Hi there,
I have been playing with the numbers for 6 / 49 to work out the odds of winning, matching 5, matching 5 plus a bonus, match 4 etc. Of course they are all here, but that is the fun of maths !
So anyway, I have some formula i can play with, and as usual you start to try and think up new things.
I have a "what if" calculation that has r-e-a-l-l-y been bugging me !
I can't work out how to do it, and was wondering if anyone has any ideas or if anyone has ever done it. Basically how I could do it on a calculator ! I'm sure there must be a way of making a formula, but I cannot do it. Just wondering if anyone has any ideas.
I am trying to work out the odds of a "NEAR MISS" By this I mean something that happens as follows.
You select your six numbers. The six numbers are then drawn. You DONT win any prize, but the numbers that are drawn are so near, either just the one above, or just the one below, your picked numbers that you think, "aww, nearly!"
EXAMPLE 1
my numbers - 5, 10, 15, 20, 25, 30.
numbers drawn - 4, 11, 14, 21, 26, 29.
I would check my ticket and every number is one above or one below, and think that wow, nearly.
EXMAPLE 2
my numbers - 5, 10, 15, 20, 25, 30
numbers drawn - 5, 10, 31, 39, 40, 49.
I would check my ticket and think I nearly won a 3ball prize, because I have two right and one which is nearly the third number.
EXAMPLE 3
Imagine this
my numbers - 5, 10, 15, 20, 25, 30
numbers drawn - 5, 10, 47, 21, 46, 29.
This time I check my ticket and I think I have nearly won a 4ball prize because I have two right and two which are "nearly" two of the other numbers.
When I say nearly, I mean that you don't win, so it would be getting none right/six near, none right/five near, none right/four near or two right/four near, two right/three near, two right/two near, but NOT three right/three near, four right/two near or five right/one near, because in those cases you didn't nearly win, you did win.
In example 1 i think the maths are as follows, but I'm not sure,
12 from 43, 10 from 42, 8 from 41, 6 from 40, 4 from 39 and 2 from 38.
This is because there are 12 near miss balls (i.e.. either one more or one less than a ball that is drawn) and six winning balls. You pick the twelve from a starting point of 43 balls because this is 49 less the 6 winning which you don't want. With one ball picked there are 48 balls left in the drum, 5 of these are winners that you don't want in this equation, so from the 43 left there are 10 near balls you can pick
In EXAMPLE 2 I think the matsh are as follows but again Im not sure.
6 from 49, 5 from 48, 2 from 43, 34 from 46, 34 from 45, 34 from 44.
This is because there are 6 winning balls, two of which I will get and 8 near miss balls of which I will get just one, and then three balls which will be neither a winning balll nor a near miss ball. From 49 i can get one of my two winning balls, from 48 I get the second of my winning balls, from 47 I get one of the 8 balls that is a near miss ball but is not one of the four remaining winning balls, from 46 I get a ball that is neither one of the four remaining winners nor one of the remaining seven near misses, from 45 I get a ball that is neither one of the four remaining winners nor one of the seven remaining near misses, from 44 i get a ball that is neither one of the remaining four winners nor one of the remaining seven near misses.
I think that makes sense.
Does anyone have any ideas how I would calculate for example,
Six near miss
Five near miss
Four near miss
Two matching and one near miss
Two matching and two near miss
Two matching and three near miss
Two matching and four near miss
One matching and five near miss
One matching and four near miss
One mathcing and three near miss
etc ?
Any ideas gratefully considered !!!!
Thanks !!!
I have been playing with the numbers for 6 / 49 to work out the odds of winning, matching 5, matching 5 plus a bonus, match 4 etc. Of course they are all here, but that is the fun of maths !
So anyway, I have some formula i can play with, and as usual you start to try and think up new things.
I have a "what if" calculation that has r-e-a-l-l-y been bugging me !
I can't work out how to do it, and was wondering if anyone has any ideas or if anyone has ever done it. Basically how I could do it on a calculator ! I'm sure there must be a way of making a formula, but I cannot do it. Just wondering if anyone has any ideas.
I am trying to work out the odds of a "NEAR MISS" By this I mean something that happens as follows.
You select your six numbers. The six numbers are then drawn. You DONT win any prize, but the numbers that are drawn are so near, either just the one above, or just the one below, your picked numbers that you think, "aww, nearly!"
EXAMPLE 1
my numbers - 5, 10, 15, 20, 25, 30.
numbers drawn - 4, 11, 14, 21, 26, 29.
I would check my ticket and every number is one above or one below, and think that wow, nearly.
EXMAPLE 2
my numbers - 5, 10, 15, 20, 25, 30
numbers drawn - 5, 10, 31, 39, 40, 49.
I would check my ticket and think I nearly won a 3ball prize, because I have two right and one which is nearly the third number.
EXAMPLE 3
Imagine this
my numbers - 5, 10, 15, 20, 25, 30
numbers drawn - 5, 10, 47, 21, 46, 29.
This time I check my ticket and I think I have nearly won a 4ball prize because I have two right and two which are "nearly" two of the other numbers.
When I say nearly, I mean that you don't win, so it would be getting none right/six near, none right/five near, none right/four near or two right/four near, two right/three near, two right/two near, but NOT three right/three near, four right/two near or five right/one near, because in those cases you didn't nearly win, you did win.
In example 1 i think the maths are as follows, but I'm not sure,
12 from 43, 10 from 42, 8 from 41, 6 from 40, 4 from 39 and 2 from 38.
This is because there are 12 near miss balls (i.e.. either one more or one less than a ball that is drawn) and six winning balls. You pick the twelve from a starting point of 43 balls because this is 49 less the 6 winning which you don't want. With one ball picked there are 48 balls left in the drum, 5 of these are winners that you don't want in this equation, so from the 43 left there are 10 near balls you can pick
In EXAMPLE 2 I think the matsh are as follows but again Im not sure.
6 from 49, 5 from 48, 2 from 43, 34 from 46, 34 from 45, 34 from 44.
This is because there are 6 winning balls, two of which I will get and 8 near miss balls of which I will get just one, and then three balls which will be neither a winning balll nor a near miss ball. From 49 i can get one of my two winning balls, from 48 I get the second of my winning balls, from 47 I get one of the 8 balls that is a near miss ball but is not one of the four remaining winning balls, from 46 I get a ball that is neither one of the four remaining winners nor one of the remaining seven near misses, from 45 I get a ball that is neither one of the four remaining winners nor one of the seven remaining near misses, from 44 i get a ball that is neither one of the remaining four winners nor one of the remaining seven near misses.
I think that makes sense.
Does anyone have any ideas how I would calculate for example,
Six near miss
Five near miss
Four near miss
Two matching and one near miss
Two matching and two near miss
Two matching and three near miss
Two matching and four near miss
One matching and five near miss
One matching and four near miss
One mathcing and three near miss
etc ?
Any ideas gratefully considered !!!!
Thanks !!!