Question

(a) A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random, and when he flips it, it shows heads. What is the probability that it is a fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. Now, what is the probability that it is a fair coin? (c) Suppose that he flips the same coin a third time and it shows tails. Now, what is the probability that it is a fair coin?

Short answer

(a) After one flip showing heads, the probability that the gambler chose the fair coin is 0.5. (b) After two flips both showing heads, the probability of choosing the fair coin goes down to \(\frac{1}{3}\). (c) Once a tails shows up, we're completely certain the fair coin was chosen; the probability that the gambler picked the fair coin is 1.

Step-by-step solution

Understand the Problem

The problem shares details of two events: (A1) choosing the fair coin and (A2) choosing the two-headed coin. There's an equal chance of choosing either coin i.e., P(A1) = P(A2) = 0.5. The result of the coin flip is the second event B. Because we don't know which coin was selected, the conditions differ for each coin. For the fair coin, getting heads P(B/A1) = 0.5 and for the two-headed coin, getting heads P(B/A2) = 1. We need to find the conditional probability P(A1/B), the likelihood that a fair coin was chosen given that heads appeared.

Use Bayes' Theorem

Bayes' theorem allows us to update our initial beliefs, based on evidence. The theorem in this scenario is: \[ P(A1/B) = \frac{P(B/A1) * P(A1)}{P(B/A1) * P(A1) + P(B/A2)* P(A2)} \] Let’s plug the known probabilities into this formula and find the solution for part (a).

Calculate for First Flip (a)

Substitute the known values into the Bayes’ formula to get: \[ P(A1/B) = \frac{0.5 * 0.5}{0.5 * 0.5 + 1 * 0.5} = \frac{1}{2} \] So, after one flip showing heads, the probability that the gambler chose the fair coin is 0.5.

Calculate for Second Flip (b)

For part (b), the coin is flipped again and it again shows heads. We then update our initial events and their probabilities. This time, the event is seeing heads twice, so the evidence related to A1 is \(P(B/A1) = 0.5^2 = 0.25\) and the evidence related to A2 is \(P(B/A2) = 1\), because the two-headed coin will always show heads. Use Bayes’ theorem again: \[ P(A1/B) = \frac{0.25 * 0.5}{0.25 * 0.5 + 1 * 0.5} = \frac{1}{3} \] So, after two flips both showing heads, the probability of choosing the fair coin goes down to 1/3.

Calculate for Third Flip (c)

For part (c), the coin is flipped again and this time it shows tails. This flip changes the probabilities because the two-headed coin cannot show tails. Therefore, \(P(B/A1) = 0.5^3 = 0.125\) (since we have 2 heads and 1 tail) and \(P(B/A2) = 0\) (since tails is impossible for the two-headed coin). Apply Bayes’ theorem: \[ P(A1/B) = \frac{0.125 * 0.5}{0.125 * 0.5 + 0 * 0.5} = 1 \] So the moment a tails shows up, we're completely certain the fair coin was chosen; the probability that the gambler picked the fair coin is 1.