Could someone please clarify the following example (it is taken from Introduction to Probability by Bertsekas):
When you enter the bank, you find that all three tellers are busy serving other customers, and there are no other customers in queue. Assume that the service times for you and for each of the customers being served are iid exponential random variables. What is the probability that you will be the last to leave?
The answer is 1 /3. Focus at the moment when you start service with one of the tellers. Then, the remaining time of each of the other two customers being served, as well as your own remaining time, have the same PDF. Therefore, you and the other two customers have equal probability 1/3 of being the last to leave.
When I think about it I get the following. Assume that the service time has PDF of Exponential(0.5). One customer has been served for 3 minutes. Then, according to Exponential(0.5), the probability of success (the probability that he leaves the bank) during the 4th minute is P(X < 4) = 0.86.
Next, my service time begins at that moment. So, I use Exponential(0.5) to calculate the probability of my success during my 1st minute, which is P(X < 1) = 0.39.
0.39 does not equal to 0.86. So, how is it possible that the other customer and I have the same probability of being the last to leave the bank?