Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
Probability problem

This is the Bertrand's Box Paradox I read about on Wikipedia:

Assume there is three boxes:

a box containing two gold coins,
a box with two silver coins
and a box with one of each.

After choosing a box at random and withdrawing one coin at random,
if that happens to be a gold coin,
the probability is actually 66% instead of 50%.

And the problem is equivalent to asking the question
"What is the probability that I will pick a box with two coins of the same color?".

No matter how hard I try, I just couldn't comprehend this..

How is the possibility of picking a gold coin the same as the probability of picking a box with two coins of the same color?

Does this imply there is a 66% chance of picking a gold coin and a 66% chance of picking a sliver coin?

If so, can we just say there is 50% chance of picking either one of them since both stand a 66% chance....?! and suddenly everything makes no sense..

[UPDATES] It is actually the probability of the remaining coin to be gold is 66% but not the probability of obtaining the gold coin is 66%.. I've misread it....

And everything makes sense now :D !

share|cite|improve this question

marked as duplicate by MJD, William, tomasz, Noah Snyder, Ayman Hourieh Oct 6 '12 at 0:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

"the probability is". The probability of what? – Hurkyl Aug 28 '12 at 14:11
If we enumerate boxes $1,2,3$ and let probability to choose either of them be $\frac13$, then the probability of choosing a gold coin is $$ \frac13\cdot 0+\frac13 \cdot\frac12+\frac13\cdot 1 = \frac12 $$ by the law of total probability. There may be some trick with a sample space, though. – Ilya Aug 28 '12 at 14:13
@Hurkyl If I'm not wrong it is the probability of the coin, that was withdrew randomly from a box which was chosen randomly, to be a gold coin. Here is the URL to the Wikipedia page. – user38927 Aug 28 '12 at 14:17
@Ilya yeah.. And I just couldn't get the trick... – user38927 Aug 28 '12 at 14:19
The point is that the version you've given (first paragraph here) is unclear, while all other formulations in the article cited refer to the conditional probability. – Ilya Aug 28 '12 at 14:19
up vote 5 down vote accepted

The result is being incompletely quoted. Perform the experiment as described. Now suppose that you end up with a gold coin. The question is: what is the probability that the other coin in the same box is gold?

This probability is $\frac{2}{3}$. Let's do an informal computation. It will be imprecise, but could be made precise by using the notion of conditional probability.

Imagine repeating the experiment $3000$ times. Then each box will be picked roughly $1000$ times. We will get a gold coin about $1500$ times. Out of these $1500$ times that we get a gold, it will have come from the two-gold box $1000$ times.

So if we restrict attention to the $1500$ times that we get a gold, about $1000$ of these times it will come from the two-gold box. So given that we got a gold coin, the probability the other coin in the same box is gold should be around $\frac{1000}{1500}$.

share|cite|improve this answer
Thanks for pointing out that! Now I understand why the problem is equivalent to asking the question "What is the probability that I will pick a box with two coins of the same color?" and I also understand why I spent 1hour on comprehending the logic behind the paradox but to no avail. I've misinterpreted the problem for so long.. – user38927 Aug 28 '12 at 14:33