2 Boxes
You encounter two boxes: Box A contains a $1 bill and a $100 bill; Box B contains two $100 bills. You randomly draw a $100 bill from a randomly chosen box. Should you switch boxes before drawing again?
Answer
Bayesian Calculation: After drawing $100, the probability your initial box is B is
\[
P(B\mid100)
=\frac{P(100\mid B)\,P(B)}{P(100\mid B)\,P(B)+P(100\mid A)\,P(A)}
=\frac{1\cdot\tfrac12}{1\cdot\tfrac12 + \tfrac12\cdot\tfrac12}
=\tfrac23,
\]
so \(P(A\mid100)=\tfrac13\).
Hence, we do not switch.
Logical Explanation: After drawing $100, three equally likely scenarios arise:
\[
\begin{aligned}
&(1)\ \text{You picked the first \$100 from Box B; the other in B is \$100 (no switch needed).}\\
&(2)\ \text{You picked the second \$100 from Box B; the other in B is \$100 (no switch needed).}\\
&(3)\ \text{You picked the only \$100 from Box A; the other in A is \$1 (you must switch).}
\end{aligned}
\]
In 2 of the 3 scenarios you would win by staying, and in 1 of the 3 you must switch to win, giving a no switching‐win probability of
\[
\frac{2}{3}.
\]