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Topic: What probability a six-number lottery ticket has 3 consecutive
numbers, the others not?

Replies: 7   Last Post: Feb 12, 2009 11:18 PM

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matt271829-news@yahoo.co.uk

Posts: 2,136
Registered: 1/25/05
Re: What probability a six-number lottery ticket has 3 consecutive
numbers, the others not?

Posted: Feb 11, 2009 9:31 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Feb 11, 8:52 am, Archie <kohmover...@yahoo.com> wrote:
> Say you can choose six numbers out of 59. What is the probability that
> three (and only three) will be consecutive? This is complicated by the
> fact that numbers appear in order from least to greatest. What if the
> six numbers contain 2 sets of 3 consecutive numbers?
>
> I'm not finding any easy formula for this.
>
> Thanks!


I'll assume that you want the probabilities after the six numbers are
sorted in ascending order. In other words, the order in which the six
are selected is irrelevant.

Let n be the number of numbers from which the six are chosen (in your
case n = 59). The total number of ways to choose the six, disregarding
order, is C(n, 6) ("n choose 6"), where C(x, y) is a binomial
coefficient calculated as C(x, y) = x!/(y!*(x - y)!).

Number of ways to choose six such that there are two separated runs of
three in consecutive order =

n - 6 ways if the lower run is 1, 2, 3
+ n - 7 ways if the lower run is 2, 3, 4
+ n - 8 ways if the lower run is 3, 4, 5
+ ...
+ 1 way if the lower run is n - 6, n - 5, n - 4

= (n - 6)*(n - 5)/2

So probability = (n - 6)*(n - 5)/(2*C(n, 6))

Number of ways to choose six such that there is exactly one run of
three and no longer run =

C(n - 4, 3) if the run is 1, 2, 3 (remaining three can be any three
from n - 4)
+ C(n - 5, 3) if the run is 2, 3, 4 (remaining three can be any three
from n - 5)
+ C(n - 5, 3) if the run is 3, 4, 5
...
+ C(n - 5, 3) if the run is n - 3, n - 2, n - 1
+ C(n - 4, 3) if the run is n - 2, n - 1, n

= 2*C(n - 4, 3) + (n - 4)*C(n - 5, 3)

BUT... every selection containing two separated runs of three will be
included twice in this count, so we need to subtract twice the answer
to the first question to get the required

Probability = (2*C(n - 4, 3) + (n - 4)*C(n - 5, 3) - (n - 6)*(n - 5))/C
(n, 6)




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