A $30 \times 30$ grid of squares contains $900$ fleas, initially one flea per square.
When a bell is rung, each flea jumps to an adjacent square at random (usually $4$ possibilities, except for fleas on the edge of the grid or at the corners).
What is the expected number of unoccupied squares after $50$ rings of the bell? Give your answer rounded to six decimal places.
This problem can be solved using the concept of expected value and the property of linearity of expectation.
First, consider a general flea. If we disregard the edges of the grid (momentarily), a flea has 4 adjacent squares to which it can jump. The probability that a flea does not remain in the initial square after 1 bell ring is 1 (Because it must jump). It can only return to its original position after an even number of jumps.
Its probability of being back after 2 jumps (2 rings) is 1/4 (it has to go back the way it came).
After 4 jumps (4 rings), we could consider the various distinct paths. Basically there are 2 scenarios, the flea makes two cycles of leaving and returning, and the flea goes to an adjacent square, travels around, and returns. Mathematically the probability is calculated as 6*(1/4)^4.
This pattern continues for additional rings.
However, given the flea must jump when the bell rings, we can say the flea is never in the original square for odd number of rings. And it’s also important to note that a flea starting at the edge has a smaller chance to return to its initial square (less options to jump), and corners have the least chance.
But since we need to find the expected number of unoccupied squares, we can simplify the problem to finding the expected number of fleas per square and subtracting this from 1.
The key is to observe the symmetry of the problem. Because the fleas all move simultaneously and independently, each square’s probability of hosting a flea is the same as every other square’s. Therefore, the expected (average) number of fleas per square after 50 rings is just the total number of fleas divided by the total number of squares, or 900/900 = 1.
Therefore, the expected number of unoccupied squares, which is just 1 minus the expected number of fleas in each square after 50 rings, is 1 – 1 = 0.
So, despite the seeming complexity of the problem, we actually expect every square to be occupied (on average) after 50 rings of the bell.
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