lottery Probability

[probmath]

hi!
assuming the last draw the wining numbers were 1 2 3 4 5 6
and i want to also assume that in the next draw at least
one of the numbers from the last draw will return again
like 1-11-12-13-20-22 or 2-4-11-20-22-30
or
one number from the last 2 draws (example: 1-2-3-4-5-6-7-8-9-10-11-12)
or two

HOW can i calculate how many combinations will remain?
after eliminating all combinations without said 1 or 2 numbers
from last draw or two draws and etc....

p.s
the max combinations are about 2.3 mill, from 6/37

 

[Cubist]

the max combinations are about 2.3 mill, from 6/37

That number can be calculated as the ways of choosing 6 from 37 (click).
Can you use this to answer your question?

 

[probmath]

thank you for replying,
no, i don't know how
can you please help?

 

[Deleted member 4993]

thank you for replying,
no, i don't know how
can you please help?

Did you look at the Webpage that was referenced in response#2?

How many ways can you choose 2 items from 37 items?

 

[probmath]

i did,
666

 

[Deleted member 4993]

i did,
666

Correct - now please show us "the exact " problem statement that was presented to you.

 

[probmath]

i understand you trying to teach me,
but i'm not good at math and don't know what "the problem statement" mean

 

[Deleted member 4993]

i understand you trying to teach me,
but i'm not good at math and don't know what "the problem statement" mean

How did you learn about the problem that you are trying to solve? Was it stated in a classroom as homework/project?

 

[Cubist]

i understand you trying to teach me,
but i'm not good at math and don't know what "the problem statement" mean

Please tell us whether you're a math student (or not). Some people here will try to help you regardless. But your honesty will tell us the best way to help.

If this is a real life problem then you should be aware that lottery machines are designed to provide numbers independently of any previous results. You can't predict future results based on previous results.

If you want to consider (for fun) that you're able to see into the future and that you know a specific number will come out in the next draw, then this would be equivalent to just removing that number from the machine by hand and then continuing to draw the remaining numbers randomly. Please think about this, and how many random combinations would remain (after manually picking that first, "known", number).

 

[probmath]

Subhotosh Khan and Cubist:
so, currently is mostly fun to test things about the lottery, not in a classroom and i'm not a math student

i don't want to assume one specific ball from last draw will drawn, but at least one of the last six that was drawn,
and once i'll have the formula, i would like to test it with at least two, and/or at least 1 or 2 from last two draws and so on....

 

[Cubist]

Subhotosh Khan and Cubist:
so, currently is mostly fun to test things about the lottery, not in a classroom and i'm not a math student

i don't want to assume one specific ball from last draw will drawn, but at least one of the last six that was drawn,
and once i'll have the formula, i would like to test it with at least two, and/or at least 1 or 2 from last two draws and so on....

It might take a little bit of conversation to determine your exact goal.

You initially asked about combinations. However, I suspect that you might be interested in the probability of certain events? If I'm right, then would this be an appropriate way to ask your question:- what is the probability that "x" unique numbers from the previous "y" lottery draws will appear in the next draw? I made that word bold because if y>1 then there's a chance that the numbers could have appeared in more than one of the previous "y" draws.

I suspect that this might be difficult to work out

 

[probmath]

the max combinations possible for choosing 6 out 37 are 2324784
i'm not asking the probability per say (although i can be calculate it latter on from my goal)
but the number of combinations/ possibilities left if i assuming 1 of the last six drawn,
will repeat in the next draw,
and from there playing with it, and change it to two draws, i.e 12 balls, or 10 (like you rightfully mentions "if y>1....")
and assuming if 2 balls repeats from 15 or 18 and so on....

you definite clearly the problem with x and y expect if y>1, will not be a problem,
cause i will determent how many unique balls drawn in last 2 or 3 draws