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Hi, I drag at math can someone please explain this to me.
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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Yes, it has 8 balls in total and to win the maximum prize you need to get all eight correect. Each time a ball is picked up of 40 other balls (it is replaced before the next ball is picked)
Can you please explain how to calculate this? I'm not so much interested in calculating it, just in knowing how to calculate it.
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You are describing a different lottery to the one on that page. Do you have 40 different numbered balls? Do you win in any order? Such as?
1,5,10,32,16,29,37,11 is this the same as 37,11,3,32,1,16,29,10?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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You are describing a different lottery to the one on that page. Do you ou have 40 different numbered balls? Do you win in any order? Such as?
1,5,10,32,16,29,37,11 is this the same as 37,11,3,32,1,16,29,10?
Yes in any order. The balls are numbered from 1 to 40
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Also it appears since it is with replacement , you have the possibility of having 2 or more numbers the same. Are you sure that the balls are replaced after each pick?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Also it appears since it is with replacement , you have the possibility of having 2 or more numbers the same. Are you sure that the balls are replaced after each pick?
Well it's not for anything real life, it's for a theatrical lottery, I just want to know how to work it out to help me learn about maths.
Can you tell what the odss are if the balls are replaced and if they are not replace please? Show me how you work it out. Thanks for all the replies so far.
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49 choose 6
I think thats right, though it depends on the number of options you have in your lottery.
Can feel it coming together.. Slowly but Surely
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49 choose 6
I think thats right, though it depends on the number of options you have in your lottery.
Thanks, but I have no idead idea what that notation means.
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Its pretty easy, though it does mean that only one of each number may be picked, and only works one game at a time.
The is a binomial coefficient, it means 49! (49 factorial) divided by 6factorial by 49 - 6 factorial.
So for a smaller example lets say
you have
or
and in general
This is the way in which Camelot calculate the odds for the UK National Lottery, well I assume so as
this method gives the same odds as they give. Around 13million or 13983816 to one.
More chance of jumping off Sears Tower and a gust of wind catching you and neatly placing you back on top,
whilst meanwhile another gust of wind grabs a beer from someone standing outside a bar and simultaneously
places it into your hand.:P
Can feel it coming together.. Slowly but Surely
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Its pretty easy, though it does mean that only one of each number may be picked, and only works one game at a time.
The is a binomial coefficient, it means 49! (49 factorial) divided by 6factorial by 49 - 6 factorial.
So for a smaller example lets say
you have
or
and in general
This is the way in which Camelot calculate the odds for the UK National Lottery, well I assume so as
this method gives the same odds as they give. Around 13million or 13983816 to one.More chance of jumping off Sears Tower and a gust of wind catching you and neatly placing you back on top,
whilst meanwhile another gust of wind grabs a beer from someone standing outside a bar and simultaneously
places it into your hand.:P
Thanks I get what the notation means now, but can you tell me where you got the numbers from?
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Hi all;
First of all since you can have duplicated numbers it is important to pick a ticket that does not have duplicates as you will have more chances to win.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi Geodude
The numbers I have used are just examples, though the 49 choose 6 is from the UK National Lottery. The rules are that you have 6 numbers to choose. and numbers 1 through 50 to choose from. The balls come out of a machine in two games on the BBC every Saturday and Wednesday. I did the very first game and in the run up made a program to decide my numbers for me, then I sat there planning what I was going to spend the money on. You won't be to amazed to hear that I didn't win.
Can feel it coming together.. Slowly but Surely
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Hi;
There are:
To figure you chances of winning , look at your ticket and do this, example
1 ,1, 2,16,16,27,31,40
You have 2 1's and 2 16's and 1 of each of the rest: Turn each multiplicity into a
factorial and put 40320 in the numerator.
Put that over:
About one chance in 650 million.
Now supposing you 2,2,2,16,17,32,40,40 you have 3 2's and 2 40's and the rest 1 of each.
Turn each multiplicity into a factorial and put 40320 in the numerator.
About 1 chance in 1 950 000 000.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi,
I would have thought that as each time a ball is picked you have a 1 in 40 chance ( if the ball is replaced ) then write this down 8 times as a fraction.>>
( 1/40 ) X ( 1/40 ) X ( 1/40 ) X ( 1/40 ) X ( 1/40 ) X ( 1/40 ) X ( 1/40 ) X ( 1/40 ) =
1 / ( 40 X 40 X 40 X 40 X 40 X 40 X 40 X 40 ) = 1 / 6553600000000
Simple a 1 in 6,553,600,000,000 chance!!
Regards,
Dr M
P.S. In a similar way to get two number 6's with a pair of dice ( 1 /6 ) X ( 1 /6 ) = 1 / 36
or a 1 in 36 chance.
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Hi DrMarten;
I looked at it this way, you have a card with 8 numbers picked, so you have 8 / 40 chances of being right on the first pick, then 7 out of 40 , etc.
8! / 40^8 , provided that all eight of your numbers on the card are different.
Anyway, welcome to the Forum!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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