Search   Memberlist   Usergroups
 Page 1 of 1 [2 Posts]
Author Message
Nigel
science forum beginner

Joined: 03 Jun 2005
Posts: 37

Posted: Mon Jun 19, 2006 1:45 pm    Post subject: 6/49 Lottery Question

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.

Thanks,

NigelH
Proginoskes
science forum Guru

Joined: 29 Apr 2005
Posts: 2593

Posted: Mon Jun 19, 2006 11:05 pm    Post subject: Re: 6/49 Lottery Question

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

 Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
 Page 1 of 1 [2 Posts]
 The time now is Sun Feb 17, 2019 1:25 pm | All times are GMT
 Jump to: Select a forum-------------------Forum index|___Science and Technology    |___Math    |   |___Research    |   |___num-analysis    |   |___Symbolic    |   |___Combinatorics    |   |___Probability    |   |   |___Prediction    |   |       |   |___Undergraduate    |   |___Recreational    |       |___Physics    |   |___Research    |   |___New Theories    |   |___Acoustics    |   |___Electromagnetics    |   |___Strings    |   |___Particle    |   |___Fusion    |   |___Relativity    |       |___Chem    |   |___Analytical    |   |___Electrochem    |   |   |___Battery    |   |       |   |___Coatings    |       |___Engineering        |___Control        |___Mechanics        |___Chemical

 Topic Author Forum Replies Last Post Similar Topics Question about Life. socratus Probability 0 Sun Jan 06, 2008 10:01 pm Probability Question dumont Probability 0 Mon Oct 23, 2006 3:38 pm Question about exponention WingDragon@gmail.com Math 2 Fri Jul 21, 2006 8:13 am question on solartron 1260 carrie_yao@hotmail.com Electrochem 0 Fri Jul 21, 2006 7:11 am A Combinatorics/Graph Theory Question mathlover Undergraduate 1 Wed Jul 19, 2006 11:30 pm