You don't get plausible deniability in any way. Turns out that the least significant bits of a pixel are not uniformly distributed. The obvious example of fail

You don't get plausible deniability in any way. Turns out that the least significant bits of a pixel are not uniformly distributed. The obvious example of failure is that you expect more pixels with value 255 than 254, since any overexposure will be stuck at 255. The rest is obvious when you look at a histogram of pixel values; as the number of LSBs used increases you see the curve become stepped. LSB replacement steganography is easily detectable at about a ratio of about 0.01 bits per pixel. The smarter plan is to add or subtract 1 randomly (or do nothing) in order to match the LSB to what you want; this symmetry is harder to detect (but still very possible, the methods are just more complicated). The reason you encrypt the data is not just for data security, it is so that the embedder does not have to worry about the data that they are embedding screwing everything up. tl;dr LSB steganography does not give you any plausible deniability because it's really easy to detect in any type of image at very low embedding levels due to inherent limitations of the process. If you want to use it, you'd be safe with a single message at perhaps a level of 0.005 bits per pixel, but then in a ten megapixel image you'd get 150kB of data which hardly seems worth it when you have to transmit losslessly. Even then, you'd still be screwed at a theoretical level with Ker's square root law.