Originally Posted by
Traveler
It seems like you mean well in your posts, but maybe you should make posts that are less mathematical heavy and more descriptive instead, that way you can get your message across better to the general public.
He's basically saying the following: suppose we grouped people into 8 different types (for example, INTx, ESFx, etc.), and had three dichotomies used for typing people (for example, I/E, T/F, N/S). If we could guess which side of each dichotomy a person was on with 90% accuracy, we could guess his type with 72.9% accuracy. That is, we might be 90% sure a person is a feeler, 90% sure that he's an extrovert, and 90% sure that he's a sensor, so we'd be 0.9^3 = 72.9% sure that all three of our guesses are correct and that he's an ESFx.
But we can also add more dichotomies to increase our chances of a correct typing; say we add four more dichotomies. These are superfluous in a sense, since they don't give us any extra information (for example, one dichotomy may have IxTxs and ExFxs on one side and ExTxs and IxFxs on the other, but we can figure out on which side of this dichotomy an individual belongs solely based on the earlier I/E and T/F dichotomies). Suppose that we can guess on which side of these new dichotomies a person lies with 90% accuracy. Now we will guess all 7 dichotomies correctly 0.9^7 = 47.8% of the time.
But even if we only guess 6 dichotomies correctly, we'll still get the type right, because of the 8 types, only one type will agree with 6 of our dichotomies, while the others will agree with 4. This depends on how the dichotomies are chosen; basically, they must be chosen so that every two types agree on exactly 3 dichotomies (see the chart with pluses and minuses for how this is possible). That way, if we get 6 out of 7 dichotomies right for a person of type X, only 3 of those dichotomies will be in common with a person of type Y; then even if we mistype on the 7th dichotomy, the person will have at most 4 dichotomies in common with type Y. So how often will this occur? For example, what are the odds of guessing the first dichotomy wrong and the rest right? That's simply 0.1 * (0.9)^6 = 5.3%. But you can also guess the second dichotomy wrong and the rest right with a probability of 5.3%; in total, then, there is a 7 * 0.053 = 37.1% of guessing exactly one dichotomy wrong, but getting the type right. So there is approximately a 37.1% + 47.8% = 85% chance of getting at most one dichotomy wrong, and therefore getting the type right.
Now what if we guess 5 dichotomies correctly? Well, then there's three types which are equally viable (for example, suppose we type someone as --+++++ using the table in the original post; then types 1, 7, and 8 are equally plausible, since they agree with our typing in 5 out of 7 dichotomies). The same is true in general; so how often does this occur? Well, there's 7C2 = 21 ways to get two dichotomies wrong. For example, say you get dichotomies x and y wrong; there are 7 choices for what x can be and 6 choices for what y can be (since x and y cannot be the same), but you must divide by 2 to avoid double-counting (e.g., getting dichotomies 1 and 6 wrong is the same as getting dichotomies 6 and 1 wrong). What are the odds of getting, say, dichotomies 1 and 2 wrong and getting the rest right? That's simply 0.1^2 * 0.9^5 = 0.59%. So the total chance of making exactly two errors is 21 * 0.0059 = 12.4%. From here, I think the original poster made a mistake. At this point, we have three types that fit our data equally well, so we can't do any better than randomly guessing amongst them. Hence, the chance of making two mistakes but guessing the correct type is 0.124/3 = 4.1%. Adding this all up, we have approximately a 85%+4.1% = 89% chance of guessing the correct type with 7 dichotomies, as opposed to a 72.9% chance with only 3 dichotomies.