Here's a solution to a more general version of the problem:

Let's say that the red envelope contains N times as much money as the blue envelope with probability p, and the blue envelope contains N times as much money as the red envelope with probability (1 - p).

Without loss of generality, N is at least 1.

If N = 1, both envelopes contain the same amount, and there is no point in switching.

If N > 1, let the variable s represent the smaller amount of money between the amounts of the two envelopes. So one envelope contains s, and the other envelope contains Ns.

Scenario 1: The blue envelope contains s, and the red envelope contains Ns. This scenario occurs with probability p.

Scenario 2: The red envelope contains s, and the blue envelope contains Ns. This scenario occurs with probability (1 - p).

Assume s is not less than 0.

The expected amount of money in the blue envelope, E(B) = sp + Ns(1 - p) = s(p + N - Np).

The expected amount of money in the red envelope, E(R) = s(1 - p) + Nsp = s(1 - p + Np).

If s = 0, both envelopes contain the same amount, and there is no point in switching.

If s > 0, compare the expectations of the amounts of money in the envelopes in terms of s:

E(B) > E(R) when

s(p + N - Np) > s(1 - p + Np)

p + N - Np > 1 - p + Np

2p - 2Np > 1 - N

2p(1 - N) > 1 - N

2p < 1, because (1 - N) < 0

p < 1/2

Similarly, E(B) < E(R) when p > 1/2, and E(B) = E(R) when p = 1/2.

The ratios of the expected amounts of money in each envelope depend only on p when s > 0 and N > 1.

Both envelopes contain the same amount of money when s = 0, N = 1, or both.

So if your degree of belief that the red envelope contains more money than the blue envelope is the same as your degree of belief that the blue envelope contains more money than the red envelope, don't bother switching, unless you need to kill time (which you already knew intuitively, but Q.E.D.)

I got the idea of using the variable s to represent the smaller of the amounts in the two envelopes from R Falk 2008: "The Unrelenting Exchange Paradox".