Probability & Statistics

Unfair Coin

One coin in 1000 is double-headed and the rest are fair. Ten straight heads multiplies the odds by 2^10 = 1024, almost exactly cancelling the 999-to-1 prior, so the posterior probability of holding the fake is barely better than an even coin toss.

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Among 1000 coins, one has heads on both sides and the other 999 are fair. You pick a coin at random and toss it ten times. It comes up heads every time. What is the probability you picked the two-headed coin?

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#Bayes

Let UU be the event that I hold the two-headed coin, with prior P(U)=11000\PP(U) = \tfrac{1}{1000}. Ten heads is certain for that coin and has chance (12)10=11024\left(\tfrac{1}{2}\right)^{10} = \tfrac{1}{1024} for a fair one. Bayes gives

P(U∣10H)=P(10H∣U) P(U)P(10H∣U) P(U)+P(10H∣fair) P(fair)=1⋅110001⋅11000+11024⋅9991000.(1)\PP(U \mid 10H) = \frac{\PP(10H \mid U)\,\PP(U)} {\PP(10H \mid U)\,\PP(U) + \PP(10H \mid \text{fair})\,\PP(\text{fair})} = \frac{1 \cdot \tfrac{1}{1000}} {1 \cdot \tfrac{1}{1000} + \tfrac{1}{1024}\cdot\tfrac{999}{1000}}. \tag{1}

Clearing the common factor 11000\tfrac{1}{1000},

P(U∣10H)=10241024+999=10242023≈0.506.(2)\PP(U \mid 10H) = \frac{1024}{1024 + 999} = \frac{1024}{2023} \approx 0.506. \tag{2}
After 10 heads in a row the chance the coin is the two-headed one is 50.6 percent. The curve crosses one half near ten heads, because ten heads is about 1024 times likelier for the two-headed coin, just edging past the 999 fair ones.

#Reading the number

Ten heads is about 10241024 times more likely from the two-headed coin than from a fair one, and that almost exactly cancels the 999999-to-11 prior against it. The evidence drags a one-in-a-thousand suspicion up to a coin toss. One more head would push it past two to one, and each further head nearly doubles the odds toward certainty.