# Find the unfair coin

I've had some fun thinking about these (and related) problems, but things get very complicated very quickly, and I constantly doubt my own work! I would like to see what others do with them. Thanks!

You have two identical-looking coins, but one is fair, and the other is unfair, coming up Heads 2/3 of the time. You flipped coin A once, and it came up Heads. You flipped coin B three times, and got two Heads and a Tail.

a) What's the probability that coin A is the unfair one?

b) Suppose you were allowed to perform one more flip, then declare which coin you believed to be unfair. Which coin should you flip? After performing this flip, which coin will you say is the unfair one (based on the result of that flip), and how likely are you to be correct?

• I don't think question a) is well-posed. I mean, either coin A is unfair or coin B is unfair. In both cases we can compute the likelihood of the flipping outcomes you describe and they may be different. – Rasmus Aug 20 '12 at 17:07

There seems to be an implicit assumption in the question that a priori either coin is equally likely to be the fair coin; without any assumptions about the a priori distribution the desired probabilities are not well-defined.

The probability for the result described if A is the fair coin is $\frac12\cdot3\cdot\frac23\cdot\frac23\cdot\frac13=\frac29$, whereas if B is the fair coin it's $\frac23\cdot3\cdot\frac12\cdot\frac12\cdot\frac12=\frac14$. Thus if both are a priori equally likely to be the fair coin, the probability for B to be fair is $\frac14/(\frac14+\frac29)=1/(1+\frac89)=\frac9{17}$.

If you flip A again, the result will be heads with probability $\frac8{17}\cdot\frac12+\frac9{17}\frac23=\frac{10}{17}$ and thus tails with probability $\frac7{17}$. If it's heads, the probabilities for the result obtained will be $\frac12\cdot3\cdot\frac23\cdot\frac23\cdot\frac13\cdot\frac12=\frac19$ if A is the fair coin and $\frac23\cdot3\cdot\frac12\cdot\frac12\cdot\frac12\cdot\frac23=\frac16$ if B is the fair coin, so you would guess that B is the fair coin and would be right with probability $\frac16/(\frac16+\frac19)=\frac35$. If it's tails, the probabilities for the result obtained will be $\frac12\cdot3\cdot\frac23\cdot\frac23\cdot\frac13\cdot\frac12=\frac19$ if A is the fair coin and $\frac23\cdot3\cdot\frac12\cdot\frac12\cdot\frac12\cdot\frac13=\frac1{12}$ if B is the fair coin, so you would guess that A is the fair coin and would be right with probability $\frac19/(\frac19+\frac1{12})=\frac47$. Your total probability of guessing right would thus be $\frac{10}{17}\cdot\frac35+\frac7{17}\cdot\frac47=\frac{10}{17}$.

If you flip B again, the result will be heads with probability $\frac9{17}\cdot\frac12+\frac8{17}\frac23=\frac{59}{102}$ and thus tails with probability $\frac{43}{102}$. If it's heads, the probabilities for the result will be $\frac12\cdot3\cdot\frac23\cdot\frac23\cdot\frac13\cdot\frac23=\frac4{27}$ if A is the fair coin and $\frac23\cdot3\cdot\frac12\cdot\frac12\cdot\frac12\cdot\frac12=\frac18$ if B is the fair coin, so you would guess that A is the fair coin and would be right with probability $\frac4{27}/(\frac4{27}+\frac18)=\frac{32}{59}$. If it's tails, the probabilities for the result obtained will be $\frac12\cdot3\cdot\frac23\cdot\frac23\cdot\frac13\cdot\frac13=\frac2{27}$ if A is the fair coin and $\frac23\cdot3\cdot\frac12\cdot\frac12\cdot\frac12\cdot\frac12=\frac18$ if B is the fair coin, so you would guess that B is the fair coin and would be right with probability $\frac18/(\frac18+\frac2{27})=\frac{27}{43}$. Your total probability of guessing right would thus be $\frac{59}{102}\cdot\frac{32}{59}+\frac{43}{102}\cdot\frac{27}{43}=\frac{59}{102}\lt\frac{10}{17}$.

So you should flip coin A again, and you would guess correctly with probability $\frac{10}{17}$.

A quicker way to do this would be to note that the current a posteriori probabilities for A or B to be fair are very close to $\frac12$, so it's plausible that your guess will be determined entirely by the last flip, and you'll guess that the last coin flipped is unfair if the result is heads and fair if it's tails. Your chance of winning by this strategy is $\frac12$ if you pick the fair coin and $\frac23$ if you pick the unfair coin, so you should pick the coin that has the slightly higher probability of being the unfair coin, A, and your winning probability is $\frac8{17}\cdot\frac12+\frac9{17}\cdot\frac23=\frac{10}{17}$, in agreement with the more systematic solution above.

Let's denote following events:

• $A: A\text{ is fair}$
• $B: B\text{ is fair}$
• $E_1: \text{ flip coin A once, and it comes up head.}$
• $E_2: \text{ flip coin B three times, and get two heads and a tail}$

It is obvious that

• $A=\overline{B}$,
• $P(A)=P(B)=\dfrac{1}{2}$,
• $P(E_1|A)=\dfrac{1}{2}$,
• $P(E_1|B)=\dfrac{2}{3}$,
• $P(E_2|A)=C_3^2\cdot(\dfrac{2}{3})^2\cdot\dfrac{1}{3}=\dfrac{4}{9}$,
• $P(E_2|B)=C_3^2\cdot(\dfrac{1}{2})^2\cdot\dfrac{1}{2}=\dfrac{3}{8}$

The first question is to compute $P(A|(E_1+E_2))$. Using Bayes' theorem, we have the following equation:

$$P(A|(E_1+E_2))=\dfrac{P((E_1+E_2)|A)P(A)}{P(E_1+E_2)}$$

Compute the each part on the right,

• $P((E_1+E_2)|A)=P(E_1|A)P(E_2|A)=\dfrac{2}{9}$

• $P(E_1+E_2)=P((E_1+E_2)|A)P(A)+P((E_1+E_2)|B)P(B)=\dfrac{2}{9}\cdot\dfrac{1}{2}+\dfrac{1}{4}\cdot\dfrac{1}{2}=\dfrac{17}{72}$

Thus we get the probability that coin A is unfair:

$$P(A|(E_1+E_2))=\dfrac{8}{17}$$

(to be continued)