Photo:
THE MIRROR
The game is a bit reminiscent of Battleships.
On the table is a square box with side length 5. Mirijana and Martin each place a square piece of paper with side length 1 in this box.
Both have their eyes closed, but make sure that the edges of their square are parallel to the edges of the square box.
Mirijana places her square first, then it's Martin's turn.
What is the probability that the two squares of Mirijana and Martin overlap?
Note: The two squares must be completely inside the box.
The probability is 49/256, which is about 19 percent.
We first simplify the task to one dimension.
Mirijana and Martin place their square with an edge length of 1 on a line with a length of 5. The center of the square must be on the line - and the square must not protrude beyond the line to the left or right.
What is the probability that both squares overlap?
As in the box above, we assume an edge on the left and right, which the square must not protrude over.
So the center of a square must be at least 0.5 from the left and right ends of the line.
The range from 0.5 to 4.5 from the left edge is available for selection.
This area has a total length of 4 - see sketch below.
We can now easily calculate the probability of an overlap if we create a sketch that visualizes the options of Mirijana and Martin at the same time.
The gridded square on the right of the sketch shows, on the x-axis, the area on the path that Mirijana can choose for the center of her square.
Its length is 4.
The y-axis shows the area of the same line that Martin can choose for the center of his square.
It also has length 4.
The square of size 4x4 shows the space of possibilities of both players.
However, the squares only overlap if the distance between the two centers is less than 1.
The area highlighted in pink represents the cases where the squares overlap.
The proportion of the pink colored area to the total area of the square is 7/16.
So there is a 7/16 chance that the squares will overlap if they are randomly placed on a line of length 5 without protruding beyond the edge of the line.
It is now easy to calculate the probability that the two squares will overlap if Mirijana and Martin put them in the square with edge length 5.
We imagine that both place a square on a horizontal line of length 5 - and a square on a vertical line of length 5. Horizontal and vertical stand for the x and y axes of the square box with the length 5
In order for the two squares to overlap, they must overlap both horizontally and vertically.
The probability of this happening is 7/16 times 7/16 – i.e. 49/256, which is about 19 percent.
I discovered this rather difficult puzzle on the math portal Mathigon.
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