Problem Analysis

We are tasked with determining the probability that boy $C$ has the ball after three throws, given the following conditions:

1. Boy $A$ always throws the ball to $B$.
2. Boy $B$ throws the ball equally likely to $A$ or $C$.
3. Boy $C$ throws the ball equally likely to $A$ or $B$.
4. The game starts with $C$ having the ball.

Steps to Solve

We will model this problem using conditional probabilities and compute the required probability using a transition matrix or recursive reasoning.

Assigning Probabilities

• The probability of $A \to B$ is $1$ (always).
• The probabilities for $B \to A$ and $B \to C$ are both $\frac{1}{2}$.
• The probabilities for $C \to A$ and $C \to B$ are both $\frac{1}{2}$.

We need to track the state of the ball (who has it) after each throw. We'll compute the probabilities after three throws.

Solution Using States

Let $P_n(X)$ represent the probability that boy $X$ has the ball after $n$ throws. At the start ($n = 0$), we have:

$P_0(A) = 0, \, P_0(B) = 0, \, P_0(C) = 1$

The transitions for each state are:

• $P_{n+1}(A) = P_n(C) \cdot \frac{1}{2} + P_n(B) \cdot \frac{1}{2}$
• $P_{n+1}(B) = P_n(C) \cdot \frac{1}{2} + P_n(A) \cdot 1$
• $P_{n+1}(C) = P_n(B) \cdot \frac{1}{2}$

Iterative Calculations

After 1 Throw:

$P_1(A) = P_0(C) \cdot \frac{1}{2} = 1 \cdot \frac{1}{2} = \frac{1}{2}$ $P_1(B) = P_0(C) \cdot \frac{1}{2} = 1 \cdot \frac{1}{2} = \frac{1}{2}$ $P_1(C) = 0$

After 2 Throws:

$P_2(A) = P_1(C) \cdot \frac{1}{2} + P_1(B) \cdot \frac{1}{2} = 0 + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ $P_2(B) = P_1(C) \cdot \frac{1}{2} + P_1(A) \cdot 1 = 0 + \frac{1}{2} = \frac{1}{2}$ $P_2(C) = P_1(B) \cdot \frac{1}{2} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

After 3 Throws:

$P_3(A) = P_2(C) \cdot \frac{1}{2} + P_2(B) \cdot \frac{1}{2} = \frac{1}{4} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$ $P_3(B) = P_2(C) \cdot \frac{1}{2} + P_2(A) \cdot 1 = \frac{1}{4} \cdot \frac{1}{2} + \frac{1}{4} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8}$ $P_3(C) = P_2(B) \cdot \frac{1}{2} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

Final Answer

The probability that $C$ has the ball after three throws is:

$\boxed{\frac{1}{4}}$

Let me know if you'd like a deeper breakdown or have additional questions!

Related Questions:

1. How would the probabilities change if $C$ always threw to $B$?
2. What happens if $B$ throws the ball with probabilities $\frac{1}{4}$ to $A$ and $\frac{3}{4}$ to $C$?
3. How can we generalize this to $n$ throws?
4. What is the probability that $B$ has the ball after three throws?
5. Can we solve this using a Markov Chain transition matrix?

Tip: Drawing a probability tree can simplify these calculations significantly!