Question
A tiny tasting event has just two seats, but five customers are booked. One customer, A, is certain to attend. B, C, D and E attend independently with probabilities 3/4, 1/2, 1/2 and 1/4 respectively. These are the only customers; non-attendees owe nothing and receive no voucher.
If more than two attend, the organiser admits customers in booking order A, B, C, D, E and denies the later bookings until exactly two remain. A denied customer receives a cash voucher of £200 for B, £120 for C, £80 for D or £40 for E. The organiser wants to budget for vouchers before anyone arrives.
Find the expected total voucher payment. Why is multiplying a single average voucher by the expected number of excess attendees generally unsafe here?
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Hint 1
For each person, find the event that this particular person attends but is denied entry; use indicator variables for individual vouchers.
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Solution
A always takes one seat. The second seat is awarded to the earliest-attending name among B,C,D,E. Thus B, when present, is always admitted and never generates a voucher.
C is turned away if both B and C attend, an event of probability \((3/4)(1/2)=\frac38\). D is turned away if D attends and at least one of B or C attends. By independence, that probability is \((1/2)[1-(1-3/4)(1-1/2)]=(1/2)(1-1/8)=\frac7{16}\). E is turned away if E attends and at least one of B, C or D attends: \((1/4)[1-(1/4)(1/2)(1/2)]=(1/4)(15/16)=\frac{15}{64}\).
Let \(I_C,I_D,I_E\) indicate these denied-boarder events. The voucher total is \(V=120I_C+80I_D+40I_E\) in pounds; simultaneous denials are allowed, so independence among these indicators is neither needed nor asserted. By linearity,
\[\mathbb E[V]=120(3/8)+80(7/16)+40(15/64)=45+35+9.375=\text{£}89.375=\text{£}\frac{715}{8}\].
This agrees with a direct enumeration of the 16 possible attendance subsets for B,C,D,E, weighting each by its independent attendance probability, admitting the earliest two and summing their denied customers' stated vouchers.
Different denied customers receive different amounts, and the priority order makes their denial probabilities unequal. The expected number denied is \(\frac38+\frac7{16}+\frac{15}{64}=\frac{67}{64}\), but multiplying that by an arbitrary unweighted 'average voucher' would not recover the expected payment. Person-specific indicators are the correct accounting units.
Assumptions
A attends with probability one; B,C,D,E are independently present or absent.
Admission is first-come by the fixed booking-priority order, with exactly two seats.
Vouchers are paid only to attendees denied a seat, with values fixed by identity; multiple payments may occur.
Final Answer
Expected compensation liability = \(\text{£}\frac{715}{8}=\text{£}89.375\).