Problem 185 Solution

Instead of 10 inches lets give the stick a length of 1. Let x be the point of the original cut. The probability given x that all four pieces are 0.1 long or greater is (x²-x+0.16)/(x²-x). Next take the integral over this probability for the larger initial cut ranging from 0.5 to 0.8. Note that if it were greater than 0.8 the smaller piece could not have two cuts more than 0.1. So we have 2 times the integral from 0.5 to 0.8 of

(1) x²/(x²-x) + ...
(2) -x/(x²-x) + ...
(3) 0.16/(x²-x)

Integral (1) works out to 0.3+ln(.4)

Integral (2) works out to -ln(.4)

Integral (3) can also be expressed as 0.16*(1/(x-1) - 1/x). The integral of 1/x from 0.5 to 0.8 is ln(0.8)-ln(0.5). The integral from 1/(x-1) from 0.5 to 0.8 = ln(x-1) from 0.5 to 0.8 = ln(-0.2)-ln(-0.5) = ln(-1)+ln(.2)-(ln(-1)+ln(.5)) = ln(.2)-ln(.5). So the initial expression equals 0.16*(ln(0.2)-ln(0.5)-(ln(0.8)-ln(0.5))) = 0.16*(ln(0.2)-ln(0.8)) = 0.16*ln(.2/.8) = 0.16*ln(1/4) = -0.16*ln(4)=-0.32*ln(2).

So the final answer is 2*(0.3+ln(.4)-ln(.4)-0.32*ln(2)) = 0.6-0.64*ln(2) = (15-16*ln(2))/16 =~ 0.1564

To check my answer I ran a simulation with the following results:

Outcome	 Number  Probability
Success:   49883201     0.156374
Failure:  269116799     0.843626
Total:    319000000     1.000000

Here is my old and much harder solution to solving the third integral...

Integral (3) is the tough one. My copy of C.R.C. Standard Mathematical Tables, which I got on Ebay, says that the integral of 1/X, where X=a+bx+cx² and q=4ac-b², is -2/sqr(-q)*tanh^-1((2cx+b)/sqr(-q)).

In the case of a=0, b=-1, c=1 the integral works out to 0.16*tanh^-1x. Given the bounds of 0.5 to 0.8 the area under the curve is 0.16*-2*tanh^-10.6

My copy of C.R.C. Standard Mathematical Tables also reminds me that tanh^-1(x) = ln(1+x)/2 - ln(1-x)/2. So (3) equals 0.16*-2*(ln(1.6)/2 - ln(0.4)/2) = -0.16*(ln(1.6)-ln(0.4)) = -0.16*ln(4)

So the final answer is 2*(0.3+ln(0.4)-ln(0.4)-0.16*ln(4)) = 2*(0.3-0.16*ln(4)) = (15-8*ln(4))/25 = (15-16*ln(2))/25 =~ 0.156386

Michael Shackleford, A.S.A, January 16, 2002