Labyrinth, 31 bytes

Not as short and neat as Sp3000's solution, but I thought I'd post this anyway as a different approach:

"" @
,, :{"
)  { $;
*"})";.
 ""

Explanation

The first loop is implysimply

""
,,

which reads in two characters at a time (the " are no-ops). After EOF, , will return -1, but only check for EOF at every second character. That means in any case the top of the stack will then be -1 and the value below is either -1 or some character code that we don't care about, because it's an unpaired coin toss.

Then )* turns the -1 and the value below into a single 0 which we need a) to get rid of those two values and b) to enter the next loop correctly. That next loop is simply

"}
""

Which shifts all the values over to the auxiliary stack. This is necessary because we want to start processing the pairs that we read first. Now the final loop:

:{"
{ $;
)";.

The ) just increments some dummy value to ensure that it's positive and the instruction pointer turns north. { pulls over the first digit of the next pair and : duplicates it. Now when we're done processing, this will have been a 0 from the bottom of the auxiliary stack. Otherwise it's either 48 or 49. In case of a zero, we exit the loop and terminate on @, otherwise, the IP turns east.

{ pulls over the other digit of the current pair. $ takes the XOR between them. If that is 0, i.e. the two are equal, the IP just continues moving south, ; discards the zero, and the IP turns west into the next iteration. If the XOR was 1, i.e. they were different, the IP turns west, discards the 1 with ; and prints the first digit with ..