Bertrand's Box Paradox

Introduction


Bertrand’s box paradox, posed by Joseph Bertrand in 1889, is a paradox of elementary probability theory. According to the paradox, there are three boxes:

  1. A box containing two gold coins,

  2. A box containing two silver coins, and

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


If a person were to choose a box at random, and from that withdraw a coin at random, which happens to be a gold coin, the paradox lies in the probability of the next coin drawn from the same box being a gold coin. At first glance, it may appear that the probability of the next coin being a gold coin is ½ as the next coin could either be gold, or silver. But it is in fact ⅔ as we will see below. 


Why is the Paradox Counterintuitive?


After a box has been chosen, but before a coin is chosen, the probability of the box having two of the same kind of coin is ⅔. If the probability of ‘choosing a gold coin’, in combination with the ‘box having two of the same kind of coin’ is ½, then the probability of ‘choosing a silver coin’, in combination with the ‘box having two of the same kind of coin’ must also be ½. And if the probability that the box has two of the same coins changes to ½ no matter what kind of coin is shown, then the probability would have to be ½ even if we hadn’t chosen a coin this way. Since we know this probability is ⅔, not ½, we have an apparent paradox.


Explanation of the Solution 

To make things simpler, let us take the three boxes as –

  1. [g, g]

  2. [s, s]

  3. [g, s]

Where ‘g’ denotes a gold coin and ‘s’ denotes a silver coin.

Now since we are only considering the event of randomly choosing one coin from a box that happens to be gold, the sample space reduces to [g, g] and [g, s]. Furthermore, intuitively speaking, this would mean that the probability of the next coin being gold or silver is equal i.e. the probability is ½ , however, this is not the case.

 Notice that the box containing two gold coins, [g, g] contains twice as many gold coins in the box which contains one silver and one gold coin [g, s] which means that there is a 2:1 probability in favour of the box [g, g]. This implies that for every three turns one plays this game and draws a gold coin the first time, there is a chance that two out of those three turns will be box [g, g]. Therefore, the probability that the second coin is also gold is 2/3.

Mathematical Solution


We know that there are three boxes with ⅓ as their probability to be chosen as shown below. The sample space for the coins chosen is {GG, SG, SS} where S denotes a silver coin, and G denotes a golden coin

(Original illustrations)


The third box is striked out because, the coin drawn first was a gold coin, but the third box only has silver coins

To find the probability of the next coin drawn being a gold coin, we must find the probability that the first coin that is drawn was from Box 1. Now, since there are three gold coins in total out of which two are in Box 1, the coin that is drawn is twice as likely to be from Box 1 than Box 2.

(Original illustrations)



Solving the problem using Bayes theorem:


Let ‘1’ denote Box 1, ‘2’ denote Box 2, and ‘G’ denote a gold coin


P(1)= (1/3) 

P(G|1)= (1)

P(2)= (1/3)

P(G|2)= (1/2)


P(1|G)=   P(1) P(G|1) ÷ P(1)P(G|1)+ P(2)P(G|2)                     


P(1|G)=  (1/3)(1) ÷ (1/3) (1) + (1/3)(1/2)


P(1|G)= (1/3) ÷ (1/2)


P(1|G)= 2/3


Therefore the probability of drawing another gold coin, given a gold coin has already been drawn, is 2/3. That is to say, the probability that the gold coin which has already been drawn belongs to Box 1, is 2/3.     

We may think that because the probability of the first two boxes being chosen was the same, the probability that the next coin is gold is (1/2). However, while thinking like this, we easily miss taking into account the evidence that the gold coin which was already drawn gives us- that it is twice as likely to be from Box 1 than Box 2.


AUTHORS:

Katherine Yendrembam

Ruhi Shahnawaz

Vandhana Nandakumar


References


Bar-Hillel, M., & Falk, R. (1982). Some teasers concerning conditional probabilities. Cognition, 11(2), 109–122. https://doi.org/10.1016/0010-0277(82)90021-x

Bertrand's box paradox. (n.d.). Retrieved July 29, 2020, from https://www.oxfordreference.com/view/10.1093/oi/authority.20110803095501915 

Maths Explained. (2017, September 9). Bertrand’s Box Paradox. [Video]. Youtube. https://www.youtube.com/watch?v=FNl7xv5ms90


Bertrand’s Box Paradox (with and Without Bayes’ Theorem) - Untrammeled Mind. (2018, November 9). Retrieved August 2, 2020, from Untrammeled Mind website: https://www.untrammeledmind.com/2018/11/bertrands-box-paradox/


Bertrand’s Box Paradox. (2014). Retrieved August 2, 2020, from Wolframcloud.com website: https://www.wolframcloud.com/objects/demonstrations/BertrandsBoxParadox-source.nb

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