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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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Stephen J. Herschkorn

Posts: 2,297
Registered: 1/29/05
Re: What probability a six-number lottery ticket has 3 consecutive
numbers, the others not?

Posted: Feb 12, 2009 11:18 PM
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Archie 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 got the following answers:

P{two disjoint triplet runs, but not a run of six} = 159/5006386 ~~
3.176 x 10^-5.
P{exactly one triplet run} = 1413877/45057474 ~~ 0.0314

Methods:

For two triplet runs, note that the highest possible run starts at 57.
Therefore, the highest lower triplet run starts at 53.
If the lower triplet run starts at n, the the higher triplet run must
start at at least n + 4, Thus, there are are [57 - (n + 4) + 1] = 54 -
n such runs. Thus, there are sum(n=1..53, 54-n) = sum(n=1..53, n) = 53
x 54/2 such sextuplets. This gives the numerator of the probability,
where the denominator is C(59, 6).

For a single triplet: If the triplet is 1 2 3 or 57 58 59, then
there are C(55,3) - 53 possible combinations for the three other
numbers. For 2 3 4 or 56 57 58, the count is C(54,3) - 52. For 3 4
5 or 55 56 57, C(54,3) - 51. For 4 5 6 or 54 55 56, there are
C(54, 3) - 50. For 5 6 7 through 53 54 55, there are C(54, 3) - 49.
So the total number of possible outcomes are 2[C(55,3) - (50 + 51 + 52 +
53)] + 55 C(54,3) - 49 x 49.

YMMV.

--
Stephen J. Herschkorn sjherschko@netscape.net
Math Tutor on the Internet and in Central New Jersey



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