Such a scheme would have two effects against an attacker trying to analyze the frequency of words and word combinations:

1. They would need more samples to differentiate between two tokens at the same level of confidence.
2. For two different tokens with similar frequency they would no longer know that the words differ.

The first means you are making the problem like that of breaking a smaller data set. Larger collections of ciphertext are still breakable. You could correct for this by making the number of different IVs depend on the total plaintext size to be encrypted, but then you make searcher even slower.

The second may even be more of a problem for an attacker. The less likely options can occur at almost the same frequency – e.g. in letters there is a significant difference between the frequency of 'e' and 't', but 'j' and 'x' are almost equally uncommon. That means even if you can deduce one token from context, you've only solved e.g. one twentieth part of the problem.

Are these insurmountable problems? No. The fix may patch some leaks, but the encryption remains much less secure than non-deterministic encryption.

Trying to extend your tweak, it might be better to use a non-uniform distribution for the different IVs. For example, illustrating only two IVs with letters, if the probability was $\frac{1}{3}$ for one and $\frac{2}{3}$ for the other, the frequency of 'e' with the first IV would become about equal to that of 'n' with the other.

Again, I emphasize that you should only encrypt information like this if you are OK with an attacker learning nontrivial facts about the plaintext.

Such a scheme would have two effects against an attacker trying to analyze the frequency of words and word combinations:

1. They would need more samples to differentiate between two tokens at the same level of confidence.
2. For two different tokens with similar frequency they would no longer know that the words differ.

The first means you are making the problem like that of breaking a smaller data set. Larger collections of ciphertext are still breakable. You could correct for this by making the number of different IVs depend on the total plaintext size to be encrypted, but then you make searches even slower.

The second may even be more of a problem for an attacker. The less likely options can occur at almost the same frequency – e.g. in letters there is a significant difference between the frequency of 'e' and 't', but 'j' and 'x' are almost equally uncommon. That means even if you can deduce one token from context, you've only solved e.g. one twentieth part of the problem.

Are these insurmountable problems? No. The fix may patch some leaks, but the encryption remains much less secure than non-deterministic, semantically secure encryption.

Trying to extend your tweak, it might be better to use a non-uniform distribution for the different IVs. For example, illustrating only two IVs with letters, if the probability was $\frac{1}{3}$ for one and $\frac{2}{3}$ for the other, the frequency of 'e' with the first IV would become about equal to that of 'n' with the other.

Again, I emphasize that you should only encrypt information like this if you are OK with an attacker learning nontrivial facts about the plaintext.

Such a scheme would have two effects against an attacker trying to analyze the frequency of words and word combinations:

1. They would need more samples to differentiate between two words at the same level of confidence.
2. For two different tokens with similar frequency they would no longer know that the words differ.

The first means you are making the problem like that of breaking a smaller data set. Larger collections of ciphertext are still breakable. You could correct for this by making the number of different IVs depend on the total plaintext size to be encrypted.

The second may even be more of a problem for an attacker. The less likely options can occur at almost the same frequency – e.g. in letters there is a significant difference between the frequency of 'e' and 't', but 'j' and 'x' are almost equally uncommon. That means even if you can deduce one token from context, you've only solved e.g. one twentieth part of the problem.

Are these insurmountable problems? No. The fix may patch some leaks, but the encryptions remains much less secure than non-deterministic encryption.

Trying to extend your tweak, it might be better to use a non-uniform distribution for the different IVs. For example, illustrating only two IVs with letters, if the probability was $\frac{1}{3}$ for one and $\frac{2}{3}$ for the other, the frequency of 'e' with the first IV would become about equal to that of 'n' with the other.

Again, I emphasize that you should only encrypt information like this if you are OK with an attacker learning nontrivial facts about the plaintext.

Such a scheme would have two effects against an attacker trying to analyze the frequency of words and word combinations:

1. They would need more samples to differentiate between two tokens at the same level of confidence.
2. For two different tokens with similar frequency they would no longer know that the words differ.

The first means you are making the problem like that of breaking a smaller data set. Larger collections of ciphertext are still breakable. You could correct for this by making the number of different IVs depend on the total plaintext size to be encrypted, but then you make searcher even slower.

The second may even be more of a problem for an attacker. The less likely options can occur at almost the same frequency – e.g. in letters there is a significant difference between the frequency of 'e' and 't', but 'j' and 'x' are almost equally uncommon. That means even if you can deduce one token from context, you've only solved e.g. one twentieth part of the problem.

Are these insurmountable problems? No. The fix may patch some leaks, but the encryption remains much less secure than non-deterministic encryption.

Trying to extend your tweak, it might be better to use a non-uniform distribution for the different IVs. For example, illustrating only two IVs with letters, if the probability was $\frac{1}{3}$ for one and $\frac{2}{3}$ for the other, the frequency of 'e' with the first IV would become about equal to that of 'n' with the other.

Again, I emphasize that you should only encrypt information like this if you are OK with an attacker learning nontrivial facts about the plaintext.