Probability in action: Could Monty Hall have made more money on the show?
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%
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)
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%
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.