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Overlapping Squares

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.


Puzzle ID:#29898
Fun:*** (2.23)
Difficulty:*** (2.98)
Submitted By:cms271828**
Corrected By:DaleGriffin




Two squares each 17cm by 17cm are drawn randomly inside a square measuring 1m by 1m.
The edges of the two smaller squares are both parallel/perpendicular to the edges of the large square.
What is the exact probability the two smaller squares overlap (or touch)?

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Apr 19, 2006

Yay! First to comment! Well, pretty good teaser, and pretty hard too. Good job.
Apr 20, 2006

Sorry too complicated for this old brain. Suspect there are seom brainiacs out there who can do these, but I am not one of them. But even so, it appears you v\gave a very good expanation of how you arrived at your answer so all in all It was a good job.
Apr 21, 2006

Grrrr that teaser wasn't easy... And y didn't you share your chocolate with me in the chat room.?? *waaahh!!!*
Apr 22, 2006

So very confusing!
Apr 22, 2006

Its not really, just follow the solution on paper and everything will become clear
Apr 26, 2006

It's a very interesting problem and a well constructed solution but I am not going to vote on this until I have thought about the solution. Certainly if the squares fall in the designated area, they will touch/overlap, but I am struggling to connect the specific areas with a general solution. I like the problem and the way the solution has been tackled!
Apr 27, 2006

Superb problem. I got it though by a round about method. Assuming the side of the small sq is A meters (0.17 in this case) and the big sq is 1 M, the answer I got is (A*(2-3A)/(1-A)^2)^2 which matches the solution.

I solved it by splitting the bigger square into three areas (a center sq of side 1-3A, 4 corner squares of side A and 4 rectangles of side A & 1-3A). The probability of the squares overlapping when the first square is in each of the above areas are 4AA/(1-A)^2, 9AA/4/(1-A)^2 and 3AA/(1-A)^2. Adding these after multiplying by first square probabilities, we can get the answer as (A*(2-3A)/(1-A)^2)^2.

The solution given by the author is definitely more elegant. Here in the xy grid of side (1-A), overlap will happen if the X coordinates of the two squares are within the center rectangle of sides sqrt(2)*A and sqrt(2)*(1-2A) or the two corner triangles. So the probability of one coordinate say X suitable for overlapping is sqrt(2)*A * sqrt(2)*(1-2A) /(1-A)^2 + AA = A(2-3A)/(1-A)^2. Square of this will give the answer.
May 14, 2006

In being one of the editors who accepted this teaser, I think it is really hard. Even I couldn't get it!
May 16, 2006

All I can say is Wow.....very hard....very good teaser though!
May 16, 2006

Thanks, I'm glad everyone liked it, my first solution was difficult, using double integrals, and was tough to understand.
But while I was lying in bed, the solution just came to me, and in the middle of the night, I got up and grabbed my calculator, tested it, and it matched my other solution.
Just to make sure it was correct, I did a simulation using a java program. I think I ran the experiment a billion times, counted the successes, and divided by a billion.
The result was practically identical (to about 4sf) to the theoretical probability.
Numerical verification isn't an absolute proof, but its good enough for me
May 18, 2006

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May 26, 2006

Thank you. My brain exploded while contemplating this question.
Sep 10, 2006

That was fun!
Mar 01, 2007
Feb 04, 2010

Are you kidding me? this is probably the most basic probability riddle ive ever seen.
Dec 04, 2011

The solution is nice and there are many more problems that can be solved using the same technique. This is a set method for problems like these.
Another problem could be like probability of whether bf and gf meet each-other in 1 hr window if the other one can only wait for 15 mins and they dont care much abt each other .
Jun 08, 2012

The squares can overlap diagonally as well in which case the distance between their centers will be \sqrt(2) times 17.

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