Bertrand’s Box Paradox

Bertrand’s Box Paradox

 

Joel George Mani (1933419)

Siddharth Sundar (1933437)

Udhaynithi V.S (1933441)

Christ University

School of Business Studies and Social Sciences

The Paradox

A paradox in probability occurs when people have wrong implications of probability or on the grounds that the stating is questionable, which leads to numerous interpretations.

Bertrand's box paradox is a paradox which falls under the umbrella of elementary probability theory. In 1889, Joseph Bertrand, a French mathematician, first brought up the paradox when he published in it his work Calcul des probabilités. The paradox can also be observed through the use of cards.

The premise of the paradox involves three boxes. All the boxes contain two coins. A box contains two gold coins, another one contains two silver coins and the remaining box contains a single coin of both gold and silver. The ‘paradox’ lies within determining the probability. Initially, after picking a box at random, we withdraw a single coin at random. By chance, if the withdrawn coin happens to be a golden one, we try to ascertain the probability of the next coin we withdraw from the same box of also being golden.

When finding the probability, we have to think carefully instead of depending on our intuition. We know we don’t have to bother about the box with the two silver coins at this point. Similarly, we assume we could chose any box at random in the hope of picking a gold coin as both boxes contain a gold coin. Our intuition might lead us to think the probability is 1/2. Unfortunately, this is incorrect. We have to pay attention to the fact that one of the boxes contains double the number of golden coins than the other box.  As a result, we are twice as likely to choose the first box. From that point of view, there is a 2:1 ratio in favour of the first box which contains two golden coins. That roughly means, on average, when we play this game thrice and pick out a golden coin first, two of those times we will pull it out of the first box. In the remaining other scenario, we will pull it out of the other box containing only one gold coin. Ergo, the probability of the second coin drawn being golden is 2/3.

Bertrand’s box paradox among other similar paradoxes is simple yet counterintuitive in nature because of which they are employed in probability as standardised examples. Their solution illustrates some basic principles, including the Kolmogorov axioms. Some problems that are very similar to the paradox include the Monty Hall problem and the Three Prisoners problem.

Mathematical Analysis

Let us now discuss this Bertrand’s Box paradox with a mathematical problem and see how this paradox leads to the popular Bayes’ theorem.

There are 3 boxes:

1)      A box containing two gold coins,

2)      A box containing two silver coins,

3)      A box containing one gold coin and a silver coin.

The Bertrand’s Box paradox lies in the probability, after choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, of the next coin drawn from the same box also being a gold coin. Here comes the question. What’s the probability the second coin you draw out from that same box is also gold?

Consider the tree diagram given above. Here, GG implies the box with two gold coins, SG implies the box with silver and a gold coin and SS implies the box with two silver coins. The probability of choosing each of the boxes is simply 1/3. We then pull a gold coin from any of these boxes. We all pull a gold coin from the 1^st ox so its probability is 1. Then, the next box contains only one gold coin out of the two coins. So the probability is ½. We cannot pull a gold coin from the 3^rd box, so we rule out the 3^rd box from consideration.

Next, to draw another gold coin, there are only two possibilities wither from box1 (GG) or from box2 (SG). As the 1^st box contains 2 gold coins, obviously the probability would 2/2 i.e. 1. As the 2^nd box contains 1 gold and 1 silver the different probabilities available from the 2^nd box are the probability of getting a gold i.e. ½ and the probability of getting a silver i.e. also ½. In this context, we don’t need the probability of getting a silver coin, so we rule that out of the equation.

Coming to the answer, the probability that the coin drawn out on the second time from that same box is also gold is 2/3. For the second drawn out to be gold, it has to come from either 1^st or the 2^nd box. In simpler terms, we add the possibilities. (1 gold coin from 2 gold coins) + (1 gold coin from 1 gold and silver coin). i.e. 2 positive outcomes from the total sample space that is 3.

There is a (1/3 * 2/2)= 1/3 probability of choosing box1 and pulling a gold coin and there is a (1/3*1/2) =1/6 probability of choosing box2 and pulling a gold coin. Now the probability can be written as the box GG given the evidence of a gold coin by setting a ratio against the sum of all the probabilities corresponding to the outcomes that meet our conditions.

 

P (choosing)*P (pulling gold having chosen GG)

P (choosing GG)*P (pulling gold having chosen)+P (choosing SG)*P(pulling gold having chosen SG)

=          1/3 * 1

     1/3*1 + 1/3*1/2

=   1/3

     ½

= 2/3

So, here we have the answer also bringing the Bayes’ theorem.

This is the basic formula for Bayes’ theorem.

We use this formula as,

Where,

 P (GG|g) implies probability of choosing box GG given the 1^st coin chosen is gold

P (g|GG) is the probability of the first coin being gold when we’ve chosen box GG.

P(GG) is the probability of choosing box GG

P(~GG) is the probability of choosing wither box SG or SS

P(g|~GG) is the probability of the first coin being gold when we’ve chosen either box SG or box SS.

That concludes our Bayes’ theorem and that is how Bertrand’s box paradox is related to Bayes’ theorem.

Solution to the Problem

This problem might look simple at first. One would think that since there are only two boxes with a gold bar/coin in it, the reason would be that the person might have picked one out of those two boxes. This is 1/2 probability that the other coin/ball in the drawer would also be gold, which is 50%. That is not the correct way to solve this. It might look right at first glance, but the answer in incorrect.

A box has two compartments, so let us assume them to be; G1, G2 to be in the first box, S1, S2 to be in the second box and G3, S3 to be in the third and final box. Noting down all the possible draws; G1G2, G2G1, S1S2, S2S1, G3S3, S3G3.

Once you have drawn a gold coin/ball, you are certain that you have chosen from either from the first box which contains two gold coins/balls or the third box which contains one gold and one silver coin/ball. Let us consider only the first box and the third box as the second box is irrelevant to us.

Something to note here is that, the first box has twice the number of gold coin/ball as the third box. Thus, the likeliness of you to have chosen first box is twice as much as choosing the third box. In simpler words to make things more understandable, there exists a 2:1 ratio which is in favour of the first box. This means that, on an average, for every three times you open a drawer of a box and pull out a gold coin/ball, two times out of three, you will get a gold coin/ball from the first box, and one time out of three you will get a silver coin/ball. Thus, it can be inferred that the probability of the second compartment of the box also containing a gold coin/ball is 2/3, which is two times out of three.

In essence, from this paradox, we can understand that we cannot trust our intuition at all times. Especially, when a problem of this sort is presented in front of us. This is why most people fail to get the right answer to this problem as they think from the surface and go with their intuition, which says there is an equal chance (50%) that the other coin/ball drawn will be a gold one. Most people do not consider first box has two separate gold coins/balls, and they take it as one, this makes them arrive at an answer which in incorrect.

 

 

References

“Bertrand’s Box Paradox”, (n.d.) Retrieved from https://www.agftutoring.com/bertrands-box-paradox/

“Bertrand’s Box Paradox: The answer is 2/3!!”, (n.d.) Retrieved from https://whyevolutionistrue.com/2018/02/20/the-answer-is-2-3/

“Bertrand’s Box Paradox (with and Without Bayes’ Theorem)”, (n.d.) Retrieved from https://www.untrammeledmind.com/2018/11/bertrands-box-paradox/