|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|
|Posted by watup90306||05/18/06|
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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|
That was fun! :D|
|Posted by McBobby1212||03/01/07|
|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.|