Problem 18. Ants on a Piece of Wood
There is a piece of wood of length 1 meter which is horizontally placed on a tree. 50 ants are located on random positions of the wood. Each ant is either directed to the right or the left with equal probabilities. Each ant starts moving at a speed of 1 meter per hour and falls down the tree when it reaches the end of the wood. If two ants intersect, both of their direction will be reversed and they both keep moving in the opposite directions with the same speed.
How long does it take on average for the last ant to fall? We assume for simplicity that the size of each ant is negligible and their locations and their directions are realized independently.
Link to the problem on Twitter: https://twitter.com/Riazi_Cafe/status/1683010953217204224
. The correct answer is equal to 50/51 hours or \(\simeq 58.82\) minutes.
Since there is no distinction between the ants, we can assume that when two ants meet each other, they bypass each other and continues their moving with the original directions. Therefore, the answer to the problem is equal to the maximum distance of an ant to the end of the wood in the direction of the ant’s movement.
Because the distance of each ant to the end of the wood in the direction of its movement is an independent random number between zero and one (in meters), the answer to the question is equal to \(\max\{a_1, a_2, a_3, ..., a_{50}\}\) where each \(a_i\) is drawn uniformly and independently at random from interval \([0,1]\). This number can be determined with the following integral: \[\int_0^1 x(50x^{49}) d_x = 50/51.\]
Clarification on the integral: The probability that one ant has distance \(d\) to the end of the wood and the rest of the ants have distance at least \(d\) to the end of the wood is equal to \(\binom{50}{1} x^{49}\).
Link to the solution on Twitter: https://twitter.com/Riazi_Cafe/status/1683959598485577728