Assume that point A is at location -1 and point B is at 1. If one source produced both A and B and the next shot is expected to be produced by the same source, then the best fit is the normal distribution with mean 0 and dispersion 1: N(0,1,x) - gives the maximum probabilities for points A and B. On the other hand, if different sources produced A and B, the best guess would be that the third shot is normally distributed with one of the sources F(x). Now since no information present about whether one or two sources produced A and B, give the same chances to either way:For F(x), give the same chances to points A and B:
F(x)=(F(-1,x)+F(1,x))/2
where
The graph for function f(x) is presented above. The minimum values are at points: -0.47 and +0.47. This means that the safest place is about a quarter distance from either end.
