THREE PRISONERS PROBLEM - A PROBABILITY PARADOX
Introduction
The three Prisoners problem is a mathematical equivalent to the Monty Hall problem with goat and car being replaced with execution and freedom, respectively. It first appeared in the mathematical games column by Martin Gardener in the popular American science magazine Scientific American in the year 1959.
What is the problem?
Here is what exactly the problem is. There are three prisoners in three different cells and all of them are sentenced to death. However, the governor has selected one among them at random to be avoided from the death sentence. The warden of the prison actually knows who is being pardoned but at the same time, he is not allowed to disclose it. For convenience, let’s take the three prisoners A, B and C to be Charles, Robin and Nick respectively. Prisoner A, that is Charles, begs to the warden to disclose the identity of one among the two who are going to be executed. "If Robin is to be pardoned, give me Nick’s name. If Nick is to be pardoned, give me Robin's name. And if I'm to be pardoned, secretly flip a coin to decide whether to name Robin or Nick."
The warden tells Charles that Robin is to be executed. Charles is pleased because he believes that his probability of surviving has gone up from 1/3 to 1/2, as it is now between him and Nick. Charles secretly tells Nick the news, who reasons that Charles' chance of being pardoned is unchanged at 1/3, but he is pleased because his own chance has gone up to 2/3. Which prisoner is correct?
Hence was the problem which gained so much attention and for which people had different kinds of explanations based on their varied perceptions and understanding of the problem.
An intuitive way of reasoning the paradox : subjective approach
After listening to the guard one of the prisoners creates an image in his mind that he has a 50:50 chance ratio for not being executed. However, he was unknown to the fact that the conversation of him with the guard was not limited to them but it was revealed to other prisoners too. Charles jumped to the conclusion that his probability of getting freedom is in the ratio of 1 is to 3. However logically, all three of them can form the same conclusion and thus have the chances of being free in the ratio of 1 is to 3. That means each one of them has the probability of 1:3 towards being free. The sum of their probabilities is equal to 1. But that is clearly impossible because the guard made his view clear that Robin will not be freed. So, therefore, this reasoning is completely wrong.
When Charles receives the information, he is not certain about his future. Charles had pointed out that his chance of being free was 1/3 whereas Robin was left with no probability at all or his probability of being free was 0. In order to get the probability as 1 or unity, Nick’s chances should be 2/3. When this type of situation occurs, we need to re-run it 100 times. Therefore, it can be said that
If Charles is set free( it will happen 50/100 times), if the coin shows head then the guard will tell Charles' name( 25/50 times) and if the coin shows a tail, then the guard will take Nick’s name ( 25/50 times). When Nick is supposed to be set free then the situation will again happen ( 50/100 times), in that case, it is expected from the guard that he will take the name of Nick’s rather than others since he has got 50/100 times which is clearly better than Charles. Let’s assume that the Guard takes Nick’s name. This leaves us with two possible explanations where the guard might come up with Robin's name and these two possibilities are as mentioned below :
If Robin then it might have occurred as Charles was set free along with it the coin has shown Heads, that is, 25 times. If the Guard names Robin then this might have happened as Nick was set free in 50 cases. The explanation of the following situation is given that the second event is twice likely to happen than that of the first event, that is, 25 times in Charles' part whereas 50 times in Nick’s part. Unlike the first condition, it does not require to be dependent on the result of the flip of a coin which ends up giving us a head or a tail to choose from. Charles had proposed a typical question to the guard which resulted in an asymmetrical situation between him and Nick. In this type of a situation, Charles should have never heard his name from the guard and as far as Nick is concerned, he could have heard his name but it is obvious that he didn't, and this is the reason why Nick received more information from the guard as compared to Charles.
Mathematical explanation
Martin Gardner had quickly risen to become a celebrated columnist who entertained the readers of his column because of his artistry in recreational mathematics which engaged his readers with the profoundness of his mastery over complex mathematics and science.Undoubtedly one of his most renowned “mathematical game “was the three prisoners problem which is sometimes regarded as his most seminal contribution to the fields of mathematics and statistics. What stood out immediately to his huge reader base about the three prisoners problem was its similarity to the infamous Monty Hall problem with many of his readers drawing a parallel between the two,hence his popularity skyrocketed when people realised that the three prisoners problem could be used as an alternative approach to counter the enigmatic Monty Hall problem .So what is the three prisoner problem ? The three prisoners problem is a rather engaging thought process that Gardener devised to satisfy the intellectual cravings of his reader which involves three prisoners A,B,C who are in the gallows awaiting their execution ,the clause being that only one among the three prisoners shall be executed the next day .It is also proclaimed that the prisoner to be executed his chosen at random . Basic probability tells us that the probability of each of the prisoners not escaping their respective execution is ⅓ . Now the interesting part of the problem is that though the guards are fully aware of the identity of the victim,they are forbidden to address any query of the prisoners regarding the state of their lives .Now let us assume that one prisoner engages with a guard and probes him regarding his mortal state.The guard tells him (let us assume that the above mentioned prisoner was A) that C is going to walk out alive the next day .On hearing this news A is crippled with anxiety ,seeing his friend in a deeply distressed state B attempts to mitigate his worry .B who was a profound mathematician consoles A by telling him that though it may appear to A that her chances of death doubled ,it was not actually so .This is the legendary Three prisoners problem .So how do we approach the question ? Let us first first raze the thicket and analyse the problem using a diagrammatic representation.
Analysing the possible scenarios:
A is free and the guard tells that B is to be executed: 1/3 × 1/2 = 1/6 of the cases
A is free and and the guard tells that C is to be executed: 1/3 × 1/2 = 1/6 of the cases
B is pardoned and the guard tells that C is to be executed: 1/3 of the cases
C is pardoned and the guard tells that B is to be executed: 1/3 of the cases
Note that the guard does not directly address the questioner’s condition
Now since the guard is choosing randomly the probability of choosing any of the three is ⅓ and also the 1/3 of the time that A is to be pardoned, there is a 1/2 chance he will say B and 1/2 chance he will say C. This means that taken overall, 1/6 of the time (1/3 [that A is pardoned] × 1/2 [that guard says B]), the warden will say B because A will be pardoned, and 1/6 of the time (1/3 [that A is pardoned] × 1/2 [that guard says C]) he will say C because A is being pardoned. This adds up to the total of 1/3 of the time (1/6 + 1/6) thereby meaning that A is being pardoned, which is accurate.
It is now clear that if the guard answers B to A (1/2 of the time of case 1, and case 4), then 1/3 of the time C is pardoned and A will still be executed (case 4), and only 1/6 of the time A is pardoned (case 1). Hence C's chances are (1/3)/(1/2) = 2/3 and A's are (1/6)/(1/2) = 1/3.
The key to this problem is that the guard may not reveal the name of a prisoner who will be pardoned. What this means is the probability of each of the three prisoners being saved and another being executed his ⅙ respectively. This means that the probability of b being alive when c is dead and vice versa is the same :1/3
P(A/B) = 1/3, P (B/A) = 2/3
A is free and the guard tells that B is to be executed: 1/3 × 1/2 = 1/6 of the cases
A is free and and the guard tells that C is to be executed: 1/3 × 1/2 = 1/6 of the cases
B is pardoned and the guard tells that C is to be executed: 1/3 of the cases
C is pardoned and the guard tells that B is to be executed: 1/3 of the cases
Group members :
ReplyDeleteVaishakh P Nair - 1933444
Vishnudath S - 1933447
Fathima A - 1933459
Pradiptya Bhattacharjee - 1933467