Posterior Probability of a Double-Headed Coin
A jar contains 12 coins: 10 are fair (equal chance of heads or tails), and 2 are double-headed (always land heads). You pick a coin uniformly at random from the jar and flip it 5 times. All 5 flips come up heads.
What is the posterior probability that the coin you picked is double-headed?
Hints
- You have two hypotheses (fair vs. double-headed) and an observation (5 heads). What theorem lets you update your belief about which coin you picked?
- Compute the likelihood of 5 heads under each hypothesis. One gives probability 1, the other gives $(1/2)^5 = 1/32$. The likelihood ratio is 32 to 1.
- Apply Bayes' theorem: the posterior is $P(D \mid H_5) = \frac{P(H_5 \mid D) \cdot P(D)}{P(H_5 \mid D) \cdot P(D) + P(H_5 \mid D^c) \cdot P(D^c)}$. Plug in the priors $2/12$ and $10/12$ and the likelihoods.
Worked Solution
How to Think About It: This is a textbook Bayes' theorem problem. You observed something (5 heads in a row) that is much more likely under one hypothesis (double-headed coin) than the other (fair coin). Before you compute anything, build intuition: a fair coin produces 5 heads with probability $1/32 \approx 3\%$, while a double-headed coin does it with certainty. So 5 heads is strong evidence for the double-headed hypothesis, but the prior odds are 10-to-2 against it. The posterior should swing heavily toward double-headed, but not all the way.
Quick Estimate: Prior odds of double-headed vs. fair: $2:10 = 1:5$. Likelihood ratio: $1 / (1/32) = 32$. Posterior odds: $32 \times 1/5 = 32/5 \approx 6.4$ to 1 in favor of double-headed. So the posterior probability is roughly $6.4 / 7.4 \approx 0.865$, or about 86-87%. That is a quick number you can give in an interview before writing anything down.
Approach: Apply Bayes' theorem directly. Define $D$ as the event that the chosen coin is double-headed, and $H_5$ as the event of observing 5 heads.
Formal Solution:
The prior probabilities are:
$$P(D) = \frac{2}{12} = \frac{1}{6}, \quad P(D^c) = \frac{10}{12} = \frac{5}{6}$$
The likelihoods are:
$$P(H_5 \mid D) = 1, \quad P(H_5 \mid D^c) = \left(\frac{1}{2}\right)^5 = \frac{1}{32}$$
The total probability of observing 5 heads is:
$$P(H_5) = P(H_5 \mid D) \cdot P(D) + P(H_5 \mid D^c) \cdot P(D^c) = 1 \cdot \frac{1}{6} + \frac{1}{32} \cdot \frac{5}{6} = \frac{1}{6} + \frac{5}{192}$$
Finding a common denominator:
$$P(H_5) = \frac{32}{192} + \frac{5}{192} = \frac{37}{192}$$
Applying Bayes' theorem:
$$P(D \mid H_5) = \frac{P(H_5 \mid D) \cdot P(D)}{P(H_5)} = \frac{1 \cdot \frac{1}{6}}{\frac{37}{192}} = \frac{\frac{1}{6}}{\frac{37}{192}} = \frac{192}{6 \times 37} = \frac{32}{37}$$
Answer: The posterior probability that the chosen coin is double-headed is $\frac{32}{37} \approx 0.8649$, or about 86.5%.
Intuition
This problem illustrates how Bayesian updating works as a tug-of-war between prior beliefs and evidence. Your prior says the coin is probably fair (5-to-1 odds), but the data -- 5 heads in a row -- is 32 times more likely under the double-headed hypothesis. The likelihood ratio overwhelms the prior, swinging the posterior from 1/6 to 32/37. This is the essence of Bayesian reasoning: extraordinary evidence (a 32x likelihood ratio) can overcome an unfavorable prior.
In practice, this kind of reasoning shows up constantly in market making and trading. Suppose you are quoting a contract and your prior says the event is unlikely, but you observe data that is far more consistent with the event occurring. The speed at which you update -- and your ability to do it in your head via odds ratios -- is what separates a good quant trader from someone who needs to pull out a spreadsheet. The odds-form shortcut (prior odds times likelihood ratio equals posterior odds) is worth memorizing: here it gives $1/5 \times 32 = 32/5$, which converts to $32/37$ immediately.