Need help with a probability problem about collisions

Question

There are $100$ worker bees inside a $1$m long closed tube randomly distributed except at the centre. A queen bee is placed at the centre of this tube. Each bee is facing either one of the two directions- left or right. At $t = 0$, all bees start flying at a speed of $1$m/s. If two bees collide, they reverse their direction. If a bee collides with the end of the tube, it bounces back. What is the probability that at $t = 1$s, the queen bee will be at the centre of the tube?

Answer 1

Let us assume that each bee has a unique flag, and when two bees collide they exchange their flags.

So,Even though the bees follow a chaotic motion, the flags follow a simple linear motion.

No matter wherever the bee goes, their flag will be at the same position,not necessarily the same as the starting position after travelling 1 m in 1 second with a different or the same bee and the relative order of bees will remain the same since the bees can't go over each other i.e. if the their are 3 bees, 1, 2 and 3 from left to right then the order 1, 2 and 3 will remain the same.

Now, let us assume that there x bees to the left of the queen and 100-x bees to the right of the queen each with their own unique flag and the queen will have her own unique flag too.

1st Case If x>50 or x<50(100-x>50),then there will be x flags to the right of centre after 1 second and 100-x flags to the left of centre after 1 second since the relative order of bees doesnt change the queen will never be at the centre, when one side has more bees than the other.

2nd Case If x=50, then after 1 second there will be an equal number of flags to left and right of centre moreover, the queen which is the 51st bee in relative order will be at the centre, since the queen flag will be the 51st flag, and definitely at the centre after 1s.

So,the answer is just the probability that the 1nd case occurs, which is $\frac{{\binom{100}{50}}}{{2^{100}}}$.