Poisson variable situation; same probability for two different outcomes

Suppose a customer service rep gets on average exactly 2 calls per minute. What are the odds of getting 1 call in the next minute?

This problem calls (no pun intended) for the Possion distribution : $$P\{X = k\} = e^{-2} \frac{2^k}{k!}$$

Here's where I'm stuck wondering. The probability of there being one call $$P\{X = 1\} = e^{-2} \frac{2^1}{1!} \approx .2707$$ is the same as the probability of getting two calls: $$P\{X = 2\} = e^{-2} \frac{2^2}{2!} \approx .2707$$

I understand that the calculations show they are the same (obviously), but conceptually, I don't see how they can be the same. And also, why should one have the same probability as the average, but not three (which has a probability of $.1804$). Unless, am I calculating it incorrectly?

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The distribution can't be symmetric, since the probability for arbitrarily high numbers of calls is non-zero, whereas the probability for negative numbers of calls is zero; thus the probabilities for $3$ and $4$ must be less than the probabilities for $1$ and $0$.
There's no contradiction in the fact that $1$ has the same probability as $2$, and in fact it's true for any integer average number of calls $n$ that $n$ calls and $n-1$ calls have the same probability, since $\mathrm e^{-n}n^n/n!=\mathrm e^{-n}n^{n-1}/(n-1)!$. In general the average and maximum of an asymmetric distribution don't coincide, so there's no reason to expect the maximum to be exactly at $n$.