MHB Why is there a probability paradox in this bag of balls?

[soroban]

A bag contains two balls.
Either ball can be black or white.

Without drawing any balls, determine the colors of the balls.Solution

The bag could contain any of three contents, each with probabiity \tfrac{1}{3}.

. . \boxed{B B} \qquad \boxed{BW} \qquad \boxed{WW}

Add a black ball to the bag.
The resulting contents are

. . \boxed{BB \,B} \qquad \boxed{BW\,B} \qquad \boxed {WW\,B}Consider the probability of drawing a black ball.

Then: .P(B) \;=\;(\tfrac{1}{3})(\tfrac{3}{3}) + (\tfrac{1}{3})(\tfrac{2}{3}) + (\tfrac{1}{3})(\tfrac{1}{3}) \;=\;\tfrac{6}{9} \;=\;\tfrac{2}{3}Since the probability of drawing a black ball is \tfrac{2}{3}.
. . there must be 2 black and 1 white ball in the bag.

Therefore, before the black ball was added,
. . the bag must have contained one black and one white ball.

 

[topsquark]

My head hurts. (Crying)

-Dan

 

[HallsofIvy]

The trick is in your first statement,

The bag could contain any of three contents, each with probabiity \tfrac{1}{3}.

.

There are four equally likely outcomes, WW, WB, BW, and BB. Each has probability 1/4.

 

[soroban]

The trick is in your first statement,
.

There are four equally likely outcomes, WW, WB, BW, and BB. Each has probability 1/4.
his makes no difference.

We still add one black ball.

\text{We have: }\;WW\!B,\;W\!BB,\;BW\!B,\;BBB

\text{Then we have:}
. . \begin{array}{cccccc}<br /> P(B|WW\!B) &amp;=&amp; \frac{1}{4}\cdot \frac{1}{3} &amp;=&amp; \frac{1}{12} \\<br /> P(B|W\!BB) &amp;=&amp; \frac{1}{4} \cdot \frac{2}{3} &amp;=&amp; \frac{1}{6} \\<br /> P(B|BW\!B) &amp;=&amp; \frac{1}{4} \cdot \frac{2}{3} &amp;=&amp; \frac{1}{6} \\<br /> P(B|BBB) &amp;=&amp; \frac{1}{4} \cdot \frac{3}{3} &amp;=&amp; \frac{1}{4} \end{array}

\text{Therefore: }\;P(B) \;=\;\tfrac{1}{12} + \tfrac{1}{6}+\tfrac{1}{6} + \tfrac{1}{4} \;=\;\frac{2}{3}

THe "proof" still holds . . . LOL!

 

[I like Serena]

Nice conundrum! :)

I believe we've found the expectation, or average if you will, of the number of white and black balls.

 

[soroban]

Hello, ILS!

You've discovered the truth behind this apparent paradox.
This is indeed a matter of expectation, not actual content.

Consider the original bag with two balls
The bag could contain any of three contents, each with probability \tfrac{1}{3}.
. . \boxed{BB}\quad \boxed{BW} \quad \boxed{WW}

Consider the probability of drawing a black ball.,

P(B) \;=\;\left(\tfrac{1}{3}\right)\left(\tfrac{2}{2}\right) + \left(\tfrac{1}{3}\right)\left(\tfrac{1}{2}\right) + \left(\tfrac{1}{3}\right)\left(\tfrac{0}{2}\right) \;=\; \frac{1}{2}

We would NOT conclude that the bag must contain one black and one white ball.
But that is exactly what we did with the bag with three balls.