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6/49 Lottery Question
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Nigel
science forum beginner


Joined: 03 Jun 2005
Posts: 37

PostPosted: Mon Jun 19, 2006 1:45 pm    Post subject: 6/49 Lottery Question Reply with quote

How many draws of a 6/49 lottery before a particular triple (eg 1-2-3)
is more than 50% probable to have been drawn?

I took the approach that there are 18424 triples possible, of which 20
appear each draw. The probability of a particular triple not appearing
after 1 draw is therefore (18424-20)/18424. For subsequent draws, raise
to the power of the number of draws. Subtract the numbers from 1 to find
the probability of a particular triple have appeared by that draw. That
gives the answer that a particular triple is more than 50% probable to
have appeared by the 639th draw.

The reason that I'm not happy with this approach is that the 20 triples
per draw are not independent of each other. Intuitively I'd expect a
sort of mild clustering effect which would raise the number of draws of
the 50% watershed slightly.

Could someone help me with the correct answer please?

Thanks,

NigelH
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Proginoskes
science forum Guru


Joined: 29 Apr 2005
Posts: 2593

PostPosted: Mon Jun 19, 2006 11:05 pm    Post subject: Re: 6/49 Lottery Question Reply with quote

nigel wrote:
Quote:
How many draws of a 6/49 lottery before a particular triple (eg 1-2-3)
is more than 50% probable to have been drawn?

I took the approach that there are 18424 triples possible, of which 20
appear each draw. The probability of a particular triple not appearing
after 1 draw is therefore (18424-20)/18424.

Hmmm. Not the "recommended way", which is:

If you have a particular triple 1-2-3, then in order to draw that
triple in a 6/49 lottery, you should count the number of 6-tuples that
contain 1-2-3:

(1) Choose the numbers 1-2-3 [1 way to do this], then
(2) Chose 3 other numbers [C(46,3) ways to do this, since order doesn't
matter].

So the probability of 1-2-3 appearing is 1*C(46,3) / C(49,6), which is
5/4606, which is the same as 20/18424.

Quote:
For subsequent draws, raise
to the power of the number of draws. Subtract the numbers from 1 to find
the probability of a particular triple have appeared by that draw. That
gives the answer that a particular triple is more than 50% probable to
have appeared by the 639th draw.

Thus solve:
1 - (4601/4606)^n = 1/2.

The answer is n=368.*, so you do need 639 draws.

--- Christopher Heckman

Quote:
The reason that I'm not happy with this approach is that the 20 triples
per draw are not independent of each other. Intuitively I'd expect a
sort of mild clustering effect which would raise the number of draws of
the 50% watershed slightly.

Could someone help me with the correct answer please?

Thanks,

NigelH
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