Consider the following three sentences:
Assume that each of the three has equal attestation. Which is the likely correct reading? To solve this problem, we need to determine the way that each reading could be derived based on the other readings. To start assume that one reading is the original reading and determine the steps that need to have been taken to arrive at the alternate readings.
If A was the original reading, then B and C were derived from A. B was derived by dropping the ending phrase (and blue). A was derived by dropping the middle phrase (red and).
If B was the original reading, then C and A were derived from B. C must have been derived from B by the word (red) being replaced with the word (blue). A would then be a combination of B and C.
If C was the original reading, then B and A were derived from C. B must have been derived from C by the word (blue) being replaced with the word (red). A would then be a combination of B and C.
To determine the probability of a reading being the correct one it is necessary to determing the probability of each of the corruption steps. For the sake of this discussion, assume that each of the readings is equally possible in terms of face value truth. Note that the probability that B or C are the correct readings are equal.
For the example case what is the probability of the following possibilities?
Each of the cases are possible. The last two are equally probable. The first seems less probable since it requires that two omissions errors be made in the same sentence. It seems likely that a scribe would combine two readings into one yielding (red and blue) from two manuscripts. Although this seems likely, there is no qualitative way to determine what the probabilities of each step are. In the end the result is still a statement of probability.
There are a number of other possible factors. These include:
There are a number of attestation factors. These are:
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Last Updated 1-1-97
