Bayesian Coin Selection and Sequential Betting

Worked Solution

How to Think About It: You observed 3 heads in a row from a coin that is either $2/3$-biased or $1/3$-biased. Three heads is $8\times$ more likely under Coin 1 than Coin 2, so the posterior heavily favors Coin 1. Once you have the posterior, every bet question is just an expected value calculation. The subtle point in part (d) is whether the negative-EV bet B can be worth taking as a hedge or information play -- but since bet B's outcome does not change the coin or affect future bet A payoffs, it cannot.

Quick Estimate: Posterior odds for Coin 1: $(2/3)^3 / (1/3)^3 = 8$. So $P(\text{Coin 1} \mid HHH) = 8/9 \approx 0.889$. Next-flip heads probability: roughly $0.889 \times 0.667 + 0.111 \times 0.333 \approx 0.593 + 0.037 = 0.630$. This is above $0.5$, so a symmetric bet on heads has positive EV.

Approach: Bayes' theorem for the posterior, then law of total probability for part (a), and direct EV calculations for the bets.

Formal Solution:

(a) Posterior and next-flip probability:

Prior: $P(C_1) = P(C_2) = 1/2$.

$$P(C_1 \mid HHH) = \frac{(2/3)^3 \cdot (1/2)}{(2/3)^3 \cdot (1/2) + (1/3)^3 \cdot (1/2)} = \frac{8/27}{8/27 + 1/27} = \frac{8}{9}$$

$$P(\text{next } H) = \frac{8}{9} \cdot \frac{2}{3} + \frac{1}{9} \cdot \frac{1}{3} = \frac{16}{27} + \frac{1}{27} = \frac{17}{27} \approx 0.630$$

(b) Bet A: win $\$2$ if heads, lose $\$2$ if tails.

$$EV_A = \frac{17}{27} \cdot 2 + \frac{10}{27} \cdot (-2) = \frac{34 - 20}{27} = \frac{14}{27} \approx \$0.519$$

Positive EV -- take the bet.

If offered repeatedly with the same coin, the edge persists. In fact, it gets stronger: each additional flip updates the posterior further. If you keep seeing heads, you become even more confident it is Coin 1, and $P(H)$ increases. If you see a tail, the posterior shifts toward Coin 2, but $P(H)$ stays above $1/2$ unless you see many tails (because the posterior already strongly favors Coin 1). The expected value of each subsequent bet A remains positive, so take it every time.

(c) Bet B: lose $\$2$ if heads, win $\$3$ if tails.

$$EV_B = \frac{17}{27} \cdot (-2) + \frac{10}{27} \cdot 3 = \frac{-34 + 30}{27} = \frac{-4}{27} \approx -\$0.148$$

Negative EV -- do not take bet B.

(d) Optimal strategy with both bets available (B at most once, A up to 100 times):

Bet A has positive EV ($\$0.519$ per play). Bet B has negative EV ($-\$0.148$). Since the bets are on separate flips (or at least independent conditional on the coin), taking bet B does not provide any information or hedging benefit that affects bet A's value. They are additive.

Optimal strategy: take bet A all 100 times, skip bet B.

Expected profit: $100 \times \frac{14}{27} \approx \$51.85$.

Answer: (a) $P(\text{next } H) = 17/27 \approx 0.630$. (b) Take bet A -- EV is $+14/27 \approx \$0.52$ per play; take it repeatedly. (c) Do not take bet B -- EV is $-4/27 \approx -\$0.15$. (d) Take all 100 bet A's, skip bet B. Expected profit $\approx \$51.85$.