How To Make Sense Of Bertrand’s Box Paradox?

Joseph Louis François Bertrand was a genius French mathematician who lived in the 19th century. By the time he was just nine years old, he had already lost his father. But even at that age, Bertrand could speak Latin fluently and was well-versed in mathematical concepts.

Between the ages of eleven and seventeen, he completed two bachelor’s degrees and a PhD in the mathematical theory of electricity. He went on to make significant contributions in the fields of number theory, probability theory, differential geometry, economics, and thermodynamics. Among many other things, he was the first to define real numbers using a notion that we now refer to as a Dedekind cut.

In 1889, Bertrand published his work titled “Calcul des Probabilités”, in which he wrote about a counterintuitive paradox involving probability. It starts with a simple puzzle.

You are presented with three identical boxes. Each box has two drawers, and inside each drawer, there is one coin. The first box contains two gold coins, one in each drawer. The second box contains one gold coin and one silver coin, one in each drawer. The third box contains two silver coins, one in each drawer.

Now, all the drawers are closed and the boxes are shuffled. You are presented with one box at random. When you open the first drawer at random (either the left-hand side drawer or the right-hand drawer), you find a gold coin inside. You take the coin out and keep it with you. Without replacing the coin that you just took out, what is the probability of finding another gold coin inside the same box (inside the second drawer)?

This is the fundamental scenario behind Bertrand’s box paradox. Why don’t you give some thought to this question? Once you have an answer in your mind, you will be able to appreciate what follows much more.

The Intuitive Answer to Bertrand’s Puzzle

To begin, we know for certain that the coin that you first drew was a gold coin. This means that the box that was presented at random was definitely not the one with two silver coins (Box 3). So, we can eliminate it from our consideration for what could be in the second drawer.

As a result, we are now left with two potential options: the second drawer could either contain a silver coin or a gold coin. In other words, the random chance leads us to only two equally probable outcomes. Based on this, it makes sense to wager that we have a 50–50 probability split.

However, Bertrand argued that this is objectively not the case. To be fair, a 50–50 split was my first consideration as well (such is the tricky nature of probability). Let us now move on to Bertrand’s reasoning behind his claim.

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The Objective Answer to Bertrand’s Puzzle

The are two mistakes/fallacies in the intuitive approach we just considered:

1. We simply eliminate the box with two silver coins (Box 3) from the probability equation altogether.

2. After reducing the problem to two boxes only, we “assume” an equal probability split (50–50).

Bertrand’s point was to consider the probabilities right from the start. We know that you were presented with a box at random and that the coin you drew from the first drawer was a gold coin.

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Since the box you were presented was at random, we originally had three equally probable boxes:

1. Box 1 — with two gold coins.

2. Box 2 — with one silver coin and one gold coin.

3. Box 3 — with two silver coins.

The fact that you drew a gold coin from the first drawer does not alter the probabilities of choosing one of these boxes retroactively. Now, considering that we had three equally probable boxes originally, AND the fact that the first coin was gold, we need to consider the following three cases:

1. Case 1 — The second draw could be the remaining silver coin from Box 2

2. Case 2 — The second draw could be one of the two gold coins from Box 1

3. Case 3 — The second draw could be one of the two gold coins from Box 1

If you are wondering why Cases 2 and 3 both refer to one of the coins from Box 1, there are two possibilities here, depending on which gold coin you drew in the first place. This fact becomes clear if we choose to number the coins as follows:

We can now compute the final probability by simply counting the contribution to a gold-coin second draw in each (equally probable) case as follows:

Probability of drawing a second gold coin after first draw = 0/3 (Case 1) + 1/3 (Case 2) + 1/3 (Case 3)

= 2/3

Therefore, the probability is, in fact, 2/3 and NOT 1/2.

Final Comments

If you are still struggling to wrap your head around why the final probability is 2/3 (and not an equal split), let us go back to the first draw. If you draw a gold coin in your first draw, you are twice as likely to have drawn from the drawer with two gold coins (2/3) as opposed to the drawer with one gold coin and one silver coin (1/3). So, the assumption that it is a 50–50 split is incorrect.

Update post publishing:

After publishing this essay, I wrote a program to simulate Bertrand’s box experiment (you can find this below at the end of the essay). I realised that the probability split of drawing a gold coin in the first draw is actually 50–50 (there are three gold coins and three silver coins at the start).

But the probability of drawing a gold coin in the second draw having already drawn a gold coin in the first draw rises to 2/3. I made a mistake in the above paragraph when I stated that the probability of drawing a gold coin in the first draw is 2/3.

I apologise for this mistake and am leaving the above paragraph untouched to keep track of the error historically and for completeness.

Probability is tricky and Bertrand knew it, which is why he made the effort to discuss this (veridical) paradox in his work. The funny thing is, more than a century later, we are arguably none the better.

If you ask me, human beings are not genetically wired to be good at probability. Any probabilistic computation we are good at is not done by actively thinking but by our subconscious subsystems. So, in my opinion, paradoxes like Bertrand’s Box are here to stay.

We need to accept our nature and continue to seek objectivity whenever we are dealing with probabilities. We cannot simply trust our intuition!

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Jupyter Notebook Code to Simulate Bertrand’s Box Experiment:

import numpy as np

# Define number of trials
num_trials = 1000000 # Change as per requirement

# Define three boxes: box0, box1, and box2
box = [0, 1, 2]

# Define the coins in each box
box0 = [1, 1] # Two gold coins
box1 = [1, 0] # One gold, one silver
box2 = [0, 0] # Two silver coins

# Initialize the counters for the first and second coins drawn
count_first_draw_gold = 0
count_second_draw_gold_given_first_draw_gold = 0

# Define the simulation loop
for i in range(num_trials):
    # Define variable for conditional probability.
    first_outcome_condition_gold = 1 # The first drawn coin must be gold
    
    # Select one of the three boxes randomly
    selected_box = np.random.choice(box)
    
    # Only box0 and box1 contain at least one gold coin.
    # If selected_box is box2, no need to check since no gold coin is present
    if selected_box == 2:
        continue

    #Draw the first coin from box0 or box1
    if selected_box == 0:
        box_outcome = np.random.choice(box0)
    else:
        box_outcome = np.random.choice(box1)

    # Check if the first draw results in a gold coin:
    if box_outcome == first_outcome_condition_gold:
        # increment count_first_draw_gold counter
        count_first_draw_gold += 1

        # Given, the first draw is gold, the second draw will ALWAYS be gold from box0
        # Given, the first draw is gold, the second draw will NEVER be gold from box1
        if selected_box == 0:
            count_second_draw_gold_given_first_draw_gold += 1

# Compute the frequentist outputs:
probability_first_draw_gold = count_first_draw_gold / num_trials
probability_second_draw_gold_given_first_draw_gold = count_second_draw_gold_given_first_draw_gold / count_first_draw_gold

print("Probability of getting a gold coin in the first draw: ", probability_first_draw_gold)
print("Probability of getting a gold coin in the second draw given a first gold draw: ", probability_second_draw_gold_given_first_draw_gold)

Probability of getting a gold coin in the first draw:  0.501006
Probability of getting a gold coin in the second draw given a first gold draw:  0.6668862249154701

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Further reading that might interest you: 

• The Unexpected Hanging Paradox – How To Solve It?
• How To Tackle The Raven Paradox?
• What Really Happens When You Divide By Zero?

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