# Something like the Monty Hall Dilemma

• Somefantastik
In summary, the jailor's reasoning is flawed because it assumes that the jailer is not going to tell prisoner A that he will be released, when in fact he will release both prisoner A and prisoner C. Therefore, the probability of prisoner A being executed is still 1/3, not 1/2 as the jailor claims.
Somefantastik
[problem]
Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner A asks the jailor to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information, since he already knows that at least one will go free. The jailer refuses to answer this question, pointing out that if A knew which of his fellows were to be set free, then his own probability of being executed would rise from 1/3 to 1/2, since he would then be one of two prisoners. What do you think of the jailor's reasoning?

[solution]

Let Ci be the ith prisoner will be executed. Ji be that jailor tells that the ith will be free. Suppose the 1st prisoner is asking. Then

P(Ci) = 1/3
P(J3|C2) = 1
P(J2|C3) = 1
P(J2|C1) = P(J3|C1) = 1/2

Compute $$P(C1|J2)$$ = $$\frac{P(C1 & J2)}{P(J2)}$$ = $$\frac{P(J2|C1)P(C1)}{(P(J2|C1)P(C1) + P(J2|C3)P(C3))}$$ = $$1/3$$

By the same token P(C1|J3) = 1/3 => Jailor's reasoning is wrong.

I don't understand how the probabilities in red were computed.

Take P(J3|C2). This means that C2 will be executed. Therefore, the jailor has to free number 3 because there's no other choice - he's not going to free prisoner A in this scheme. So P(J3|C2)=1. The same applies to the next one. As for the last one, it's prisoner A to be executed, so each of the others will be released with probability 1/2.

Somefantastik said:
[problem]
Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner A asks the jailor to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information, since he already knows that at least one will go free. The jailer refuses to answer this question, pointing out that if A knew which of his fellows were to be set free, then his own probability of being executed would rise from 1/3 to 1/2, since he would then be one of two prisoners. What do you think of the jailor's reasoning?

[solution]

Let Ci be the ith prisoner will be executed. Ji be that jailor tells that the ith will be free. Suppose the 1st prisoner is asking. Then

P(Ci) = 1/3
P(J3|C2) = 1
P(J2|C3) = 1
P(J2|C1) = P(J3|C1) = 1/2
The notation is a bit confusing. You referred to the prisoners as "A", "B", and "C" but here they are being referred to as 1, 2, and 3. I'm going to call them "A1", "B2", and "C3"!Also remember that the jailor is certainly NOT going to tell A1 that he will be set free.
P(J3|C2) is the probability that the jailor tells him that prisoner C3 will be set free given that prisoner B2 will be executed. If prisoner B2 is to be executed, and the jailer will NOT tell A1 that he is going to be released, he MUST answer that C3 is going to be released: P(J3|C2)= 1.
If, instead, it is C3 who will be executed, and the jailer will NOT tell A1 himself that he is going to be released, then he MUST anwer that B2 is going to be release: P(J2|C3)= 1.

However, if it is, in fact, A1 himself who is going to be executed, then the jailer can answer either B2 or C3- and they are equally likely: P(J3|C1)= P(J3|C1)= 1/2.

Compute $$P(C1|J2)$$ = $$\frac{P(C1 & J2)}{P(J2)}$$ = $$\frac{P(J2|C1)P(C1)}{(P(J2|C1)P(C1) + P(J2|C3)P(C3))}$$ = $$1/3$$

By the same token P(C1|J3) = 1/3 => Jailor's reasoning is wrong.

I don't understand how the probabilities in red were computed.

dhris said:
Take P(J3|C2). This means that C2 will be executed. Therefore, the jailor has to free number 3 because there's no other choice - he's not going to free prisoner A in this scheme.
That's not quite correctly stated. He is in fact going to free both prisoner C and prisoner A- he just isn't going to tell prisoner A that he is going to be freed!

So P(J3|C2)=1. The same applies to the next one. As for the last one, it's prisoner A to be executed, so each of the others will be released with probability 1/2.

Yes, poor wording on my part. It may not make any difference to the calculation, but it's a huge difference if you happen to be Prisoner A!

Last edited:

## 1. What is the Monty Hall Dilemma?

The Monty Hall Dilemma is a probability puzzle named after the host of the game show "Let's Make a Deal", Monty Hall. It involves a contestant choosing one out of three doors, one of which has a prize behind it. After the contestant makes their initial choice, the host opens one of the remaining doors to reveal that it does not have the prize. The contestant is then given the option to switch their choice to the other unopened door. The dilemma lies in whether it is more advantageous to stick with the original choice or switch to the other door.

## 2. Why is the Monty Hall Dilemma so controversial?

The Monty Hall Dilemma is controversial because it goes against our intuition and common sense. Many people believe that the contestant's chances of winning are 50/50 after one door is revealed, but in reality, the chances are actually 2/3 if the contestant switches their choice. This concept is difficult for some people to grasp and accept, leading to disagreement and debate.

## 3. How does the Monty Hall Dilemma relate to probability?

The Monty Hall Dilemma is a classic example of conditional probability. It demonstrates how the probability of an event can change based on new information. In this case, the revelation of one door being empty changes the probability of the other doors containing the prize.

## 4. Is there a mathematical explanation for the Monty Hall Dilemma?

Yes, there is a mathematical explanation for the Monty Hall Dilemma. The key to understanding the dilemma lies in understanding the concept of conditional probability and using the Bayes' theorem. By calculating the probability of the different scenarios, it becomes clear that switching doors is the more advantageous choice.

## 5. Are there real-life applications of the Monty Hall Dilemma?

While the Monty Hall Dilemma may seem like a theoretical puzzle, there are actually many real-life applications of this concept. It can be used to explain the advantage of changing strategies in games such as poker or blackjack. It can also be applied in decision-making scenarios where new information affects the probabilities, such as in the field of medicine or finance.

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