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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?

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    $\begingroup$ Your example has nothing to do with the memoryless property. The property states that given an event the time to the next event still has the same exponential distribution. In your case with parameter $\lambda$ =0.5. $\endgroup$ – Michael R. Chernick Dec 30 '17 at 0:45
  • $\begingroup$ @MichaelChernick you're wrong, the memoryless property absolutely explains how to approach this problem. As long as the tellers waiting times are exponential and iid, the answer is 1/3 regardless of what the actual value of $\lambda$ may be, and the memoryless property explains this. See my answer below. $\endgroup$ – David Marx Dec 30 '17 at 1:18
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The memoryless property here means that if the expected waiting time for a new arrival is $\lambda$ and it has already been $t$ since the last event, the expected waiting time is still $\lambda$ and not $\lambda - t$.

In the case of your specific example, the "event" that you are waiting for is for a given teller to become available. To translate this into the context of the memoryless property, it means that regardless of how long each of those individual customers has already been talking to their teller prior to your arrival, your estimate is that it will take $\lambda$ time for a teller to become available. When that happens, we have a new observation for the process, you step up to be served and the clock starts all over again for all three tellers. It doesn't matter how long they've already been working with that particular client, all we know is that their window still isn't open and it takes on average an additional $\lambda$ amount time for any teller to become available regardless of how long they've already been at work on a particular customer. Therefore, our estimate of how long it will take for any of the three current customers to get served is the same, so each has an equal chance of being the next one served and, equivalently, the last one to leave.

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    $\begingroup$ What you say in the first two lines of your answer is true but is implied by what I stated. So why do you say that I am wrong. Also the OP was completely misunderstanding the concept. $\endgroup$ – Michael R. Chernick Dec 30 '17 at 1:35
  • $\begingroup$ You said "your example has nothing to do with the memoryless property", but this property is the key to the entire problem. $\endgroup$ – David Marx Dec 30 '17 at 2:49
  • $\begingroup$ I meant that the OP was computing probabilities that did not relate to the memoryless property. $\endgroup$ – Michael R. Chernick Dec 30 '17 at 3:00
  • $\begingroup$ Could you please clarify if the probability to leave the bank would change if someone would tell me the entire history of the process? In this case, if I knew how much time each customer has already been served by the time of my arrival, would my calculations in the question be correct? $\endgroup$ – Anna Dec 30 '17 at 5:58
  • $\begingroup$ No. Your calculations fundamentally misunderstand the memoryless property. If someone has been served for 3 minutes, the probability they will leave after less than a total of 4 minutes is really "the probability that they have less than 1 minute of additional service remaining," i.e. p(x<1), not p(x<4). Please review the first sentence of my post. This is exactly the same issue as assuming a roullete table spinning "black" many times means "red" is more likely: the geometric distribution is also memoryless. en.wikipedia.org/wiki/Memorylessness $\endgroup$ – David Marx Dec 30 '17 at 19:38

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