Bertrand's Box Paradox

There are three ways you can get to the position posed by the question, all equally likely: -

• You draw 1st gold coin from box containing two gold coins. Gold coin remains in box.
• You draw 2nd gold coin from box containing two gold coins. Gold coin remains in box.
• You draw gold coin from box containing one gold coin and one silver coin. Silver coin remains in box.

The second coin is drawn from the same box . . . two scenarios lead to it being gold, one scenario leads to it being silver.

Thus the overall probability that the other coin in the same box is also gold is $\boxed{\dfrac{2}{3}}$ .

When I solved the problem in my head I took a shortcut. I visualized a single "black box" of all the information given: - One gold coin drawn from four coins, for I knew that the silver-only box must be discounted. This leaves two gold coins and one silver coin in my "black box". Thus, the chance that the next coin is gold is $\frac{2}{3}$ .

$\begin{aligned} P(\text{two gold coins}\mid\text{first coin gold}) &= \frac{P(\text{first coin gold}\mid\text{two gold coins})P(\text{two gold coins})}{P(\text{first coin gold})} && \color{#3D99F6}\text{Bayes' theorem}\\ &= \frac{1\times\frac{1}{3}}{\frac{1}{2}} \\ &=\boxed{\dfrac{2}{3}} \end{aligned}$

There are six coins in total, all equally likely to be selected. Since you know a gold coin was taken, only three of the six possibilities remain. In two of these, the gold coins come from the same box. So, the probability that the other coin is also gold equals $\boxed{\frac{2}{3}}$ .

You are twice as likely to pull a gold coin out of the first box than the third.

We had three boxes with coins given that we have already found one gold coin the problem reduces to finding the probability that a coin draw would result in gold when we had 1 silver and 2 gold coins because of that extra information.

Two thirds of the gold coins are in the box with two gold coins.

Once one gold coin is pulled out, there are two boxes in question. Since the odds of a gold coin being pulled at of a box with two gold coins is greater then a silver and gold box. The odds are over 50%. The only option over 50% is 2/3 so I picked that.

There are 2 types of coin, of which 3 are silver and 3 are gold.

There are 3 types of box All Gold (1g) All Silver (1s) Half Gold/Half Silver ((1/2)g + (1/2)s)

(1g) + (1s) + ((1/2)g + (1/2)s) = 3 Boxes

We can minus Box (1s) as can't be both silvers. We have picked a gold. We can see our remaining options are of two boxes, all gold and half Gold & half silver.

Which can be written in half boxes to illustrate we are yet to determine which ((1/2)g) we picked from what chest .

(2 (1/2)g) + (( 1 / 2)g + ( 1 / 2)s) = 2 Boxes left

Now we minus the gold coin in our hand ((1/2)g ) and are left with:

(1/2)g +(1/2)g +(1/2)s = 100% of what's left

All half quantities are equally likely (33.3%) as a ratio (2:1) two of them are gold to one silver.

Therefore there is a 66.6% probability the next coin is gold.

If anyone has trouble grokking this puzzle, here's another way to think of it.

Instead of the coins being in boxes, supposed that they are attached in pairs with long strings. You take all six coins, drop them in a hat, and stir so that the strings are too mixed up for you easily to see which ones connect. You grab one of the three gold coins. When you follow its string to the attached coin, what are the chances you expect to find another gold coin on the other end? Of the three gold coins, two of them connect to another.

The solution, as many posters have demonstrated, is 2/3. Thirty years on from the Monty Hall Problem we still have mathematicians denying basic probability. It's depressing.

For the answer that you're "supposed" to get imagine playing this game 300 times. 150 of those times you end up with a gold coin. 100 of those situations occur in the box with two gold coins. Of course once you have chosen a box there is no probability involved. You either have a gold coin or you don't. So the correct answer to the question as given is 1 or 0. In this situation this is pedantry but poor phrasing leads to a lot of confusion in this subject and is a good way of hiding paradoxes for other questions.

Simple Bayes Theorem.