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 Squares Probability III (Posted on 2010-08-19)
Consider a 6 by 6 grid of random, base 10 integers, uniformly distributed between 1 and 99, such as in "Another Square". Construct a second grid where the content of each cell in the second grid is the sum of the orthogonal neighbors of its corresponding cell from the first grid.

What is the probability that a given number in the second grid is a perfect square?

If the grids were infinite, what is the probability that a given number in the second grid would be a perfect square?

 No Solution Yet Submitted by Larry No Rating

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 re(2): solution Comment 3 of 3 |
(In reply to re: solution by Harry)

You are right. The respective probabilities for the totals of two, three and four random integers are:

541/99^2
41798/99^3
3524977/99^4

Then, weighted with the 4/36, 16/36 and 16/36 proportions of these types of square, the overall probability becomes 35954257/864536409 ~= 0.0415879037894863257.

The infinite case is just the final 3524977/99^4 = 3524977/96059601 ~= 0.0366957281032220818.

 Posted by Charlie on 2010-08-20 17:22:50

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