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Nigel
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 Posted: Mon Mar 21, 2005 7:24 am Post subject: Positional Limits Question. |
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Just on the off-chance someone might know the answer to this off the top
off their head, or have a suitable program that can be easily adapted to
calculate it:
For a 6/49 lottery, for all combinations which sum to 150, what
percentage include at least one number from the set {7,14,21,29,36,43},
and is this different to the percentage over all combinations?
Ta muchly,
Evil Nigel |
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Nigel
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| Posted: Mon Mar 21, 2005 8:36 pm Post subject: Re: Positional Limits Question. |
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Gerry wrote:
| Quote: | "Nigel" <[email protected]> wrote in message
news:[email protected]...
Just on the off-chance someone might know the answer to this off the top
off their head, or have a suitable program that can be easily adapted to
calculate it:
For a 6/49 lottery, for all combinations which sum to 150, what
percentage include at least one number from the set {7,14,21,29,36,43},
and is this different to the percentage over all combinations?
Ta muchly,
Evil Nigel
1 = 67,942
2 = 22,704
3 = 3,152
4 = 233
5 = 0
6 = 1
Data courtesy Lotwin
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Thanks Gerry.
You have to remember that I was never the sharpest knife in the drawer,
and age has made me even blunter (but a blunt knife is more dangerous
than a sharp one!).
I'm guessing those numbers are the number of combinations summing to 150
that include 1,2,3,4,5 or 6 of my specified set. To work out percentages
I also need to know how many combinations include 0 from that set.
The comparison percentages I should be able to work out using
hypergeometric calculations.
The reason I posed the question (even though you didn't ask) is that I
have a variant of my North American Lottery Forecasts system which shows
an above average tendency to get a sum of rankings very close to 150,
and the closer it gets to 150, the more numbers seem to be drawn that
are on the positional expectations within the rankings. I was wondering
whether this is a fluke or a mathematical tendency related to the sum of
the rankings.
Evil Nigel |
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