# Probability in action: Could Monty Hall have made more money on the show?

Tavish Srivastava 18 Apr, 2015

Monty Hall could have lost 66.7% of times in the show, if contestant consistently took the best strategy. Could he reduce these losses and have made more money for the show?

Apparently, it boils down to how good he was with probablities!

Monty Halparin, also known as Monty Hall, hosted a game show in 1960’s called “Lets make a deal”. I have no doubt that the show minted billions of dollars over the years. The game has two levels – In the first level the maximum probability you can win is 33% which becomes 66% in the second round. Just imagine, that if audience knew the best strategy to approach this problem, host will lose 66% of times.

In this article, we will not touch up on the derivation of the Monty Hall’s problem as this is something thousands of people have studied and published. This article will serve as food for thought of digging even deeper into the same problem.

What happens if we add a door (making a total of 4 doors) and open 2 of them? This certainly makes the audience believe that their chances of winning increases further making the show even more interesting. But does the host lose more than 66% of times, if audience takes up the best strategy consistently. If not, probably the director should add a door in their show making it more interesting, making the deals longer and losing less number of times even against the best strategies.

[stextbox id=”section”] Actual Monty Hall’s problem :  [/stextbox]

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car (Ferrari, let’s say); behind the others – goats. You pick a door, say 1, and the host, who knows what’s behind the doors, opens another door, say 3, which has a goat. He then says to you, “Do you want to pick door 2?” Is it to your advantage to switch your choice?

[stextbox id=”section”] Simplistic solution  : [/stextbox]

Following are the events :

[stextbox id=”grey”]

A : door 1 has car                         B : door 2 has car                          C : door 3 has car

Co : Event that host opens door 3 and shows a goat

P(Co/C) = 0  (as host will never show the door with a car)

P(Co/A) = 0.5 ( as the contestant choose a door with car, host can show any of the doors 1 or 2 )

P(Co/B) = 1 ( host is left with no choice but open door 3)

P(A) = P(B) = P(C) =1/3 (all the events are equally likely)

Using Bayes theorem

P(Co) = 1/3*(0+1+0.5) = 0.5

P(A/Co) = (0.5*1/3) /0.5 = 1/3 = 33.34%

P(B/Co) = (1*1/3) / 0.5 = 2/3 = 66.67%

`[/stextbox]`

Hence, the contestant should switch the door raise his chances of winning from 33.34% to 66.67%. But notice, if contestant applies the best strategy the host would have lost 66.67% of times.

[stextbox id=”section”] Modified Monty Hall’s problem Part 1 :[/stextbox]

You’re given the choice of four doors: Behind one door is a car; behind the others, goats. You pick a door, say 1, and the host, who knows what’s behind the doors, opens another door, say 2, which has a goat. He then says to you, “Do you want to pick door 3 or 4?” What is your best strategy (lets say this choice 1)?

[stextbox id=”section”] Finding the best strategy to choice 1:[/stextbox]

Following are the events :

[stextbox id=”grey”]

A : door A has car                         B : door B has car                          C : door C has car               D : door D has car

Bo : Event that host opens door B and shows a goat

P(Bo/C) = P(Bo/D) = 1/2  (as host can show any of the door which does not have the car)

P(Bo/A) = 1/3 ( as the contestant choose a door with car, host can show any of the doors 1, 2 or 3 )

P(Bo/B) = 0 ( host cannot show the door with car behind it)

P(A) = P(B) = P(C) = P(D) = 1/4 (all the events are equally likely)

Using Bayes theorem

P(Bo) = 1/4*(0+0.5+0.5 + 0.333) = 1/3

P(A/Bo) = P(Bo/A)*P(A) / P(Bo) = (1/3 * 1/4) / (1/3) = 1/4 = 25%

P(C/Bo) = P(D/Bo) = 1/2(1-1/4) = 3/8 = 37.5% (By symmetry the two probabilities will be equal)

`[/stextbox]`

Hence, the best strategy as of now will be to shift to either door C or door D.

[stextbox id=”section”] Modified Monty Hall’s problem Part 2 :[/stextbox]

The host further opens up door D and shows a goat again. You are again asked if you want to make a shift back to door A or choose to stay at door C.

Finding the best strategy to choice 1:

[stextbox id=”grey”]

Do : Event that host opens door D and shows a goat

P(Do/  A  /Bo) = Probability of host opening door D given that B is already open and A has the car

= 1 (Host has no other choice)

P(Do/ C /Bo) = 0.5 (Host has a choice between A and D)

P(Do/ D /Bo) = 0

P(Do/Bo) = P(Do/ A / Bo)*P(A/Bo) + P(Do/C/Bo)*P(C/Bo) =1*1/4 + 3/8*1/2 = 7/16

P(C/Do/Bo) = (3/16) / (7/16) = 3/7 = 42.8%

P(A/Do/Bo) = 57.2%

`[/stextbox]`

Again, we see that the chances of winning a car is significantly increased if we make a switch from door C to door A.  Note that we still have not explored the best strategy if host would have opened door A. If you do similar type of calculations in this revised scenario, you will see that the best strategy exist if you switch to door D. The chances of winning in this scenario becomes 66.67%.

On comparison of these two scenarios to original Monty Hall problem we see that the maximum of the probability of host losing in 4 door scenario is equal to 3 door scenario.Let’s think of the scenario in which the host is forced to open door A (which is a loss making proposal). The only case when host is bound to open door A is when participant originally chooses door A.

P(anchor losing in 4 door-2 open) = 0.25 * 66.67% + 0.75*57.2% =  59.6%

P(anchor losing in 3 door-1 open) = 66.67%

Note that we have left a scenario in which the participant sticks to his original door in the first scenario. This time the probability to win will come out higher than before but we will not touch up on this case. This is because this case assumes that the candidate is already aware that the host will open the second door after the participant chooses to stay with his first choice. In original Monty Hall’s problem the participant does not make his choice based on the knowledge that the host will open on of the other two doors. Imagine that you choose door A and host shows you a goat in door B. Now hosts asks you if you want to switch, and you choose to stay with the door A. Next moment, the host says “Looks like you really love door A, so let’s show you what is behind door A.” You obviously have only 25% of probability to win in this case.

[stextbox id=”section”] Summary of findings :[/stextbox]

We see that the probability of anchor losing is much lower in a 4 door scenario as compared to 3 door scenario. We have no intention to give recommendation to this legacy game show but will want to spark a discussion on the following : “If 4 door – 2 open Monty hall scenario gives a lesser probability of anchor losing compared to 3 door – 1 open scenario, certainly will excite the audience even more than the traditional Monty hall as it intensifies the illusion of winning on the wrong door and obviously consumes more time per contestant, then why did the show director restricted themselves to only 3 doors?” If the answer lies outside the world of probability, then its not worth initiating a debate on the same but if you have any inputs within the scope, do let us know your thoughts in the box below.

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Tavish Srivastava 18 Apr, 2015

Tavish Srivastava, co-founder and Chief Strategy Officer of Analytics Vidhya, is an IIT Madras graduate and a passionate data-science professional with 8+ years of diverse experience in markets including the US, India and Singapore, domains including Digital Acquisitions, Customer Servicing and Customer Management, and industry including Retail Banking, Credit Cards and Insurance. He is fascinated by the idea of artificial intelligence inspired by human intelligence and enjoys every discussion, theory or even movie related to this idea.

Jaap Karman 08 Apr, 2014

all depends on the related conditions. In the original 3-door problem this condition is: - the question is always done, the presentator has no personal interest of the result. Imagine the question would have been in 33,3% of the cases knowing the major price is behind that door. In the revised 4-door problem you did not mention the conditions. - Are the 2 questions always done, the presentator has no personal interest of the result. You can stick the first round and than switch. -> 75% is this allowed? You could switch the first round and than... Two options opening the original door or one of the first tree others. (not worked) out, interesting thoughts. The original 3 door is already having that much discussion as being counter intuitive. For sure the best action would switch at some moment.

Tavish 08 Apr, 2014

Jaap, Here you do have an interesting thought and this is something I wanted to cover in the article in detail as well(however this is left as a note in the main article). But this would have made the article long and complicated. But because you brought this up, there are two scenarios in the 4 door puzzle. 1. You switch for the first time because this seems as that best solution at this point, and switch again in the second time : Probability to win is 59.6%. 2. You stick with the same door for the first time : Probability becomes higher in this case but in this case we are assuming that the participant knows that the host will show 2 doors with a goat after my first choice. Even in the original Monty Hall's the participant does not make his first choice based on the act that the host will show you a door after our choice. When you are given a choice for the first time, you should switch to increase your probability to win. An interesting thing you will notice here is that the first solution is a mixture of two local maxima which actually does not sum up to global maxima. Or simply put, solution 1 seems to be the best solutions in two different instances but 2 leads the participant to higher chances of winning. But still we will not consider case 2 as it assumes that the host will open one more door after your choice. Appreciate you bringing this up. Hope you enjoyed this exercise. Tavish

Fredrik 10 Apr, 2014

Tavish, I am not sure if your calculations are correct. If I understand correctly, the contender has four choices: 1. Stick with the door chosen initially, 2. Switch door after first opening and then stick with that door, 3. Stick with the door after first opening and then switch door, 4. Switch door after both openings. In case 1 the probability of a win is 0.25. (You win if you choose the car inititally). In case 2 you win if you choose a goat initially and then switch to the car, so the probability of a win is 0.75*0.5=0.375. In case 3 you win if you choose a goat initially, because when the second door opens, the car must be behind the door which was not chosen at first, so the probability of a win is 0.75. In case 4 you win if you choose the car initially, since after the first opening you switch to a goat and after the second opening you must switch back to the car. So if you chose the car initially you win with probability 1. If you choose a goat initially you win if you choose a goat after the first opening, because in that case you have to switch to the car after the second opening. So, if you first choose a goat, the probability of switching to another goat after the first door is opened is 0.5. Altogether, you win the car with probability 0.25*1+0.75*0.5=0.625. The best strategy is number 3 and it give you a higher probability to win than in the 3-door game. That might be the reason why they only had 3 doors in the show.

Tavish 10 Apr, 2014

Fredrick, You are right in saying that we have 4 scenarios to consider. However following are my take on each of the scenarios : Case 1 : You are right. The probability of winning is 25% indeed. Case 2 : You have taken the two events as independent of each other and multiplied the two directly. In this case the two events are not independent. The probability of winning (based on the calculation done in the article) in this case will be 0.25*33.3% + 0.75*42.8% = 40.43%. Case 3: You have made an error making this case. If you already know that you switched to a car, then your probability to win is 1 and not 0.75 . But there is no way you will know that you switched to a car. You have stated case 2 well in the initital 4 pointers "Switch door after first opening and then stick with that door". Let us know in case I have misinterpretted your case 3. Hope this helps. Tavish

Fredrik 10 Apr, 2014

I am sorry, but I do not think you got mine case 2 and 3 right: When you do your first pick you either choose a door with a goat behind it or a car (but you don't know which). You choose a goat with probability 0.75 and a car with probability 0.25. Let us consider case 2: If there is a car behind the first door you choose you will lose with probability 1 since you will switch door one time and there is only one car (and you decided to only make a switch after the first door is opened). On the other hand, if there is a goat behind the first door you choose, then there will be two doors to which you can switch, one with a car behind it and one with a goat. Given that their were a goat behind our first door, we win with probability 0.5. So we have the probability of wininng: P(win) = P( win | "car behind first door") * P("car behind first door") + P(win | "goat behind first door") * P("goat behind first door") = 0*0.25+0.5*0.75 = 0.375 Now to case 3: You choose to keep your first pick after the first door is opened but switch after the second door is opened. This is the same scenario as the following: We have 4 doors and after your first pick two doors with goats behind them are opened. If you initially picked the door with a car behind it you will loose (because you switch) BUT if you picked a door with a goat behind it you will win (because the the doors with the other two goats are opened). So the probability of winning: P(win) = P( win | "car behind first door") * P("car behind first door") + P(win | "goat behind first door") * P("goat behind first door") = 0 * 0.25 + 1 * 0.75 = 0.75 I do not assume that you know what is behind the closed doors, (if I did the game would not be that interesting). Best/Fredrik

Jaap Karman 10 Apr, 2014

Agree with Fredrik the best strategy case 3 should deliver 0.75 probability It is as counter-intuitive as the original case.

Tavish Srivastava 10 Apr, 2014

Fredrik, Let's consider your case 2 first. Say you choose door A and the host opens up door B. You have two options, either to switch to C or D. You make a shift to C (both C and D will have equal chances of winning). Now the host opens and shows door D (or A) . In your solution you have evaluated P(win) which is P(car in C) or say P(C). In this question you have two additional information and you need to calculate P(C | D is opened | B is opened) instead of P(C). To calculate P(C | D o | Bo) you need to apply Baye's theorem : P(C|Do|Bo) = { P (Do | C | Bo) * P(C | Bo) } / { P (Do | C | Bo) * P(C | Bo) + P (Do | A | Bo) * P(A | Bo) } = 3/7 Similarly P(C | Do | Bo) comes out to be 1/3. In your solution, you assume that the P("car behind the first door") remains constant at 0.25 even after you make a switch door for the first time. This probability has changed because there are two additional facts i.e. door A is opened and door D is opened. In case 3 : I understood your case wrongly last time. You are right, the probability to win is 75% indeed. But this case assumes that the contestant knows after choosing to stay with the door for the first time that the host will open up a second door and then he will shift to the alternate door. Original Monty Hall's puzzle does not assumes the contestant makes his first choice knowing host will show him a door before he finally opens the chosen door. Hence, the best strategy available for the candidate is to switch both the time even if the global maxima exist if he stays with a door for the first time and switched the next time. Thanks for bringing this up. Will be happy to debate if you are still not convinced. Tavish

Pravin 31 Aug, 2015

Intuitive explanation what I found is 1st Scenario were there 3 doors and monty opens one- If the audience select one door(lets say A) he/she has 33% chance of winning. Here the remaining 66% of probability of winning a car is distributed among door B and C. If Monty opens door C which has a goat, we can strike door C, now door B has 66% chance of having a car whereas door A remains with same 33% chance. If I apply this logic to scenario 2 i get 75/2 which is 37.5 % chance of winning car by switching door. If I apply the same logic to scenario 3 I am getting 75% chance of winning a car by switching door? Can you correct me if I am wrong? Thank you