From a colleague (who wishes to remain anonymous) comes this poser:
Suppose that we have a box containing three objects: one red ball and two identical blue cubes. One object is randomly selected from the box, but we do not get to see or touch it. The object is put into a (low-quality) color-measuring device, which tells us that there is a 50% chance that the object is red and a 50% chance that the object is blue. What is the chance that the object is a ball?
I’ll give you some hints.
1. The words “randomly selected” are equivalent to just “selected”. This change allows you to deduce the chance of pulling out any object. I.e., since there is no information about how the objects are selected, particularly in how the objects’ characteristics (color, shape, size) might influence this selection, we are not justified in including any of this in our premises and thus we can say with what probability any object is selected.
2. The main question is “What is the chance that the object is a ball?” but since the ball is red and the cubes blue, and the measurement device has no premises about the shape of the object, we can rephrase the main question to, “What is the chance that the object is red?” Can that be answered?
Busy day here at Briggsville, so this is a much help as you get.
Update Forgot to say: I’ll post my solution tomorrow. Easy answer, difficult derivation.
All probability is conditional (as are all statements of logic). We want
(1) Pr(R | Machine says probability of R = 1/2, E)
were E are the details in the paragraph which were given to us and R means red, B means blue, and “Machine says probability of R = 1/2″ is the information that we have. This is a standard measurement error problem. Shorten the latter to MSPRH, since it’s easier to write. Now (1) is
(1′) Pr(R | MSPRH, E) = Pr(R, MSPRH | E) / Pr(MSPRH | E)
and that equals
(2) Pr(MSPRH| R, E) x Pr(R | E) / Pr(MSPRH | E)
and we expand the denominator
(3) Pr(MSPRH| R, E) x Pr(R | E) / [Pr(MSPRH| R, E) x Pr(R | E) + Pr(MSPRH| B, E) x Pr(B | E)]
Because of course Pr(MSPRH | E) = Pr(MSPRH , R| E) + Pr(MSPRH , B| E), etc.
(4) Pr(MSPR = 1/2 | R, E) = 1 – Pr(MSPR n.e. 1/2 | R, E)
where “n.e.” means not equal. And this may be expanded
(4′) Pr(MSPR = 1/2 | R, E) = 1 – Pr(MSPR = p1 | R, E) – Pr(MSPR = p2 | R, E) – … – Pr(MSPR = pn | R, E)
where every term on the r.h.s. except “MSPR = 1/2″ appears. The pi, i = 1, 2,…,n are just those values which the machine can display. These are, of course, discrete and finite (imagine the dial or print out). But we can even soften that requirement below.
We are not given any information about the machine’s capabilities, except that it can at least spit out “the probability of R = 1/2″. We inferred from this that it could have said, “the probability of R = pi.” And from this, since we have no other information, and because of the symmetry of individual constants, we can say that
(5) Pr(MSPR = pi | R, E) = Pr(MSPR = pj | R, E)
So this gives us
(6) Pr(MSPR = 1/2 , R| E) = 1 – (n-1)/n = 1/n
since there are n different values the machine may display. Plus these into (3) and we get
(7) [ (1/n) x (1/3) ]/ [(1/n) x (1/3) + (1/n) x (2/3)]
since Pr(R | E) = 1/2 and Pr(B | E) = 2/3, via E. If n is finite, then (7) = 1/3; or if we take (7) to the limit in n, it is still equal to 1/3. This also assumes the same argument holds for B in (5); that is:
(8) Pr(MSPR = pi | B, E) = Pr(MSPR = pj | B, E).
The other way to go at this is to think that the machine always spits out the same probability when confronted with an R or B object. That is
(9) Pr(MSPR = 1/2 | R, E) == pR = (1, 0)
where “==” means equivalent, and
(10) Pr(MSPR = 1/2 | B, E) == pB = (0, 1)
and pR and pB are fixed at those values, not variable, then (7) becomes
(7′) [ pR x (1/3) ]/ [pR x (1/3) + pB x (2/3)]
which may be re-written as
(11) 1/ [1 + 2 x pB/pR ]
Now, we deduce that pR = pB = 0 cannot happen simultaneously. After all, the machine did say “MSPR = 1/2″ and the object has to be R or B. And that means the probability of Pr (R | MSPR = 1/2, E) = 0, 1, or 1/3, depending on if pA = 0 & pB = 1; if pA = 1 & pB = 0; or if pA = 1 & pB = 1 respectively.
The only information we have on pA and pB is that they may be 0 or 1, so the probability they take those values (given this information and via the statistical syllogism) is 1/2 each. Thus Pr (R | MSPR = 1/2, E) = 0 x (1/4) + 1 x (1/4) + (1/3) x (1/4) = 1/4 + 1/12 = 4/12 = 1/3; where the (1/4) comes from the probability that Pr(pA = i , pB = j | E) = Pr(pA = i | pB = j | E) x Pr(pB = j | E), where i,j = 0,1. If i = j = 0 then Pr(pA = i | pB = j | E) = 0. Otherwise, Pr(pA = i | pB = j | E) = Pr(pA = i | E) since we have no information that knowing pB = j tells us anything about pA = i.
So the answer is 1/3, no matter how you slice it. To do better, we need more information about the vicissitudes of the machine.