A disk that is 1 cm in diameter is tossed onto a 6x6 grid of squares with each square having side length 1.25 cm.
A disk is in a winning position if no part of it touches or crosses a grid line otherwise it is in a losing position
The disk is tossed and it lands in a random position so that no part of it is outside the grid.
What is the probability that the disk is in a winning position?
(In reply to
Three answers (spoiler) by Steve Herman)
" you win if the center of the disk is in an area of (6*.75)^2 = 20.25"
The .75 in the above, I assume, is from 1.25  .5. But the center can't be within .5 of the left side of the small square, nor within .5 of the right side; the same applies in the other dimension, within neither .5 cm of the top nor the bottom. That takes a full centimeter off the 1.25 cm of each side, so it should be 1.25  1, or .25 cm making the total winning disk area (6*.25)^2 = 2.25.
In fact I now see an error in my own arithmetic (strange as I used a calculator, but must have mistyped). The 6.25 in my first numerator should be the 2.25 mentioned above. I will fix that; but it is indeed for a radius of .5 cmit has to limit placement on all four sides of each square, and my interpretation was your # 1).

Posted by Charlie
on 20160708 20:46:20 