## Monty Hall Problem

Monty Hall is back, for one last time, to host the famous show from the 1960s ‘Let’s Make a Deal’. You are his final guest and the prize is really desirable – a red Ferrari 458 Italia. Monty is no generous host; he could be over 90 years old, but he is going to make you work for your prize. So here is the deal, there are three closed doors (A, B, and C) and behind one of these doors is the Ferrari and the remaining doors have a goat each. You need to pick a door and can take home whatever is behind that door – you really hope it’s the Ferrari. After doing your calculation you picked door A – you know that was a wild guess. Monty being a generous host tries to help you out by opening one of the remaining two doors i.e. door B and displays the beautiful goat behind the door. By the way, Monty knows the location of the car and will never open the door concealing it. He asked you for the last time if you would like to switch from door A to door C. This is where it gets really difficult – whether to stick or switch. The decision is particularly difficult when a huge audience (studio and television) is watching you live. Despite the pressure, you have decided to do the long calculation for this problem using the Bayes’ theorem. I think you have made a wise choice.

## Bayes’ Theorem to Solve Monty Hall Problem

You are aware of the difficulty of this problem. The solution to this problem is completely counter-intuitive. Marilyn Vos Savant was asked to solve the same problem by a reader in her column ‘Ask Marilyn’ in Parade magazine. Marilyn, by the way, is listed as the person with ‘Highest IQ’ by the Guinness Book of World Records. She recommended switching as that will increase one’s chances of winning the car by a factor of two. Her answer created a major furor with one of the readers (with a Ph.D.) writing her the following:

“[Marilyn] You blew it, and you blew it big! Since you seem to have difficulty grasping the basic principle at work here, I’ll explain. After the host reveals a goat, you now have a one-in-two chance of being correct. Whether you change your selection or not, the odds are the same. There is enough mathematical illiteracy in this country, and we don’t need the world’s highest IQ propagating more. Shame!” – Scott Smith, Ph.D. University of Florida |

You know it is not going to be easy but Bayes’ Theorem will guide you well. We have discussed Bayes’ Theorem a couple of times in previous articles on YOU CANalytics, you may want to refer to those articles Bayesian Inference – Made Easy and O.J. Simpson case.

You have scribbled down your solution to the problem on a piece of paper. The following is a version of your scribbling

It seems Marlyn was right or was she? You want to solve it one more time to be doubly sure. So you start all over again, to begin with, you need to find the prior probability i.e. the chances of having the car behind each door. Since there is no other information available the prior probabilities is the same for all the doors. This was fairly easy as shown in the formula below.

There is not much to choose and hence you have chosen door A. Now comes the difficult part, Monty Hall has revealed a goat behind door B this has infused some information into the above problem. This needs us to create a second layer of probabilities post the event i.e. probability of Monty to open door B when you have chosen door A. Remember that Monty knows the location of the car and will never open the door concealing the car.

This requires us to calculate the conditional probability or chances of the event given the position of the car. What are the chances for Monty to open door B if the car is behind door A?

Monty could have opened either 'door B' or 'door C'. As You have chosen 'door A' (He cannot open this door)

Second question: What are the chances for Monty to open door B if the car is behind door B?

Monty will never open door B if it was concealing the car.

Final question: What are the chances for Monty to open door B if the car is behind door C?

Monty has to open door B, he has no other choice. You have already chosen 'door A' and door C is concealing the car.

Having this knowledge leaves you with one final task to calculate the posterior probabilities. One of the calculations is shown below.

where,

Let us revisit your scribbling from the previous calculation.

The chances of winning the car are indeed 2 times higher i.e. 2/3 when you switch than when you stick i.e. 1/3. Marilyn was right! Going by your calculation you have decided to switch to door C.

#### Sign-off Note

Monty opens the door you chose i.e. door C and reveals your prize – a goat! Hard luck. If Monty Hall plays this game with you 1000 times you are most likely to end up with 667 Ferraris – that’s how probability works. You have decided to donate the goat to a charity and feeling good about it (probably better than driving a Ferrari).

Formally, the Monty Hall problem can be generalized by increasing the number of doors or the number of people (players).

The 2-person Monty Hall problemThere are four closed doors (A, B, C and D) and behind one of these doors is a prize and the remaining doors are empty. Monty knows the location of a prize. There are two players, Adam and Eve. The sequence of the game is as follows:

First step– Adam and Eve choose the door (let’s assume that they do it independently, at the same time). They know what they did. Their first decisions are common knowledge.Second step– Monty opens an empty door.Third step– Eve changes her initial choice. We know (from the classic Monty Hull problem) that a change in the decision is a rational strategy. This means that Eve is a rational player.Fourth step– Adam (knowing Eve second choice) also changes his initial decision. He is also a rational player.What is the probability of winning a prize by Adam?Does the probability of success depends on the assumption of independence (in the first step of the game)? Does the sequencing of decision in the first step can increase the chances of final success? [Answer: Yes] Does the assumption of sequentiality in the fourth step is important? [Answer: No]. What we can model this 2-Person Monty Hall in a Bayesian framework?

Thanks Jan for sharing the generalized form of Monty Hall problem. In my opinion the beauty of science is in generalizing special cases to form a holistic theory.

Solution to N door problemIf one would increase the number of doors from 3 to N, and repeat the above Monty Hall experiment as it is, then the individual probability for each door (not chosen by the player) will increase from 1/N to (N-1)/((N)*(N-2)). Monty Hall problem is of course a special case of this with N = 3. My intuition behind this is that probabilities of the other (N-1) doors get realigned / redistributed with the reduction in a door (revealed by Monty) i.e. initially (N-1)/N probability that was with other the (N-1) doors gets redistributed to the remaining (N-2) doors with the reduction of a door i.e. (N-1)/((N)*(N-2))

Many Players Game, I really liked your extension of the problem with Adam and Eve. There are so many possibilities in this version. Would really like to explore this one further.Thanks again Jan

Your solution is not complete.

Let N be the number of doors. Monty can open m empty doors, m=1,2,…,N-2.

So probability of win prize (sucess) after change your first decision is:

P(S/m)=(N-1)/ (N*(N-m-1)) (*)

For standard Monty Hall N=3, and m=1. P(S/change)=(3-1)((3*(3-1-1)=2/3.

If N=4, we Monty has more freedom: he can send m=1 (one message), m=2 (and no more). If m=1, then P(S/m=1)=(4-1)/(4*(4-1-1))=3/8. If m=2, then P(S/m=2)=(4-1)(4*(4-2-1)=3/4. If we assume that Mony uniformly distribute his choices over m, i.e., p(m=1)=p(m=2)=…=1/(N-2), then P(S) is an expected value of P(S/m).

Note that m=0, change of your first decision has no effect on the probability of sucses, i.e. P(S/m=0)=1/N. Formula (*) is correct way to calculate conditional probability of winning prize, and shows that messages send by Monty has value. If Monty sent maximum information (i.e. maximally reduces uncertainty of Adam m=N-2), then the probability of winnig prize is maximal 1-1/N, i.e. (N-1)/N.

In 1975 D.L. Ferguson presented a generalization of this problem for the case of n boxes, in which Monty Hall opens p boxes. In this situation, the probability the contestant wins when he switches boxes is (n-1)/[n(n-p-1)]. This is formula (*) I presented in my last post using m instead p. I did not know about Ferguson result. If Monty distribute his choices over p the probability the contestant wins is expected value of this conditional probability over Hall sujective probablity inclination to give m messages to contestant.

Thanks Jan for generalizing my solution for N doors with m or p doors revealed by Monty. I really enjoyed this discussion, thanks for that.

Nothing new in this !!

If you have solved it before then I guess so. However, this was one of my favorite problems about probability while growing up – simple and counter-intuitive at the same time. Had to retell this story.

And now we all know you’re very clever.

Interesting. Give this logic, in a three point selection case, this implies that one should always switch from your initial selection, to the other selection. In all cases. And thus implying that your initial selection is alway the one with the lower probability. So the Bayes math is indicating that the first selection is always the lower probability? Math aside, this is highly doubtful.

Yes it may seem doubtful but trust me it is not. Let me try to present a non-mathematical perspective to solve this problem. I will divide the problem/solution into two parts:

Part 1:All doors are equally likely to have the car behind them, so there is nothing to choose (you chose door A).Part 2:In this part after you have picked a door (Door A), Monty Hall, who knows about the location of the car, is giving you a clue about the location of the car. In another word he is informing you about the doors that were not picked (i.e. door B & C in our case) by opening one of them. Since you have blocked door A, he is not providing any information about door A (remember he is not allowed to open door A). You have more information about Door B & C but no additional information about Door A . This asymmetry of information is causing all the confusion.Does this make sense?

Part 1. Players are independent. And they are indifferent to the presented options A, B, C. From these assumptions we have uniform joint probability distribution. The probability that Player pick the door with prize is the sum of probabilities of elementary events:

AA + BB+ CC, i.e. (1/3×1/3) + (1/3×1/3)+(1/3×1/3)=1/3. In this case, x, the probability of win prize is 1, becouse the prize is behind the door. It is conditional probability P (sucess/×)=1. P(×)=1/3. Of course the total probability of win in the game is P (sucess/x)×P (×)=1×1/3=1/3. But the probability that the Player pick incorect door is 1- P (×)=P(y), i.e 2/3. So, probability of win under y is P (sucess/y)=1. But you may win the prize if you change your first selection. If you do not have any new information the total probability of win is P (sucess/y)×P (y)×1/2=1/3. You have 2 additional options and no information, so you must divide your expectation by 1/2.

Part 2. But if you receive additional information from Monty and do not ignore it the probability of win is P (sucess/y)×P (y)×1=2/3. Now you know from Monty that all conditional probability of win P (sucess/y) is behind only one door.

There is absolutly no probability puzzle or paradox in Monty Hall problem if you think on joint probability distribution! Independency and indifference in Part 1 constitute joint distribution by multiplication principle, and we have two situations x and y. The essence of the game is an asymtetry of information. Monty first decision is private (ony he knows where the prize is). The Player first decision is public. In the Part 2 Monty decision depends on public and his private infirmation. If Player in Part 2 changes first selection he utilise Monty private information embodied in Monty message (behavior). The problem is that in Part 1 Monty and Player decisions are independent. And in Part 2 are dependent. This “independence ilussion” is the core problem of the Monty paradox.

Totally makes sense!

From a pure logic standpoint, it Does make sense for the contestant to always switch! Initially, with three points to select from, the initial choice has a 1/3 chance of selection. So if the contestant stays with their initial choice (A in example), it will match 1/3 of the time, on average.

The other two choices hold a combined 2/3 chance of selection. (B, C in example) If one of the two remaining options is removed or otherwise identified as a bad option, that 2/3 chance still remains for the two combined items – again, B & C in example.

Since the contestant has been told to avoid B, that it does not have the prize, C now has on average a 2/3 chance of having the prize.

An excellent article.. Well explained.. Will recommend to others

Maybe I am the dumbest person in the room, but I don’t quite get it. P(Open B | Car @ C) = 1 holds true only if there’s an implicit assumption that there’s a goat @A. Monty cannot open the door at A (maybe the car is actually at A). If the above is true, P(Open B | Car @ C) = 0 and not 1. So assuming it to be 1, and not 1/2 – isn’t that a false assumption? If both door B and door C had goats, Monty can choose either one – so not sure why is the conditional probability 1??? Is there some reference material I can read, because even after solving this multiple times, I would take the conditional probability to be 1/2 and not 1.

Monty Hall problem refuses to die down! Kisalay, don’t be so critical about yourself, this problem has an extremely counterintuitive solution. Let me try to help you out:

P(open B| car@C) means probability of opening door B when the event car behind door C has already happened. Hence, with the assumed case that car is behind door C, we know that door A will certainly have a goat behind it. In this event Monty has no choice but to open door B. Hence, P(open B| car@C)=1.

Additionally, when we solved this problem (look at the first branch of the tree diagram above), we have taken all three cases i.e car@A, car@B, and car@C (sum of these 3 probabilities is 1). This way we have created a complete set of possibilities, and based on this complete set we are evaluating probabilities for Monty’s actions. Hope this helped. Let me know if you still have questions.

Well… Thanks a lot for the wonderful article.

My 2 cents…

The probability of selecting a particular door is definitely 1/3. However, you initially know that one of the wrong door would be filtered out.So the probability of finding the car remains 1/2 only.

so there are 2 different things now…

Probability of selecting a particular door : 1/3

Probability of finding the car: 1/2

Thank you so much …

Here’s the first Monty Hall simulator where you can change the rules to better understand your odds:

Play a few rounds with 20 doors. Have Monty open 18 of them. The simulator lets you play up to 10,000 rounds at a time.

Oops clarification – the

advanced Monty Hall simulatoris at http://igotmontyhallproblems.usHello Roopam,

I found out one interesting video on this https://www.youtube.com/watch?v=rZn-WADQpcc

Thanks