Question
A raffle box contains ten physical tickets: four belong to Amina, two to Ben, one to Chen and three to Dia. One ticket is drawn uniformly at random.
After seeing its owner, the organiser may either award the prize to that person or award nobody. The decision uses a separate randomization, independent of the ticket draw, with an acceptance probability chosen for each owner. There is no redraw.
Choose the four acceptance probabilities so that all four people have the same unconditional chance of winning and a prize is awarded as often as possible. What is that maximum award probability, and what chance does each person have of being the winner conditional on a prize being awarded?
Get hint
Hint 1
If an owner has \(m\) of the ten tickets and is accepted with probability \(a\), their unconditional chance of winning is \(ma/10\).
Check your answer
Solution
Let \(a_A,a_B,a_C,a_D\) be the four acceptance probabilities, each between 0 and 1. Equal unconditional winning chances require a common value \(c\) with
\[
\frac4{10}a_A=\frac2{10}a_B=\frac1{10}a_C=\frac3{10}a_D=c.
\]
Only the owner of the one selected ticket can win, so the total probability of an award is \(4c\).
Chen's winning probability is \(a_C/10\le1/10\). Thus equality of the four winning chances forces \(c\le1/10\), giving the upper bound
\[
P(\text{award})=4c\le\frac4{10}=\frac25.
\]
The bound can be attained. Set \(c=1/10\) and solve the four equations:
\[
(a_A,a_B,a_C,a_D)=\left(\frac14,\frac12,1,\frac13\right).
\]
Every value is a valid acceptance probability, and each owner's resulting winning probability is exactly \(1/10\). A prize is therefore awarded with probability \(2/5\) and nobody wins with probability \(3/5\). No other allowed rule can award a prize more often while preserving equal unconditional chances, because it would require Chen to win with probability greater than the chance of drawing Chen's ticket.
Finally,
\[
P(\text{owner }i\text{ wins}\mid\text{award})
=\frac{1/10}{2/5}=\frac14
\]
for each owner. Rejecting every draw would also give equal unconditional chances, but would not maximize the award probability. The single-draw rule is essential: rejection ends the raffle rather than starting another draw.
Assumptions
All ten physical tickets are equally likely to be drawn, with ownership counts \((4,2,1,3)\) for Amina, Ben, Chen and Dia.
There is exactly one draw. Rejection results in no winner and no further draw.
Acceptance depends only on the observed owner, using that owner's chosen probability and a randomization independent of the ticket draw.
The fairness condition is equality of the four unconditional winning probabilities; the objective is the largest overall probability of awarding a prize.
Final Answer
Accept Amina, Ben, Chen and Dia with probabilities \(1/4,1/2,1,1/3\), respectively. Each then wins with unconditional probability \(1/10\); the maximum probability of awarding a prize is \(2/5\). Conditional on an award, each person has probability \(1/4\) of being the winner.