If I exchange infinitely many digits of $\pi$ and $e$, are the two resulting numbers transcendental?

If, as is commonly believed, $\pi$ and $e$ are normal numbers, then one can use a counting argument (or entropy argument) to show that no possible transposition of $\pi$ and $e$ can produce a rational number. Indeed, if there was a rational number that could be made this way, then its digit expansion would eventually be periodic with some period $q$; by repeating this period, one can make $q$ large and even. If one looks at a given $q$-digit block of this periodic expansion, then either $\pi$ would have to share at least $q/2$ of its digits with this fixed block, or $e$ would. But if $\pi$ is normal, the former happens with density at most $\binom{q}{q/2} 10^{-q/2}$ among all the $q$-blocks, and if $e$ is normal, the latter happens with density at most $\binom{q}{q/2} 10^{-q/2}$. For $q$ large enough, the two densities sum to less than 1 (here we use the fact that the base is at least $4$ - not sure which way things will go in base $2$ or base $3$), and so one cannot actually match the given rational number.

[There ought to be some slick information theoretic way to reformulate the above argument, perhaps using the Shannon entropy inequalities, but I was not able to locate one.]

Settling the problem unconditionally looks to be at least as hard as making some major breakthrough on the normality of $\pi$ and $e$. Even ruling out a terminating decimal (i.e. that for all sufficiently large $k$, either the $k^{th}$ digit of $\pi$ or the $k^{th}$ digit of $e$ vanishes) is probably out of reach of current technology.

Nice question, Erin. Here is one quick easy thing to say.

If $\pi$ and $e$ disagree in infinitely many digits, then there are continuum many choices of the particular subset of those digits to swap, and so we get continuum many different numbers this way. Since there are only countably many algebraic numbers, it would follow that most of the time, yes, you do get transcendental numbers by doing this.

I'm unsure, however, whether one can say that all the resulting reals are transcendental. Perhaps we'll have to wait for some number theory experts to answer.

Lastly, if it happens (as seems unlikely) that all but finitely many digits of $\pi$ and $e$ are the same, then $\pi-e$ would be rational, and furthermore swapping the digits doesn't actually do anything except on those finitely many digits of difference, and so this won't affect transcendentality. In this case, there are only finitely many possible reals resulting, but they are all differing from the original reals by only finitely many digits, and so yes, they are all transcendental.

A variant of my previous answer. It is commonly believed that all irrational algebraic numbers are normal. If this is the case, then there can be at most two algebraic numbers (up to shifts by rationals) that can be obtained by transposing digits of $e$ and $\pi$.

To see this, suppose for sake of contradiction that there are three algebraic numbers $\alpha,\beta,\gamma$, no two of which differ by a rational, that can all be attained by transposing digits of $e$ and $\pi$. By the pigeonhole principle, we see that for each natural number $k$, at least one of the pairs $(\alpha,\beta)$, $(\alpha,\gamma)$, $(\beta,\gamma)$ agree at the $k^{th}$ digit. By the pigeonhole principle again, this means that one of these pairs agrees on a set of digits of (upper) density at least $1/3$. Without loss of generality we can assume that the pair $(\alpha,\beta)$ has this property, and that $\beta > \alpha$. But then, by long subtraction, the difference $\beta - \alpha$ will have digits $0$ or $9$ on a set of digits of upper density at least $1/3$, which contradicts the normality of $\beta-\alpha$. (Now I need the base to be at least seven!)

It might be possible to upgrade "up to shifts by rationals" in the above claim by "up to shifts by terminating decimals", but I have not strenuously attempted to do this. It's also worth noting that this is an example of an ineffective argument, in that no bound whatsoever is provided on the height of the algebraic numbers that might still be obtainable by transposing digits of $e$ and $\pi$, even if one had some quantitative normality bound on algebraic numbers depending on the height.

p.s. We can combine the two answers: if we assume that the sum of $\pi$ and an algebraic number, or the sum of $e$ and an algebraic number, is always normal, then the answer to the original question is positive: every transposition of $\pi$ and $e$ is transcendental. For if there was an algebraic number $\alpha$ that was achievable as a transposition, then it would have to share at least half its digits with $\pi$ or $e$, and hence one of $|\pi-\alpha|$ or $|e-\alpha|$ would have digits $0$ or $9$ on a set of upper density at least $1/2$, contradicting the normality of these numbers. (Now I need base at least five.)