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Forum index » Science and Technology » Math » Probability
Formula for repeats
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Stig Holmquist
science forum beginner


Joined: 30 Apr 2005
Posts: 48

PostPosted: Mon Sep 26, 2005 9:44 am    Post subject: Formula for repeats Reply with quote

Regular lotto games have the matrix n/N andn numbers are drawn
without replacement. But consider instead that each number is replaced
before the next is drawn. What then would be the probability of
finding a duplication amongst the n numbers after completing the
drawing?

I generated 50 random numbers with replacement and printed them
in 10 columns and 10 rows and then looked at sets of 5 numbers
at a time going down the columns, because it is ease to spot
duplications at a glance. I found 22 pairs in 100 sets.

I seem to be unable to find a binomial table that allows me to
predict the result I have, so I must assume the problem is
more complex.

What probability formula should I use for the set of 5 out of 50
and what general formula do I need to make a similar
prediction for any "lotto" with replacement, such as 6/49.

Bertil
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Duncan Smith
science forum beginner


Joined: 29 Apr 2005
Posts: 21

PostPosted: Mon Sep 26, 2005 11:59 am    Post subject: Re: Formula for repeats Reply with quote

Stig Holmquist wrote:
Quote:
Regular lotto games have the matrix n/N andn numbers are drawn
without replacement. But consider instead that each number is replaced
before the next is drawn. What then would be the probability of
finding a duplication amongst the n numbers after completing the
drawing?

I generated 50 random numbers with replacement and printed them
in 10 columns and 10 rows and then looked at sets of 5 numbers
at a time going down the columns, because it is ease to spot
duplications at a glance. I found 22 pairs in 100 sets.

I seem to be unable to find a binomial table that allows me to
predict the result I have, so I must assume the problem is
more complex.

What probability formula should I use for the set of 5 out of 50
and what general formula do I need to make a similar
prediction for any "lotto" with replacement, such as 6/49.

Bertil

The probability of *no* pairs for a 6/49 would be

1 * (48/49) * (47/49) * (46/49) * (45/49) * (44/49)

more generally

n!/((n-k)!*n^k)

for a k/n lottery.

Subtract from 1 for the probability of at least one pair.

Duncan
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