Probability question in an M/M/2/4 queue

Consider the following scenario:

A queue with two servers in which service times are exponentially
distributed with mean 1/m. They're both currently servicing packets,
and in addition there are two packets waiting in the queue (P_1 will
be the first to be served, P_2 second.)

What's the probability that P_2 will finish its service before P_1? I
have an answer, but is my reasoning okay?

There are two ways this can happen: (1) Server 1 can finish its job
and get P_2 followed by Server 2 finishing its job and receiving P_2,
and then Server 2 can finish servicing P_2 before Server 1 finishes
with P_1. Or, (2) the same deal with Server 1 and 2 replaced.

Since both servers have exponentially distributed service times with
the same mean, it's easy enough to show that there's a 50/50 chance
that Server 1 will finish its original job before Server 2 does, and
by the memoryless property of the exponential distribution it seems to
me that the final answer should be (1/2)^3 + (1/2)^3 = 1/4, but this
seems too easy.

Thanks,
Mark
.

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