Unfair Heads #2

Question

You have 11 coins, 10 of them are fair coins, and the 11th coin has two heads.

If you pick one of the coins at random, toss it three times and record a HHH (it landed three times on heads), what is the probability that the randomly picked coin is a two-heads coin?

Accepted Answer

If you pick 1 coin randomly out of 11, you have a 10/11 probability of choosing a fair coin, and 1/11 to choose the two-heads coin.

 	If you toss a fair coin, the chances that you have 3 heads in a row are 12.5% (0.5 * 0.5 * 0.5).
 	If you choose the two-heads coins, these chances are 100% (1 * 1 * 1).

So, all in all, the probability of recording HHH with a fair coin is equal to 10/11 * 1/8 = 10/88, and the probability of recording HHH with a two-heads coin is equal to 1/11 * 1 = 1/11.

Therefore, the probability that one of them is a two-heads coin is 1/11 / 18/88 = 18/88 = 8/18 = 4/9.

Alternative Approach Using Conditional Probability

To determine the probability that the randomly picked coin is the two-heads coin, given that it landed heads three times in a row (HHH), we can use Bayes' theorem.

Let's define the events as follows:

 	$A$: The event that the coin picked is the two-heads coin.
 	$B$: The event that three tosses of the coin result in heads (HHH).

We need to find $P(A | B)$, the probability that the picked coin is the two-heads coin given that it landed heads three times in a row. Bayes' theorem states: \begin{equation}P(A | B) =  \frac{P(B | A) P(A)}{P(B)} \end{equation} First, we determine each component of the equation:

 	$P(A)$: The prior probability of picking the two-heads coin.
Since there are 11 coins and only one is the two-heads coin:
\begin{equation}P(A) = \frac{1}{11}\end{equation}
 	$P(B | A)$: The probability of getting HHH given that the picked coin is the two-heads coin.
Since the two-heads coin always lands heads:
\begin{equation}P(B | A) = 1 \end{equation}
 	$P(B)$: The total probability of getting HHH regardless of which coin was picked.
This can be found using the law of total probability:
\begin{equation}P(B) = P(B | A) \cdot P(A) + P(B | A^c) \cdot P(A^c) \end{equation}

where $P(A^c)$ s the event of picking a fair coin. We already know $P(A) = \frac{1}{11}$ and $P(B | A) = 1$. Now we need $P(B | A^c)$ and $P(A^c)$.

 	$P(A^c)$ is the probability of picking a fair coin:
\begin{equation}P(A^c) = 1 - P(A) = 1 - \frac{1}{11} = \frac{10}{11} \end{equation}
 	$P(B | A^c)$ is the probability of getting HHH given that the picked coin is a fair coin.
For a fair coin, the probability of landing heads three times in a row is:
\begin{equation}P(B | A^c) = (\frac{1}{2})^3 = \frac{1}{8} \end{equation}

Putting these together: \begin{equation}P(B) = P(B | A) \cdot P(A) + P(B | A^c) \cdot P(A^c) \end{equation} \begin{equation}P(B) = 1 \cdot \frac{1}{11} + \frac{1}{8} \cdot \frac{10}{11} \end{equation} \begin{equation}P(B) =  \frac{9}{44} \end{equation} Finally, apply Bayes' theorem: \begin{equation}P(A | B) =  \frac{P(B | A) P(A)}{P(B)} = \frac{1 \cdot \frac{1}{11}}{\frac{9}{44}} = \frac{4}{9}\end{equation} Simulation in Python
Sometimes it's good to simulate probability questions in Python for coding exercise purposes, especially if you're applying for a quant/ systematic role.
import random

def flip_coin(coin_type):
    """
    Simulates flipping a coin.
    :param coin_type: 'fair' for a fair coin, 'two_heads' for the two-heads coin
    :return: 'H' for heads, 'T' for tails
    """
    if coin_type == 'fair':
        return 'H' if random.random() < 0.5 else 'T' elif coin_type == 'two_heads': return 'H' 

def simulate_trial(): 
    """ Simulates picking a coin at random and flipping it three times. :return: (is_two_heads_coin, result_is_hhh) """ 
    # Pick a coin at random 
    coin = 'two_heads' if random.randint(1, 11) == 1 else 'fair' 
    # Flip the coin three times 
    result = ''.join(flip_coin(coin) for _ in range(3)) 
    # Check if the result is HHH 
    return (coin == 'two_heads', result == 'HHH') 

def estimate_probability(trials): 
    """ Estimates the probability of picking the two-heads coin given that HHH was observed. 
    :param trials: Number of trials to simulate 
    :return: Estimated probability 
    """ 
    hhh_count = 0 
    two_heads_and_hhh_count = 0 
    
    for _ in range(trials): 
        is_two_heads_coin, result_is_hhh = simulate_trial() 
        if result_is_hhh: 
            hhh_count += 1 
            if is_two_heads_coin: 
                two_heads_and_hhh_count += 1 
    return two_heads_and_hhh_count / hhh_count if hhh_count > 0 else 0

# Run the simulation with a large number of trials
trials = 1000000
estimated_probability = estimate_probability(trials)
print(f"Estimated probability: {estimated_probability}")

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