I'm talking about the whole sequence; you're talking about the number of heads (or) tails in the sequence. The number of heads is a sufficient statistic, so we

I'm talking about the whole sequence; you're talking about the number of heads (or) tails in the sequence. 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.