Three positive integers are chosen at random without replacement from 1,2,....,64.
What is the probability that the numbers chosen are in geometric sequence?
Order of choice doesn't matter. For example, 4-1-2 would qualify as
numbers in geometric sequence.
Generalise this result (in terms of n) covering the situation where three positive integers are chosen at random without replacement from 1,2,.....,n2.
First note that there are C(64,3)=41664 unordered ways of choosing 3 numbers from the 64. The real question is how many of these triplets are in geometric sequence.
If the common ratio of terms r=2 there are 64/2^2 = 16 triplets
if r=3 there are 64/3^2=7.111 so 7 triplets
if r=4 there are 4 triplets
if r=5 there are 2
if r=6, 7 or 8 there is only 1 each.
The total is 32 unordered triples.
***EDIT: THIS IS WRONG. I DID NOT CONSIDER NON-INTEGER RATIOS***
32/41664 = 1/1302 ≈ .000768
Bonus: You can basically count up the same way there are C(nē,3) = nē(nē-1)(nē-2)/3 unordered ways of choosing 3 numbers.
Then count the unordered triples as before
Sum [nē/rē] for all r from 2 to n
Then divide the two.
I'll work on the closed form for this.
***WHICH MEANS THIS IS ALSO WRONG***
Edited on February 28, 2013, 10:57 am
Posted by Jer
on 2013-02-28 10:54:38