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Grex Science Item 110: Lotto !
Entered by sholmes on Wed Feb 25 02:38:21 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 )

4 responses total.

#1 of 4 by sholmes on Wed Feb 25 02:47:38 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

#2 of 4 by rcurl on Wed Feb 25 06:29:35 2009:

"If I select N numbers randomly from 1..45 ."  With or without replacement?
(i.e., can selected numbers be duplicated?).

#3 of 4 by sholmes on Wed Feb 25 07:12:21 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

#4 of 4 by rcurl on Wed Feb 25 21:26:32 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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