You flip a possibly-biased coin 20 times and get half heads, half tails, e.g. "THHHTTTTTHTHTTHHHTHH". Under the model where the bias is 0.5—a fair coin—the probability of that sequence is (0.5)^20 or about one in a million. In fact, the probability of any sequence you could observe is one in a million. Under the model where the bias is 0.4 the probability is (0.4)^10 × (0.6)^10 or about one in two million. That is, t…
Not quite; the probability of n/2 successes in n trials is given as Binomial(n,p) not p^n. p^n is correct for a single sequence but there are many possible sequences that result half heads, half tails and so you have a factor of "N choose X" or the so called "Binomial Coefficient". > (0.4)^20 × (0.6)^20 and I think you mean (0.4)^10 × (0.6)^10 or more generally p^x*(1-p)^n-x.
The number of heads is a sufficient statistic, so we'll get the same likelihood ratios out, but the likelihood values themselves will be larger.
You could make a similar point about the original CrossValidated Normal(0, 1)^N example by summarizing the data with the mean and sum of squares.
This doesn't work if the data were Cauchy(0, 1)^N instead.