Two Coins: One Fair, one Biased


Two Coins: One Fair, one Biased, problem


Assuming the fair coin is the one that was tossed, then the probability of it showing heads and of the other coin showing two heads is given by

$\displaystyle \frac{1}{2}\cdot{3\choose 2}\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^{1}=\frac{2}{9}.$

On the other hand, if the biased coin was tossed once then the probability of the observed result is

$\displaystyle \frac{2}{3}\cdot{3\choose 2}\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{1}=\frac{2}{8},$

telling us that more likely it was the biased coin that was tossed once.

Solution 2

Without any a-priori knowledge, let us assume that the first coin is equally likely to be either of the two coins. Let $D$ denote the observation and $B$ the event that the first coin is biased. Thus, $P(B)=P(\overline{B})=1/2$,

$\displaystyle\begin{align} P(D|B)&=\left[\frac{2}{3}\right]\cdot \left[C(3,2)\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)\right]=\frac{1}{4}, \\ P(D|\overline{B})&=\left[\frac{1}{2}\right]\cdot \left[C(3,2)\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)\right]=\frac{2}{9}. \end{align}$

Using Bayes theorem,

$\displaystyle\begin{align} P(B|D)&=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|\overline{B})P(\overline{B})} \\ &=\frac{1/4}{1/4+2/9}=\frac{9}{17}>\frac{1}{2}. \end{align}$

Thus, the observation makes the first coin more likely to be biased.

Solution 3

Prior (relative) odds are even, while likelihood ratio is

$\displaystyle\begin{align} LR&=\frac{Pr(E|B,F)}{Pr(E| F,B)}= \frac{\displaystyle\frac{2}{3}\cdot\left(\frac{1}{2}\right)^3}{\displaystyle\frac{1}{2}\cdot\left(\frac{2}{3}\right)^2\cdot\frac{1}{3}}= \frac{\displaystyle\frac{2}{3}\cdot\frac{1}{8}}{\displaystyle\frac{1}{2}\cdot\frac{4}{27}}\\ &= \frac{9}{8}. \end{align}$

Odds in favor of the first coin being biased and the second coin being fair are $9:8,$ i.e. $\displaystyle prob=\frac{9}{17}.$


This is problem 1989-4 from the A Friendly Mathematics Competition by Rick Gillman (ed.)

Solution 2 is by Amit Itagi; Solution 3 is by Joshua B. Miller.


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