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Category: | Probability |

Submitted By: | cms271828 |

Fun: | (2.23) |

Difficulty: | (2.99) |

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)?

Show Answer |

Posted by Psychic_Master | 04/19/06 |

Yay! First to comment! Well, pretty good teaser, and pretty hard too. Good job. :wink: | |

Posted by vlerma | 04/20/06 |

:oops: 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. | |

Posted by BrownEyes | 04/21/06 |

Grrrr that teaser wasn't easy... And y didn't you share your chocolate with me in the chat room.?? *waaahh!!!* :cry: :cry: | |

Posted by Moondancer | 04/22/06 |

So very confusing! :( | |

Posted by cms271828 | 04/22/06 |

Its not really, just follow the solution on paper and everything will become clear | |

Posted by Jimbo | 04/26/06 |

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! :D | |

Posted by garul | 04/27/06 |

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. | |

Posted by Jim1029 | 05/14/06 |

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

Posted by kayleeskitties | 05/16/06 |

All I can say is Wow.....very hard....very good teaser though! :D | |

Posted by cms271828 | 05/16/06 |

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 :D | |

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Posted by udoboy | 05/26/06 |

Thank you. My brain exploded while contemplating this question. | |

Posted by love_____xoxo | 09/10/06 |

whoa. That was fun! :D | |

Posted by McBobby1212 | 03/01/07 |

my.....brain.....is.....turning......to.......mush :-? | |

Posted by gnarldroot | 02/04/10 |

Are you kidding me? this is probably the most basic probability riddle ive ever seen. | |

Posted by bug-e | 12/04/11 |

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 :P. | |

Posted by rachit138 | 06/08/12 |

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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