Choose a random number that is different from a bunch of other secret numbers
I think this would work, although whether it's practical is another matter. For large $x$ it won't be. It's basically an application of mental-poker.

First, choose a secure commutative encryption algorithm that is not vulnerable to known plaintext or chosen ciphertext attacks.


Everyone generates a random encryption key.
Everyone but Alice uses their key to encrypt their secret unique number.
Alice uses her key to encrypt every item in the list $[0,...,x]$, except her own number. She then shuffles the list.
Everyone passes their encrypted secret number/list to the next person (e.g. clockwise), who encrypts it, shuffling the list when they have it. Repeat until all the numbers and the list have been passed around. At that point both Alice's list and everyone else's number has been encrypted by everyone.
Remove all the encrypted numbers from the list. Since the encryption is commutative, each encrypted number will match itself in the list, but because everyone has shuffled the list, they don't know what those numbers originally were.
Pick the first number from the list. Have everyone decrypt it with their key and pass the result around, until they've made a whole round. Now everyone should have the same random number, which they can verify by comparing their results.

(If the number would only be decrypted once, the final decrypter could cheat and pick any number, hoping it doesn't match the others'.)




I.e. with your example values:


Alice: key $a$, Bob: $b$, Eve: $e$
Bob: $E_b(52)$, Eve: $E_e(193)$
Alice: $[E_a(0), ... E_a(17), E_a(19), ... E_a(200)]$, shuffle.
Alice $\mapsto$ Bob: $[E_a(i), ...]$, Bob: $[E_b(E_a(i)), ...]$, shuffle.

Bob $\mapsto$ Eve: $E_b(52)$, Eve: $E_e(E_b(52))$.

Eve $\mapsto$ Alice: $E_e(193)$, Alice: $E_a(E_e(193))$.

Alice $\mapsto$ Bob: $E_a(E_e(193))$, Bob: $E_b(E_a(E_e(193)))$.

Bob $\mapsto$ Eve: $[E_b(E_a(j)), ...]$, Eve: $[E_e(E_b(E_a(j))), ...]$, shuffle.

Eve $\mapsto$ Alice: $E_e(E_b(52))$, Alice: $E_a(E_e(E_b(52)))$.
Now the list, $E_a(E_e(E_b(52)))$ and $E_b(E_a(E_e(193)))$ can be revealed and the Bob and Eve's numbers removed. Since the encryption is commutative, $E_a(E_e(E_b(52)))$ will match $E_e(E_b(E_a(52)))$ somewhere in the list, as will $E_b(E_a(E_e(193)))$.
Everyone takes the first number in the list, say $E_e(E_b(E_a(42)))$. They each decrypt it, e.g. Bob: $D_b(E_e(E_b(E_a(42)))) = E_e(E_a(42))$. They pass them around like in step 4, so that eventually e.g. Bob gets $E_b(42)$, which he can decrypt and announce, comparing to Alice and Eve's results.