Two reds in a row, fair spinner or rigged spinner?

Question

You have two spinners: a fair one that lands on red half the time, and a rigged one that lands on red $90\%$ of the time. You pick one at random and spin it twice. Both spins land on red.

What is the probability you used the rigged spinner?

Accepted Answer

Each spinner starts with prior 1/2. Compare the chance of two reds under each. - Fair spinner: (1/2)^2 = 1/4 = 0.25. - Rigged spinner: (0.9)^2 = 0.81. Weight by prior and likelihood: P(	ext{rigged} \mid RR) = \frac{0.5 \cdot 0.81}{0.5 \cdot 0.81 + 0.5 \cdot 0.25} = \frac{0.81}{1.06} \approx 0.764 = 76\%. Two reds is more than three times as likely from the rigged spinner, so belief moves from 50\% to about 76\%. Takeaway. Only two spins already gives clear evidence; more red spins would keep pushing the posterior toward the rigged spinner, since a fair one rarely produces long red streaks.

Two reds in a row, fair spinner or rigged spinner?

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