11:36 am

Quote: Originally posted by fallingfirst on Apr 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 every combination, just the winning ones *
fallingfirst

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Apr 19, 2012, 12:28 pm

Quote: Originally posted by RJOh on Apr 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.
RJOh

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Apr 19, 2012, 6:54 pm

Quote: Originally posted by fallingfirst on Apr 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 every combination, just the winning ones *
nickbrownsfan

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Apr 22, 2012, 1:47 pm

Quote: Originally posted by fallingfirst on Apr 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)

42*41*40/3*2*1=11480

11480*455=5223400

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