Bertrand's Box Paradox Calculator

Explore the counterintuitive probability puzzle with three boxes and coins

About the Bertrand's Box Paradox Calculator

Bertrand's Box Paradox is a classic probability puzzle that challenges our intuition regarding conditional events. The setup involves three identical boxes: one contains two gold coins (GG), one contains two silver coins (SS), and one contains one of each (GS). If you choose a box at random and draw one coin that turns out to be gold, what are the odds that the remaining coin in that same box is also gold? Most people instinctively guess 50%, assuming they are either in the GG box or the GS box, making it a coin flip.

This calculator demonstrates why that intuition is incorrect. By applying Bayesian probability, we see that the information of drawing a gold coin first makes it twice as likely that you picked the GG box compared to the GS box. This tool is frequently used by statistics students, logic enthusiasts, and math teachers to visualize why the actual probability is 2/3. It serves as a fundamental lesson in defining a sample space correctly rather than looking at the remaining boxes in isolation.

Formula

P(G2 | G1) = P(G1 and G2) / P(G1) = (1/3) / (1/2) = 2/3

The formula uses conditional probability (Bayes' Theorem). P(G2 | G1) is the probability that the second coin is Gold given that the first coin drawn was Gold. P(G1 and G2) is the probability of choosing the box with two Gold coins (1/3). P(G1) is the total probability of drawing a Gold coin from any box, which is 1/2 (since 3 out of the 6 total coins are Gold). Dividing 1/3 by 1/2 yields 2/3.

Worked examples

Example 1: A student selects one of three boxes, pulls out a gold coin, and wants to know the odds of the second coin being gold.

1. Identify total coins: 3 gold, 3 silver. 
2. Probability of picking a gold coin initially P(G1) = 3/6 = 0.5. 
3. Probability of picking the GG box P(GG) = 1/3. 
4. Apply Bayes' Theorem: P(GG | G1) = [P(G1 | GG) * P(GG)] / P(G1). 
5. P(GG | G1) = [1.0 * (1/3)] / 0.5 = 0.6667.

Result: 2/3 (approximately 66.67%). The user is twice as likely to have the GG box as the GS box.

Common use cases

Teaching introductory statistics students about the differences between physical boxes and the probability of individual coin faces.
Preparing for quantitative finance interviews where brain teasers involving conditional probability are common.
Demonstrating the concept of 'sample space' in a high school mathematics curriculum.

Pitfalls and limitations

Confusing the number of remaining boxes with the number of possible successful outcomes.
Failing to account for the fact that the GG box has two 'ways' to provide a gold coin first, while the GS box only has one.
Assuming that because the SS box is eliminated, the remaining two boxes must have equal weighting.

Frequently asked questions

why isn't the probability of the second gold coin 50 percent?

No, the third box (Silver/Silver) is irrelevant because the premise of the problem requires you to have already drawn a Gold coin. Since the Silver/Silver box cannot yield a Gold coin, it is eliminated from the sample space of possible starting boxes.

is bertrand's box paradox a conditional probability problem?

Total probability considers all possible outcomes, while conditional probability limits the scope to a specific event that has already occurred. In this paradox, the 'condition' is that the first coin is Gold, which disproportionately favors the box that contains two of them.

who invented the bertrand box paradox?

Joseph Bertrand introduced this problem in 1889 in his work 'Calcul des probabilités'. He used it to demonstrate how easily the human mind miscalculates odds when dealing with restricted choice scenarios.

is bertrand's box paradox the same as the monty hall problem?

The Monty Hall Problem involves a host who knows what is behind doors and makes a choice to reveal them, whereas Bertrand's paradox is a static selection problem. However, both rely on the same counterintuitive logic regarding updated information and sample spaces.

what if i pick a silver coin first in the bertrand box?

If you draw a Silver coin first, the probability that the second coin in the same box is also Silver is 2/3. The logic is perfectly symmetrical to the Gold coin scenario.