Then using the discounting scheme described, we have P(damnbritish) = 1/(x + 6) P(damn) = 2/(x+6) P(british) = x/(x + 6) and We seek an x for which P(damn)P(british) > P(damnbritish).
we get 2x/(x + 6) 2 > 1/(x q-6) ~ x > 6 The actual scores for MBDP-1 when presented with damnbritish for a second time are -logP(damnbritish) = 2.8 and -logP(D&m) -logP(brItIS) = 1.8 + 0.9 = 2.7.