Solution
Third in Line at the Bank
Show the problem again
A bank has two tellers. When Carol walks in, Alice is being served by one teller and Bob by the other, and nobody else is waiting. Carol will be served by whichever teller becomes free first. All service times (Alice's remaining time, Bob's remaining time, and Carol's full service time) are independent exponential random variables with the same mean. What is the probability that Carol is the last of the three to leave the bank? Give a reduced fraction.
Worked solution
The answer is 1/2. Carol cannot leave before one of Alice and Bob has finished, since she has not yet started. At the moment the first of them leaves, Carol begins service and the other customer is still being served. By memorylessness, that customer's remaining service time is a fresh exponential with the same mean as Carol's, and the two are independent. Two identical, independent exponential clocks started together are equally likely to ring first, so Carol outlasts the remaining customer with probability exactly 1/2. The intuition "Carol started last, so she probably finishes last" is wrong precisely because exponential service has no memory of time already spent.
Source: Classic exponential-distribution exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.