Knickerbocker Photo (Update)

[Snowman]

Originally Posted by benjulmag

Snowman,

Not sure what you are saying. But to be clear, if each individual has a 90% chance of being a Knickerbocker, the probability of all six individuals being Knickerbockers is 0.9^6 which equates to 53%. The probability of only one of the members being a Knickerbocker is 1 - (0.1^6) which equates to the 99.9999% you cite. For it to be a Knickerbocker group photo, the relevant probability would be the 53%, not the 99.9999%.

We're not quite saying the same thing. Your math is correct if the question is phrased as "what is the probability that all 6 men are indeed a match for the 6 people Steve thinks they are?"

But that's not quite the same question I was answering above. If this is indeed a Knickerbockers photo, then the subjects in the photo are not independent of one another (independent in the statistical sense). In other words, if one of them is indeed a Knickerbocker, then that increases the likelihood that a second person is also a Knickerbocker. And if 2 are known to be Knickerbockers, then again, it increases the likelihood that a 3rd is, etc. Knickerbockers are likely to be photographed together. So my framing of the question "what are the odds that this is a Knickerbockers photo?" approaches it with that dependence structure in mind. It basically calculates what the odds are of all of his 90% Knickerbocker matches to be wrong rather than what the odds are for each one to be correct independently. My approach allows for, say, 5 of his 6 matches to be correct but him mistaking the identity of the 6th one, thus still making the photo a "Knickerbockers" photo.