Probability parabox

S

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,586

I posted this problem a few years ago.
As I recall, no one supplied a satisfactory explanation.


Problem

There are two balls in a covered box.
Each ball is either black or white.

Without looking or sampling,
. . determine the color(s) of the balls.



Solution

There are three equally likely cases:

. . \(\displaystyle \boxed{\bullet\;\bullet} \quad \boxed{\bullet\;\circ}\quad \boxed{\circ\;\circ}\)


Add one black ball to the box.

. . \(\displaystyle \boxed{\bullet\;\bullet\;\bullet} \quad \boxed{\bullet\;\bullet\;\circ} \quad \boxed{\bullet\;\circ\;\circ}\)


Consider the probability of drawing a black ball.

. . \(\displaystyle P(\text{black}) \;=\;\tfrac{1}{3}(1) + \tfrac{1}{3}\left(\tfrac{2}{3}\right) + \tfrac{1}{3}\left(\tfrac{1}{3}\right) \;=\;\tfrac{2}{3}\)

\(\displaystyle \text{Since }P(\text{black}) =\tfrac{2}{3}\text{, the box }must\text{ contain }\begin{Bmatrix}\text{2 black} \\ \text{1 white} \end{Bmatrix}\)


\(\displaystyle \text{Therefore, before the black ball was added,}\)
. . . . . . . . .\(\displaystyle \text{the box must have contained: }\:\begin{Bmatrix}\text{1 black} \\ \text{1 white} \end{Bmatrix}\)

 
soroban said:

I posted this problem a few years ago.
As I recall, no one supplied a satisfactory explanation.
OK I'll be a sucker and bite.

Problem

There are two balls in a covered box.
Each ball is either black or white.

Without looking or sampling,
. . determine the color(s) of the balls.

Solution

There are three equally likely cases: And the equality of these probabilities is known how? Ahh, the principle of insufficient reason, OF COURSE.

. . \(\displaystyle \boxed{\bullet\;\bullet} \quad \boxed{\bullet\;\circ}\quad \boxed{\circ\;\circ}\)

Add one black ball to the box.

. . \(\displaystyle \boxed{\bullet\;\bullet\;\bullet} \quad \boxed{\bullet\;\bullet\;\circ} \quad \boxed{\bullet\;\circ\;\circ}\)


Consider the probability of drawing a black ball.

. . \(\displaystyle P(\text{black}) \;=\;\tfrac{1}{3}(1) + \tfrac{1}{3}\left(\tfrac{2}{3}\right) + \tfrac{1}{3}\left(\tfrac{1}{3}\right) \;=\;\tfrac{2}{3}\)
Actually, you have calculated P(Black | Principle of Insufficient Reason), which tells us nothing about P(Black). Even if we accept that principle as a law of the universe that graciously eliminates our ignorance, you have calculated P(Black | Equal Probabilities of Zero, One, and Two Blacks Originally in the Box). So the EXPECTED number of black balls originally in the box was 1, but an expectation is not a certainty.
How am I doing?
Click to expand...
 
edit:nevermind

I'm not so sure this is any different than concluding P(black)=1/2 implies one is black :?
 

The fallacy in the "arguement" is in the probability of drawing a black ball
. . and the subsequent conclusion.

Suppose there were three balls in the box, any of them either black or white.

Then there are four equally likely cases:

. . \(\displaystyle \boxed{\bullet\;\bullet\;\bullet} \quad \boxed{\bullet\;\bullet\;\circ} \quad \boxed{\bullet\;\circ\;\circ} \quad \boxed{\circ\;\circ\;\circ}\quad\)

Add one black ball to the box:

. . \(\displaystyle \boxed{\bullet\;\bullet\;\bullet\;\,\bullet} \quad \boxed{\bullet\;\bullet\;\bullet\;\,\circ} \quad \boxed{\bullet\;\bullet\;\circ\;\,\circ} \quad \boxed{\bullet\;\circ\;\circ\;\,\circ}\quad\)


Consider the probability of drawing a black ball.

. . \(\displaystyle P(\text{black}) \:=\:\tfrac{1}{4}(1) + \tfrac{1}{4}(\tfrac{3}{4}) + \tfrac{1}{4}(\tfrac{2}{4}) + \tfrac{1}{4}(\tfrac{1}{4}) \;=\;\tfrac{5}{8}\)


\(\displaystyle \text{Since }P(\text{black}) = \tfrac{5}{8},\,\text{ the box contains: }\:\begin{Bmatrix}\text{5 black}\\ \text{3 white} \end{Bmatrix}\) . What?

The probability is correct . . . The inference is wrong.

 
I think your answer is a much more graphic way of elucidating what I said. Thanks
 
You must log in or register to reply here.
Top