Barbara writes a sequence of integers starting with the number 12. Each subsequent integer she writes is chosen randomly with equal probability from amongst the positive divisors of the previous integer (including the possibility of the integer itself). She keeps writing integers until she writes the integer 1 for the first time, and then she stops.

An example of one such sequence is 12, 6, 6, 3, 3, 3, 1.

What is the expected value of the number of terms in Barbara’s sequence?

12 can go to 12, 6, 4, 2, 1 each with probability 1/6

So from each of these we need the expected number of additional terms.

A 1 gives 0 additional terms

A 2 gives a 1/2 chance of 1 additional term and a 1/2 chance of being a 2 again.   If we get a 2 there will be at least one extra turn beyond what we had.
x = 1/2*1 + 1/2(1+x)
x = 1 + 1/2 x
1/2 x = 1
x = 2 additional terms.

A 3 also gives 2 additional terms.

A 4 gives
a 1/3 chance of 0+1=1 additional term,
a 1/3 chance of 2+1=3 additional terms, and
a 1/3 chance of being a 4 again.
x = 1/3*1 + 1/3*3 + 1/3(1+x)
x = 5/3 + 1/3 x
2/3 x = 5/3
x = 5/2 additional terms.

A 6 gives
a 1/4 chance of 0+1=1 extra term,
a 1/4 chance of 2+1=3 extra terms,
a 1/4 chance of 2+1=3 extra terms, and
a 1/4 chance of being a 6 again.

x = 1/4*1 + 1/4*3 + 1/4*3 + 1/4(1+x)
x = 2 + 1/4 x
3/4 x = 2
x = 8/3 additional terms.

A 12 gives
a 1/6 chance of 0+1=1 term
a 1/6 chance of 2+1=3 terms
a 1/6 chance of 2+1=3 terms
a 1/6 chance of 5/2+1 = 7/2 terms
a 1/6 chance of 8/3+1 = 11/3 terms and
a 1/6 chance of being a 12 again

x = 1/6*1 + 1/6*3 + 1/6*3 + 1/6*7/2 + 1/6*11/3 + 1/6(1+x)
x = 79/36 + 1/6 x
5/6 x = 91/36
x = 91/30 terms after the first 12

Add 1 for this first 12 and the solution is 121/30 or 4.03333... terms

Posted by Jer on 2010-04-09 15:08:16