Two containers hold several balls. Once a second, one of the balls is chosen at random and switched to the other container. After a long time has passed, you record the number of balls in each container every second. In $10,000s$, you find $80$times when all the balls were in one container (either one) and the other container was empty.

a. How many balls are there?

b. What is the most likely number of balls to be found in one of the containers?

Given :

In $10,000s$, number of times all the balls were in one container : $80$times

Theory used :

Because there are two directions and they are chosen at random, the probability of a second ball moving in the same direction as the first is just $1/2$. It's also worth noting that if we start with two boxes each containing $N/2$balls, the first draw will be random, while the rest must all be in the same direction.

That is to say, all $\frac{N}{2}-1$draws must be in the same direction.

Probability of getting an empty box is :

$p={\left(\frac{1}{2}\right)}^{\frac{N}{2}-1}$ ...(1)

a. Starting from a state where the total number of balls $N$is evenly distributed between the two boxes, we must switch$N/2$balls in one way, one after the other, to arrive at a state where one box is full and the other is empty.

From equation (1), we have :

$$\begin{gathered} \frac{N}{2}-1={\log }_{0.5}p \\ \Rightarrow N=2(1+{\log }_{0.5}p) \end{gathered}$$

We must now calculate the probability. We know that the state we're looking for happened a certain number of times in a certain number of trials. As a result, the probability will be $p=\frac{80}{10000}=0.008$

Substituting this result, we can find the number of balls to be

$N=2(1+{\log }_{0.5}0.008)=\boxed{\mathbf{16}}$

b. The number of balls will tend to be distributed evenly because the procedure is random and has been repeated for a long time; that is, the most likely number to be discovered in the boxes will be$\boxed{\mathbf{8}}$