Dear Cristian-Augustin,
There seems to be a mistake in your formula
P = (2*a^3 + 2*b^3 - 4*a^2 - 4*b^2 - 4*a*b + 3*a + 3*b + 2)
/ (2 -a - b) / 3
for, as you point out, a=0, b=1 should give P=2/3. My own
calculations indicate that your first two terms 2*a^3+2*b^3
should be a^3+a^2*b+a*b^2+b^3 instead. Then the expression
simplifies to P = 1 - ((1-a)^2 + (1-b)^2) / 3
If a<length1<b and c<length2<d the expression is a bit more
complicated, so we introduce the following notation:
AC4 = (1-a-c)^4, if (1-a-c) is positive, otherwise AC4=0
and BC4, AD4, BD4 analogously. The final formula is
P = 1 - (AC4-BC4-AD4+BD4)/3((1-a)^2-(1-b)^2)((1-c)^2-(1-d)^2)
The derivation is straight-forward. First compute the
probability P(L,M) that two randomly placed intervals of fixed
lengths L and M will intersect. This is easy:
P(L,M) = 1 - (1-L-M)^2 /(1-L)(1-M), if (1-L-M) is positive,
otherwise P(L,M)=1. Next compute the probability density f(L)
of the length of a random interval, under the condition that
this length lies between a and b. This is even easier:
f(L) = 2(1-L)/((1-a)^2-(1-b)^2), if a<L<b, zero otherwise.
The desired probability is the very simple integral
P = int f(L)f(M) P(L,M) dL dM over 0<L<1, 0<M<1.
/Henrik Eriksson, Stockholm
.
.
=================================================================
Instructions for joining and leaving this list, remarks about the
problem of INAPPROPRIATE MESSAGES, and archives are available at:
. http://jse.stat.ncsu.edu/ .
=================================================================
