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Probability Paradox


Well-known member
Feb 2, 2012

I saw this "proof" many years ago.
.I thought you might enjoy it.

A bag contains two marbles.
Either can be Black or White.
Determine the colors of the marbles.

Answer: one Black marble and one White marble.


[tex]\text{There are three equally likely situtations.}[/tex]
. . [tex]\text{The bag contains: }\:BB,\,BW,\,WW.[/tex]

[tex]\text{Add one White marble to the bag.}[/tex]

[tex]\text{Then we have:}[/tex]

[tex][1]\;P(BBW) \,=\,\tfrac{1}{3}[/tex]
. . [tex]P(W\,|\,BBW) \:=\:\left(\tfrac{1}{3} \right)\left(\tfrac{1}{3} \right) \:=\:\tfrac{1}{9}[/tex]

[tex][2]\;P(BWW) \,=\,\tfrac{1}{3}[/tex]
. . [tex]P(W\,|\,BWW) \:=\:\left(\tfrac{1}{3}\right)\left(\tfrac{2}{3} \right) \:=\:\tfrac{2}{9}[/tex]

[tex][3]\;P(WWW) \,=\,\tfrac{1}{3}[/tex]
. . [tex]P(W\,|\,WWW) \:=\:\left(\tfrac{1}{3}\right)\left(\tfrac{3}{3} \right) \:=\:\tfrac{3}{9}[/tex]

[tex]\text{Hence: }\:p(W) \:=\:\tfrac{1}{9}\,+\,\tfrac{2}{9}\,+\,\tfrac{3}{9} \:=\:\tfrac{6}{9}\:=\:\tfrac{2}{3}[/tex]

[tex]\text{The probability of drawing a White ball is }\tfrac{2}{3}.[/tex]
. . [tex]\text{The bag }must\text{ contain 2 White balls and 1 Black ball.}[/tex]

[tex]\text{Therefore, before the White ball was added,}[/tex]
. . [tex]\text{the bag had one White ball and one Black ball.}[/tex]

Last edited:


Apr 4, 2014
Are we supposed to point out the fallacy?