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Stig Holmquist science forum beginner
Joined: 30 Apr 2005
Posts: 48
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Posted: Mon Sep 26, 2005 9:44 am Post subject:
Formula for repeats
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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
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Posted: Mon Sep 26, 2005 11:59 am Post subject:
Re: Formula for repeats
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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
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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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