On a regular two dimensional coordinate plane, you have a square with side length 1 unit.

Pick a point within the square at random, and from there travel a random but straight direction .5 units.

What is the probability that you end up still within the square?

(I don't know why the previous message was crash)


I defined the square in the range
- 0.5 < x < 0.5 and
- 0.5 < y < 0.5

I can divide the square in four sub squares,
one in each cuadrant.

I will find the probability that the end point
lies on the original square given that the start
point lies on the first quadrant

finally i must multiply this probability by four
since there are four quadrants.

On the first quadrant two types of points are
identified:

- The one that lies under the curve
0.5-sqrt(0.5*0.5-(x-0.5)(x-0.5))
- The one that lies over the same curve.

the diference between these two types of point is:


Type 1: If we draw a circle with center in a
point of this type, the circle intercept the
original square 4 times.

Type 2: The circle intercept the original Square
only 2 times

The importance of knowing how many times the
square is intercepted is that one can know
exactly from what angle to what angle a end
point is outside of the original Square.

In Type 1 there are two angle zones where the
end point is outside in Type 2 there is only one
zone

To find the probability (already multiplied by four):

Type 1:

The integral from x=0 to x=0.5 of
the integral from y=0 to 0.5-sqrt(0.5*0.5-(x-0.5)*(x-0.5)) of

(
2*pi
- 2*arcsin( sqrt(0.5*0.5-(0.5-x)(0.5-x))/0.5)
- 2*arcsin( sqrt(0.5*0.5-(0.5-y)(0.5-y))/0.5)
) /2/pi * dx * dy /0.25

Type 2:

The integral of x=0 to x=0.5 of
the integral of y=0.5-sqrt(0.5*0.5-(x-0.5)(x-0.5)) of

(
2*pi
-pi/2
- arcsin( sqrt(0.5*0.5-(0.5-y)(0.5-y))/0.5)
- arcsin( sqrt(0.5*0.5-(0.5-x)(0.5-x))/0.5))
) /2/pi * dx * dy /0.25


These two integrals has three parts each that
give the result

(4-pi)/4

(pi*pi-12)/16/pi

(pi*pi-12)/16/pi

3Pi/16

(-pi*pi-4)/32/pi

(-pi*pi-4)/32/pi

Adding these 6 parts gives the result of:

1-7/4/pi or aproximately 0.4429576992



Pablo Meraz

Posted by Pablo on 2004-01-29 15:25:48