In a lottery drawing 6 numbers from 1 to 50, what is the probability that the drawing contains at least one...
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Problem
In a lottery drawing 6 numbers from 1 to 50, what is the probability that the drawing contains at least one pair of consecutive numbers? Down here in Texas we have one of those take from the poor to make one rich schemes--the State lottery. It involves picking 6 numbers from 1 through 50 without replacement. I was surprised to note that 10 of the last 16 drawings had consecutive numbers. A fluke? What are the odds of one or more pairs of consecutive numbers in such a drawing (assuming this is a fair, i.e. random drawing)? -- Brian Hall <brian@gebeg.win.net>
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My calculations put the odds of one or more consecutive numbers to be
519217/1059380 exactly. Does anyone else care to confirm this
calculation?
-- Dave Radcliffe <radcliff@alpha2.csd.uwm.edu>
Sorry, I can't confirm. The probability is:
1 - binomial(45,6)/binomial(50,6) which is equal to
18442
-----
37835
That is about 48.7 percent.
-- Mark van Hoeij <hoeij@sci.kun.nl>
If you take a drawing without consecutive numbers and subtract 5 from
the biggest number, 4 from the next biggest, ..., 0 from the smallest,
you get 6 distinct numbers between 1 and 45. This is a 1-1
correspondence of the two sets involved, so the probability of having
consecutive numbers in a drawing is 1-C(45,6)/C(50,6) = 18442/37835 ~=
0.48743.
-- David Moews <dmoews@xraysgi.ims.uconn.edu>
In article <34v9r0INNhlp@uwm.edu> radcliff@alpha2.csd.uwm.edu (David G
Radcliffe) writes:
>My calculations put the odds of one or more consecutive numbers
>to be 519217/1059380 exactly. Does anyone else care to confirm
>this calculation?
Six unequal numbers from 1 to 50 with no two consecutive is equivalent
to six unequal numbers from 1 to 45 (namely a0,a1-1,a2-2,a3-3,a4-4,a5-5
if the original six were a0,a1,a2,a3,a4,a5 in increasing order). Since
each of the 50!/44! configurations of six numbers is assumed to have
equal probability, the probability of having no two consecutive is thus
(45!/39!) / (50!/44!) = 19393/37835 or about 51.257%, so some two
consecutive numbers occur with complementary probability
1 - 19393/37835 = 18442/37835 or about 48.743% . How did you get
519217/1059380 (roughly 49.011%) ?
-- Noam D. Elkies <elkies@zariski.harvard.edu>, September 1994