Question
You reserve a shared taxi with a fixed $80 fare for the journey. You will definitely travel. Four other people have been invited; each independently accepts with probability 1/2, and nobody else can join. Just before departure everyone knows who accepted, and the $80 is split equally among the people who actually ride, including you.
How much should you expect to pay? Why is it incorrect to divide $80 by the expected number of passengers? Give an exact answer rather than a simulation estimate.
Get hint
Hint 1
The number of people in the taxi is one plus the number of accepted invitations. Which of those passengers is certain to be there?
Check your answer
Solution
Let \(N\) be the number of optional riders. The four responses are independent \(\operatorname{Bernoulli}(1/2)\), so \(N\sim\mathrm{Binomial}(4,1/2)\). Because you are guaranteed to ride, the actual occupancy is \(1+N\), never zero, and your fare is \(80/(1+N)\).
By the binomial distribution,
\[
E\!\left[\frac{80}{1+N}\right]
=\frac{80}{16}\left(\frac{\binom40}{1}+\frac{\binom41}{2}
+\frac{\binom42}{3}+\frac{\binom43}{4}+\frac{\binom44}{5}\right).
\]
The bracket is \(1+2+2+1+1/5=31/5\), and \(80/16=5\). Thus the expected payment is \(5\times31/5=\boxed{\$31}\).
For an independent analytic check, use \(1/(k+1)=\int_0^1x^k\,dx\):
\[
E\!\left[\frac{1}{1+N}\right]
=\int_0^1\left(\frac{1+x}{2}\right)^4\!dx
=\frac{1-(1/2)^5}{5(1/2)}=\frac{31}{80}.
\]
Multiplying by $80 again gives $31.
The average passenger count is \(E[1+N]=1+4(1/2)=3\). Dividing by that mean gives \(\$80/3\approx\$26.67\), not the average of the actual fare shares. In general, \(E[1/K]\ne1/E[K]\) when occupancy \(K\) varies. Low-occupancy rides produce disproportionately large shares, which raises your expected bill.
Assumptions
The focal traveller always rides, and four invitations are accepted independently with probability 1/2 each.
The total fare is fixed at $80 regardless of the number of passengers, and all actual riders—including the focal traveller—split it equally.
All participation decisions are observed before splitting the fare; none is conditioned on the fare outcome.
Final Answer
$31 expected payment. If \(N\sim\mathrm{Binomial}(4,1/2)\) counts the optional riders, the fare share is \(80/(1+N)\); \(E[80/(1+N)]=31\), not \(80/E[1+N]=80/3\).