ChatterBank12 mins ago
Lottery odds
Can anyone explain & demonstrate the maths involved in working out the odds of winning the lottery. A few of us have tried to work it out but get nowhere near the 14 million to one Camelot claims it to be. I think we have missed something, but don't know what.
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Geezer beat me to it, but I have added a bit to exlain why you do what you do.
Firstly, think how many lots of two numbers can be pulled from a hat containing four. The possibilities are:
1 & 2
1 & 3
1 & 4
2 & 3
2 & 4
3 &4
a total of six.
To work this out without counting them, you multiply from 4 downwards by 2 numbers, and 1 upwards by 2 numbers. (This is because 4 is the number of balls in the hat, 2 is the number in the �sets� you want to count.)
This gives:
4 x 3
-----
1 x 2
= 12 divided by 2 = 6.
Now extend this to see how many �sets� of 6 there are in 49:
49 x 48 x 47 x 46 x 45 x44
----------------------------------
1 x 2 x 3 x 4 x 5 x 6
= 10,068,347,520
---------------------
720
= 13,983,816
There are thus 13,983,816 possible combinations of six numbers which can be drawn by the machine, so your chances of winning are a little under 14m to 1.
You can use this to establish all betting probabilities. A "Yankee" consists of six doubles, four trebles and a fourfold.
Doubles from 4:
4 x 3
------
1 x 2
=6
Trebles from 4:
4 x 3 x 2
-----------
1 x 2 x 3
= 4
Fours from 4 (no need to calculate, but the maths works just the same):
4 x 3 x 2 x 1
---------------
1 x 2 x 3 x 4
= 1
S'easy when you know how!
Firstly, think how many lots of two numbers can be pulled from a hat containing four. The possibilities are:
1 & 2
1 & 3
1 & 4
2 & 3
2 & 4
3 &4
a total of six.
To work this out without counting them, you multiply from 4 downwards by 2 numbers, and 1 upwards by 2 numbers. (This is because 4 is the number of balls in the hat, 2 is the number in the �sets� you want to count.)
This gives:
4 x 3
-----
1 x 2
= 12 divided by 2 = 6.
Now extend this to see how many �sets� of 6 there are in 49:
49 x 48 x 47 x 46 x 45 x44
----------------------------------
1 x 2 x 3 x 4 x 5 x 6
= 10,068,347,520
---------------------
720
= 13,983,816
There are thus 13,983,816 possible combinations of six numbers which can be drawn by the machine, so your chances of winning are a little under 14m to 1.
You can use this to establish all betting probabilities. A "Yankee" consists of six doubles, four trebles and a fourfold.
Doubles from 4:
4 x 3
------
1 x 2
=6
Trebles from 4:
4 x 3 x 2
-----------
1 x 2 x 3
= 4
Fours from 4 (no need to calculate, but the maths works just the same):
4 x 3 x 2 x 1
---------------
1 x 2 x 3 x 4
= 1
S'easy when you know how!
The 10billion answer you came up with are the number of combinations of six from 49 drawn in the correct order.
The bottom bit of the formula (dividing by 720) makes the order irrelevant.
So if you had to forecast the draw, and get it in the same order as the numbers are drawn by the machine, the odds would be 10bn (approx.) to one.
Going back to my "two from four" example, the cominations in the correct order are doubled (each of the six possibilities reversed). So the formula is simply 4 x 3.
The bottom bit of the formula (dividing by 720) makes the order irrelevant.
So if you had to forecast the draw, and get it in the same order as the numbers are drawn by the machine, the odds would be 10bn (approx.) to one.
Going back to my "two from four" example, the cominations in the correct order are doubled (each of the six possibilities reversed). So the formula is simply 4 x 3.
Also, beware when you try to calculate the odds against, say, getting three numbers correct (the �10 prize).
The number of threes in 49 is:
49 x 48 x 47
---------------
1 x 2 x 3
= 110,544/6 = 18,424
So you would think the odds against winning �10 are 18,423 to 1.
However:
In pulling out six numbers, the machine is actually selecting 20 sets of 3 numbers. [Remember (6 x 5 x 4)/(1 x 2 x 3) = 20]. This means you have 20 lots of 3 each of which can win.
In addition, when you buy a lottery ticket you choose six numbers. Although you are only paying for one go at getting all six correct, you are actually buying 20 goes at getting three correct [again, (6 x 5 x 4)/(1 x 2 x 3) = 20].
So the probability of matching three is actually reduced by a factor of 20 for each of the reasons outlined above, so they become (18,424/400) = 45 to 1 (approx).
Many people see this as a bit rough, since Camelot only pays 9 to 1 for matching three numbers. However (as usual) all is not as it seems because for your �1 you are actually getting one go at matching all six, six goes at matching five numbers, fifteen goes at matching four and twenty goes at matching three. And this, of course, does not include the �five plus bonus ball� winner.
The number of threes in 49 is:
49 x 48 x 47
---------------
1 x 2 x 3
= 110,544/6 = 18,424
So you would think the odds against winning �10 are 18,423 to 1.
However:
In pulling out six numbers, the machine is actually selecting 20 sets of 3 numbers. [Remember (6 x 5 x 4)/(1 x 2 x 3) = 20]. This means you have 20 lots of 3 each of which can win.
In addition, when you buy a lottery ticket you choose six numbers. Although you are only paying for one go at getting all six correct, you are actually buying 20 goes at getting three correct [again, (6 x 5 x 4)/(1 x 2 x 3) = 20].
So the probability of matching three is actually reduced by a factor of 20 for each of the reasons outlined above, so they become (18,424/400) = 45 to 1 (approx).
Many people see this as a bit rough, since Camelot only pays 9 to 1 for matching three numbers. However (as usual) all is not as it seems because for your �1 you are actually getting one go at matching all six, six goes at matching five numbers, fifteen goes at matching four and twenty goes at matching three. And this, of course, does not include the �five plus bonus ball� winner.
and of course judge, I think only about 40p of the �1 is for the prize fund.
interesting curiosities. �10 for 3 numbers is the only guarenteed prize so it is in theory possible to get less for 4/5/6 than it is for 3, unlikely I know but Camelot have a special clause that if there are so many �10 winners that the fund is eaten up they can reduce the prize even for 3 numbers in such a case the payout for 4/5/6 would be 0, scary!
interesting curiosities. �10 for 3 numbers is the only guarenteed prize so it is in theory possible to get less for 4/5/6 than it is for 3, unlikely I know but Camelot have a special clause that if there are so many �10 winners that the fund is eaten up they can reduce the prize even for 3 numbers in such a case the payout for 4/5/6 would be 0, scary!
You can increase your probabilities if you narrow your selections to the obvious. There is a flaw with probability theory even in tossing a coin. Each toss will be 0.5 even if a succession of heads or tails have come up beforehand. but you know after many tosses it will even out 50/50.
This strategy will allow you to ignore consecutive balls, to ignore 6 numbers within each section 1-10, 21-30, etc. The theorists will say it is possible but common sense tells you otherwise.
This strategy will allow you to ignore consecutive balls, to ignore 6 numbers within each section 1-10, 21-30, etc. The theorists will say it is possible but common sense tells you otherwise.
Not so, sp!
Every combination of six numbers has an equal chance of being drawn. If numbers 1-6 were drawn it would obviously be declared "a miracle". That's only because it would be noticeable. The combination of numbers 2, 25, 38, 41, 26 and 49 is equally miraculous, but not so noticeable.
Reducing your choices in the way you describe does not increase your chances of success.
Every combination of six numbers has an equal chance of being drawn. If numbers 1-6 were drawn it would obviously be declared "a miracle". That's only because it would be noticeable. The combination of numbers 2, 25, 38, 41, 26 and 49 is equally miraculous, but not so noticeable.
Reducing your choices in the way you describe does not increase your chances of success.
An easy way to explain why no particular set of six numbers is any more or less likely to be drawn is this:
The function of the numbers is to make the balls uniquely distinguishable. Instead of numbers, you could use differently coloured balls. Then the absurdity of stating (for example) red, green, black, pink, orange, blue is any more or less likely than any other colour combination becomes instantly apparent.
The function of the numbers is to make the balls uniquely distinguishable. Instead of numbers, you could use differently coloured balls. Then the absurdity of stating (for example) red, green, black, pink, orange, blue is any more or less likely than any other colour combination becomes instantly apparent.
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