I think you selected the right solution to the wrong problem, I've underlined the culprit.
In our case this translates to the 6 witnesses all selecting the same person in the lineup but NOT restricted to Oswald.
The best analogy I can come up with is putting four numbered balls in a jar and let 6 persons in turn pick a ball, announce the number and puts the ball back. In your example Adam picks first and the rest must comply with Adam's choice, all ones, twos etc. There are four valid outcomes. Probability is 1/4^5 = 1/1024.
In the correct solution one red ball (Oswald) and three blue balls (the fillers) are put in a jar. Each person in turn picks a ball, announces the color and puts the ball back. Only one outcome is valid: all red. No numbering of each person in the lineup is needed as we're only interested in wheather red (Oswald) is picked or not. Probability is 1/4^6 = 1/4096.
Mr. BELIN. Now, when you signed it--what I want to know is, before you went down, had they already put on there a statement that the man you saw was the No. 8 man in the lineup?
Mr. WHALEY. I don't remember that. I don't remember whether it said three or two, or what.
Mr. BELIN. Did they have any statements on there before you went down to the lineup?
Mr. WHALEY. I never saw what they had in there. It was all written out by hand. The statement I saw, I think, was this one, and
that could be writing. I might not even seen this one yet. I signed my name because they said that is what I said. Mr. BELIN. Well, Mr. Whaley---
Mr. WHALEY. I know, sir, but I don't think you can understand what I had to put up with that afternoon.
Whaley was illiterate and had no idea what they had written...
"I don't remember whether it said three or two, or what.
