math-prob-01

# The Vanishing Coins

You start with **3 fair coins** on the table.

A **round** consists of flipping every coin that is currently on the table (all at once). After the flips, remove every coin that landed **tails**; every coin that landed **heads** stays on the table for the next round. You then play another round with whatever coins remain.

The game ends at the moment the table is empty (no coins remain).

Let $R$ be the number of rounds played until the table is empty. Compute the expected value $\mathbb{E}[R]$.

Give your answer as an exact fraction in lowest terms, written as `a/b` (for example, `7/3`).

To find the expected number of rounds \( R \) until all coins are removed, we start by analyzing the process step by step.

1. **Initial Setup**: We begin with 3 fair coins.
2. **Round Dynamics**: In each round, each coin is flipped independently. If a coin lands heads, it stays; if it lands tails, it is removed. The number of coins remaining after each round follows a binomial distribution.

Let's denote the number of coins remaining after \( k \) rounds as \( X_k \). Initially, \( X_0 = 3 \).

### Step-by-Step Analysis:

1. **Round 1**:
   - Each coin has a \( \frac{1}{2} \) probability of landing heads and \( \frac{1}{2} \) probability of landing tails.
   - The number of heads (and thus the number of coins remaining) after the first round follows a binomial distribution \( X_1 \sim \text{Binomial}(3, \frac{1}{2}) \).
   - The expected number of coins remaining after the first round is \( \mathbb{E}[X_1] = 3 \cdot \frac{1}{2} = 1.5 \).

2. **Round 2**:
   - The remaining coins after the first round are \( X_1 \), which is a random variable with expected value \( 1.5 \).
   - Each of these \( X_1 \) coins again has a \( \frac{1}{2} \) probability of landing heads.
   - The number of coins remaining after the second round is \( X_2 \), which follows a binomial distribution \( X_2 \sim \text{Binomial}(X_1, \frac{1}{2}) \).
   - The expected number of coins remaining after the second round is \( \mathbb{E}[X_2] = \mathbb{E}[X_1] \cdot \frac{1}{2} = 1.5 \cdot \frac{1}{2} = 0.75 \).

3. **General Case**:
   - After \( k \) rounds, the expected number of coins remaining is \( \mathbb{E}[X_k] = 3 \cdot \left( \frac{1}{2} \right)^k \).

We need to determine the expected total number of rounds \( R \) until all coins are removed. This occurs when \( X_k = 0 \).

### Summing the Rounds:

The total expected number of rounds \( R \) is the sum of the expected number of rounds needed for each stage until no coins remain:
\[ R = \sum_{k=1}^{\infty} \mathbb{E}[T_k] \]
where \( T_k \) is the number of rounds needed to remove all coins after \( k \) rounds.

Since \( X_k = 3 \cdot \left( \frac{1}{2} \right)^k \), we need to find the expected number of rounds until \( X_k = 0 \).

The expected number of rounds to remove all coins can be computed as follows:
\[ R = \sum_{k=1}^{\infty} \frac{1}{\mathbb{E}[X_k]} \]
\[ \mathbb{E}[X_k] = 3 \cdot \left( \frac{1}{2} \right)^k \]
\[ \frac{1}{\mathbb{E}[X_k]} = \frac{1}{3 \cdot \left( \frac{1}{2} \right)^k} = \frac{2^k}{3} \]

Thus,
\[ R = \sum_{k=1}^{\infty} \frac{2^k}{3} = \frac{1}{3} \sum_{k=1}^{\infty} 2^k \]

The series \( \sum_{k=1}^{\infty} 2^k \) is a geometric series with the first term \( a = 2 \) and common ratio \( r = 2 \). The sum of this infinite geometric series is:
\[ \sum_{k=1}^{\infty} 2^k = 2 + 4 + 8 + \ldots = \frac{2}{1-2} = -2 \]

However, this series diverges, indicating that the expected number of rounds grows exponentially. To correctly compute the expected value, we need to consider the process more carefully.

### Correct Approach:

Instead of summing the series directly, we recognize that each round reduces the expected number of coins by half. We can use the concept of expected value for geometric series in a different way.

The expected number of rounds to remove all coins can be computed using the concept of the geometric series and the fact that each round halves the expected number of remaining coins.

The expected number of rounds to remove all coins is given by:
\[ R = 3 \sum_{k=1}^{\infty} \frac{1}{2^k} \]

The series \( \sum_{k=1}^{\infty} \frac{1}{2^k} \) converges to 1:
\[ \sum_{k=1}^{\infty} \frac{1}{2^k} = 1 \]

Thus,
\[ R = 3 \cdot 1 = 3 \]

Therefore, the expected number of rounds \( R \) until all coins are removed is:
\[ \boxed{3} \]