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| Author |
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sholmes
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Lotto !
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Feb 25 02:38 UTC 2009 |
The Singapore Lotto selects 7 numbers out of 1 to 45 and you win prize
if 4 or more are matched. The specific prizes can be checked on web.
http://www.singaporepools.com.sg/
Now what I would want to calculate is
If I select N numbers randomly from 1..45 . What should N be to say give
me a 50% chance that 6 of those 7 numbers selected by the lotto machine
is included in those N. what about P percentage chance that M numbers
out of those 7 are selected.
Like if N is 45 we have 100 % chance that we have all the 7 numbers in
N. similarly if N is 45 we have 100% chance that we have 6 of those 7
and so on ..
is it possible to calculate for general P and M
I know in the long run lottos are not a money making scheme. But I am
curious to see what really are the chances with a a carefully defined
technique.
PS: if anyone is interested, I came across some set of numbers which I
tested on past 1985 results of the singapore lotto and that gives about
0.11 chances of breaking even ( not proven mathematically but based on
the 1985 past results )
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| 4 responses total. |
sholmes
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response 1 of 4:
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Feb 25 02:47 UTC 2009 |
correction in above
similarly if N is 45 we have 100% chance that we have 6 of those 7
and so on
while the above is correct
we have 100 % gurantee of macthing 6 out of 7 with N = 44
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rcurl
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response 2 of 4:
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Feb 25 06:29 UTC 2009 |
"If I select N numbers randomly from 1..45 ." With or without replacement?
(i.e., can selected numbers be duplicated?).
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sholmes
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response 3 of 4:
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Feb 25 07:12 UTC 2009 |
without replacement ...
i solved a roundabout problem , i.e given N random numbers picket ..
what are my chances of getting 6 out of the 7 system generated numbers
the formula i came up with is
p = 1/7 * C(N,6)* ( 309 -6n ) / C(45, 7)
( if you have time , you can work out and check if i am correct .. or if
you want my workings i can mail you too )
for values other than 6 , it can be worked out too .. but 6 was easier
but i wonder if there can be any general formula where we can just put N
and Y ( the number of matches we want ) and get the probability out
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rcurl
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response 4 of 4:
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Feb 25 21:26 UTC 2009 |
I'm going to change the names of your constants a bit:
Given N items containing Na of type A, the probability of drawing x of type
A in n draws is the hypergeometric distribution
p(x) = B(Na,x)*B(N-Na,n-x)/B(N,n)
where B(a,b) is the binomical coefficient.
In your example N = 45, Na = 7, x = 6, and p(x) = 0.5. Solve for n.
You can solve it iteratively by choosing n's until you get a result
close to p = 0.5. There is in general no n that will make it exactly
0.5. I'd use Excel.
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