Answer
Yes — take the bet. The con man’s reasoning is wrong.
Work Step by Step
Assume he picks a coin uniformly at random and (independently) shows you a random side of that coin. The possible coins are HH, TT, HT (each picked with probability 1/3. Given that the face shown is heads, TT is impossible. Now count the equally likely coin–side outcomes that could produce a head:
- If he picked HH: there are 2 sides that would show heads.
- If he picked HT: there is 1 side that would show heads.
- If he picked TT: 0 sides.
So the number of head-showing outcomes =
2 (from HH)
+ 1 (from HT)
= 3..
Of those 3 outcomes, 2 correspond to the HH coin. Hence
$P(HH|\text{shown head})=\frac{2}{3}$
Expected value for an even-money bet on HH (win $1 if HH, lose $1 otherwise):
$EV=P(HH)\cdot 1+(1-P(HH))\cdot (-1)=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}>0$
So the probability it’s the two-headed coin is
2/3, not 1/2. An even-money bet on it being the two-headed coin has positive expected value for you — take it.
(If, however, the con man is not selecting a side at random but is actively choosing to show you a head whenever possible, the analysis changes — you must then model his strategy. The classic statement assumes a random side is shown.)
