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Two points are picked at random uniformly from the border of a square with side length l.
What is the probability the distance d between the points is larger than l?
As usual, watch the video for a solution.
Probability Random Points On Square Sides Longer Than Side Length
Or keep reading.
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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.
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Answer To Probability Random Points On Square Sides Longer Than Side Length
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
Without loss of generality, suppose the first point is on the bottom side of the square (if not we can rotate the square to force it). There are 4 equally likely cases for the second point.
A) The second point is on the same side. The two points cannot be greater than l in distance, so the probability is 0.
B) The second point is on the opposite side. The two points must at least be l apart in distance, so the probability is 1.
C) The second point is on the adjacent right side.
D) The second point is on the adjacent left side.
By symmetry, the last two cases have the same probability and we need to figure it out.
Adjacent sides
Let the first point be (x, 0) and the second (0, y). Since x and y are uniformly between 0 and l, we know the midpoint M(x/2, y/2) will be uniformly distributed between [0, l/2] × [0, l/2]. This is a square with side length l/2.
Suppose the two points have a distance of exactly l. What is the locus of the midpoints M? This will exactly be a quarter circle.
To see why, the distance line segment can be broken down into two congruent right triangles with sides y/2 and x/2, and the midpoint divides the distance line segment into two lengths of l/2. Since we have a right triangle we have:
(x/2)2 + (y/2)2 = (l/2)2
For non-negative x and y this is precisely the quarter circle.
If the midpoint is inside this quarter circle and the two axes, the distance between the two points will be less than l. If the midpoint is outside of this quarter circle boundary, then the distance will be larger than l.
The small square has area (l/2)2, and the quarter circle has area π/4(l/2)2.
Pr(d < l)
= (quarter circle area)/(square area)
= (π/4(l/2)2)/(l/2)2
= π/4
Pr(d > l)
= 1 – Pr(d < l)
= 1 – π/4
Final calculation
There are 4 equally likely cases.
A) The second point is on the same side. Probability = 0.
B) The second point is on the opposite side. Probability = 1.
C) The second point is on the adjacent right side. Probability = 1 – π/4.
D) The second point is on the adjacent left side. Probability = 1 – π/4.
Each case has a chance of 1/4, so the total probability is 1/4 times the sum of the probabilities in each case:
(1/4)(0 + 1 + (1 – π/4) + (1 – π/4))
= 3/8 – π/8
≈ 0.357
Source
Reddit Askmath
https://www.reddit.com/r/askmath/comments/14t49sd/trivia_question_about_probability/
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