An Investigation into the Counterintuitive Monty Hall ProblemSunset High SchoolIB Mathematics HLMs. BertiJared ThompsonJanuary 19, 2018Jared ThompsonBerti Period 110/17/2017Monty HallRelatively easy riddles that challenge the top math professors of our society intrigue me on the highest levels. In the early 1990’s, Marilyn Vos Savant was a popular female mathematician who was often regarded as the smartest woman alive. She was the first to popularize a riddle that we now know as the Monty Hall problem, as shown on the TV show hosted by Monty Hall, “Deal or No Deal.” The public sent in their answers convinced they were right on this seemingly easy problem, and the responses were all along the same lines of reasoning which claimed that the chance of winning the prize was 50%. When Savant published her answer, the public responded with outrage.Many professors were quick to discard Marilyn’s answer because it appeared to be absolutely wrong and “Maybe women look at math problems differently than men.

” (Don Edwards, Sunriver, Oregon). To their shame, these professors didn’t take the time to logic the problem out; once they did and realized their mistake, they hurriedly retracted their letters. Even with the thousands of resources online proving this riddle, it continues to produce discussion between mathematicians of all strengths to this day.The Monty Hall riddle provides a great opportunity for observing how our intuition can often lead us astray, because the correct answer may not always be obvious nor intuitive.The basic scenario of the Monty Hall scenario goes as such:”A contestant on a game-show is given the choice of three doors: Behind one door is a car; behind the other two doors are goats. After the contestant picks a door, the host, who knows what’s behind all the doors, opens one of the unchosen doors and reveals a goat. He then says to the contestant, “Do you want to switch to the other unopened door?” Is it to the contestant’s advantage to switch doors?” (Stay or Switch, Jeremy Jones)Instead of using cars and goats, throughout my examples I will be using a smartphone as the prize and flip phones as the duds because they are a much more tangible way of representing items that high schoolers do and do not desire.Response from Ph.

D.’s’s in 1990’sThe Ph.D.’s’s looked at the riddle in two parts: before and after the host revealed a flip phone.Participant chooses Door A Host reveals flip phone in Door B In this case, the Ph.D.

‘s’s assumed that when Door B is revealed to be a flip phone, the entire problem is essentially ‘reset.’ However, this process completely ignores the initial setup of the three doors where the contestant does not know where any smartphones or flip phones are.The Ph.D.

‘s’s knew that in the initial problem, there were two flip phones and a smartphone behind the three doors. Therefore when the contestant picks Door A, the probability that a smartphone was behind their chosen door was ?, so P(Smartphone) = .3. Once Door B (look above) was revealed to have a flip phone behind it, the Ph.D.’s’s removed this option from the equation (look below) assuming that since there was no longer any chance that there was a smartphone behind the door, there was no reason to include it.

They now saw that the probability of a smartphone being behind one of the two doors was an equal split 50/50 chance, since there is a flip phone behind one door and a smartphone behind another. Thus, they assumed that the probability of winning a smartphone by staying and by switching is ½, so now P(Smartphone) = .5 or 50%The Ph.D.’s’s eliminated Door B Only two doors remaining The important concept that these professors excluded is the fact that the host knows from the start where the flip phones are, so even without choosing a door, the participant knows that the host will consequently reveal a flip phone. No matter what door the contestant chooses, the host will always reveal a flip phone. Thus, there should be no changes in probability after the host opens a door.

Marilyn vos SavantSavant was the first person to popularize the Monty Hall problem, but not the first to propose or even solve the riddle. Once publicly published, it became popular because so many professors were quick to go with the Ph.D.’s problem solving technique, which was the intuitive way to solve the problem, so when Savant revealed her answer, there was public outrage. I will detail Ms. Savant’s technique in solving this riddle which plays on counter-intuitive thinking but relies on fundamental probability.Let’s start with the initial problem again. This time, instead of splitting the problem into two parts, let’s keep them together and therefore look at the problem holistically.

Participant chooses Door A Host reveals flip phone in Door B The Monty Hall problem can primarily be solved with the use of Bayes’ Theorem, which is:P(Win | Flip Phone) =P(Flip Phone | Win) ? P(Win)P(Flip Phone)Where ‘Win’ is the probability the contestant wins a smartphone if they switchAnd ‘Flip Phone’ is the probability the host reveals a flip phoneThis can also be read as “The probability that the contestant will win a smartphone if they switch given the host has already revealed a flip phone behind a door is equivalent to the probability that a flip phone will be revealed by the host given the contestant won by switching, multiplied by the probability of winning a smartphone by switching, all divided by the probability of the host revealing a flip phone.” P(Flip Phone | Win) may seem unnecessary because “the probability that a flip phone will be revealed by the host given the contestant won by switching” isn’t actually a possibility, but it’s necessary for solving.The host knows where the two flip phones and smartphone are and has been instructed to open one of the two remaining, unchosen doors to reveal a flip phone, not a smartphone. This means that the probability of the host revealing a flip phone is 100%, or P(Flip Phone) = 1. Regardless of whether or not the participant ends up winning the smartphone, the host will always have revealed a flip phone before the contestant choose their final door. Thus, the probability that the host will choose a flip phone when the contestant wins is also 100%, or P(Flip Phone | Win) = 1.

In the beginning, it is clear that there is a ? chance that whatever door the contestant chooses will have the smartphone and a ? chance that it will contain a flip phone. If we were to take the three possible variables, flip phone 1, flip phone 2 and smartphone, we could only arrange them in 6 different ways (seen on page 5). In 4 out of 6 of these variations, the contestant chooses a flip phone and is able to switch to the smartphone after seeing what the host reveals.

Therefore, P(Win) = 4/6 = 0.7.Now we can insert these probabilities into Bayes’ Theorem and find the probability that the contestant will win when they switch assuming the host revealed a door with a flip phone behind it:P(Win | Flip Phone) =P(Flip Phone | Win) ? P(Win)P(Flip Phone)P(Win | Flip Phone) =1 ? 0.71P(Win | Flip Phone) =0.7There is a 67% chance that the contestant will win a smartphone if they switch doors after a flip phone has been revealed.This situation assumed that the host knew where the items were and that the contestant was allowed to switch. There are slightly different probabilities when looking at scenarios that do not follow these two rules.My Further Investigation into Monty Hall’s RiddleLocation of Smartphone is Unknown to AllUnder this assumption, the contestant must still choose a door and the host must choose another door, but this time the host is choosing a door randomly from the two remaining doors.

Referring back to Bayes’ Theorem:P(Win | Flip Phone) =P(Flip Phone | Win) ? P(Win)P(Flip Phone)Where ‘Win’ is the probability the contestant wins a smartphone if they switchAnd ‘Flip Phone’ is the probability the host reveals a flip phoneThe probability that the host will reveal a flip phone behind their chosen door is ?, because there is a ? chance that there is a smartphone under each door and thus a ? chance that it’s a flip phone, P(Flip Phone) = 0.7. This part is the same for both of the following scenarios, which depart from each other for P(Flip Phone|Win) and P(Win). Scenario 1: Contestant Can Switch to Either DoorThere are six different outcomes of this setup (keep in mind there are 2 different flip phones):Contestant picks Door A with flip phone 1, host reveals Door B with flip phone 2, Door C has smartphoneContestant picks Door A with flip phone 2, host reveals Door B with flip phone 1, Door C has smartphoneContestant picks Door A with flip phone 1, host reveals Door B with smartphone, Door C has flip phone 2Contestant picks Door A with flip phone 2, host reveals Door B with smartphone, Door C has flip phone 1Contestant picks Door A with smartphone, host reveals Door B with flip phone 1, Door C has flip phone 2Contestant picks Door A with smartphone, host reveals Door B with flip phone 2, Door C has flip phone 1The probability of winning when switching occurs in outcomes 1-4 because in 3-4 the contestant would switch to the door opened by the host and in 1-2 the contestant would switch to Door C with the smartphone behind it. Thus, the probability of winning when switching is ?, so P(Win) = 0.

7.For the last probability, P(Flip Phone | Win) represents the chance that the host chooses a flip phone when the contestant wins. Only outcomes 1-4 are potential variations where P(Flip Phone | Win) can be true because these are the outcomes in which the contestant wins. Given that the contestant wins, only 2 out of these 4 outcomes, outcomes 1-2, have the host choosing a flip phone. Therefore, the chance that the host chooses a flip phone given that the contestant won is ½, so P(Flip Phone|Win) = 0.5.These values are then put back into Bayes’ Theorem:P(Win | Flip Phone) =P(Flip Phone | Win) ? P(Win)P(Flip Phone)P(Win | Flip Phone) =0.5 ? 0.

70.7P(Win | Flip Phone) =0.5When the host does not know where the smartphone or flip phones are and the contestant is allowed to switch to any door, they have a 50% of winning the smartphone if they switch after a flip phone has been revealed.Scenario 2: Contestant cannot Switch to the Door that Host Reveals as SmartphoneThis next scenario is a bit more complicated. Assuming, like before, the host doesn’t know where the items are and chooses one of two doors randomly.

However, if they happen to open a door with the smartphone behind it, the contestant is not allowed to switch to that door.Now, referring to the same outcomes as used previously, the probability of winning when switching is only ?, where only outcomes 1-2 work because, in outcomes 3-4, the host reveals the smartphone so the contestant automatically loses. Therefore, P(Win) = 0.3.Of the two winning outcomes for this scenario, outcomes 1-2, the probability that the host chooses a flip phone given that the contestant won is 100% because in both outcomes 1-2, the host chooses a flip phone. Thus, P(Flip Phone | Win) = 1.Finally, these values are entered back into Bayes’ Theorem:P(Win | Flip Phone) =P(Flip Phone | Win) ? P(Win)P(Flip Phone)P(Win | Flip Phone) =1 ? 0.30.

7P(Win | Flip Phone) =0.5It turns out that even if the contestant is not allowed to switch to a door that was revealed by the host to be a smartphone, they will still have a 50% of winning the smartphone if they switch after a flip phone is revealed.Further ExtensionLife is not always so simple as the Monty Hall problem seems to imply; in fact, life is much more complicated than just three options. This riddle can be expanded into higher levels of thinking by increasing the number of doors and examining the relationship between the many variables.In Savant’s original presentation of the riddle, there are three doors with a prize and two duds behind them, and after the participant chooses a door the host opens another door with a dud behind it.

Looking back at the setup of Bayes’ Theorem, we can substitute in variables instead of numbers to generalize the riddle and apply it to multiple doors:P(Win | Flip Phone) =P(Flip Phone | Win) ? P(Win)P(Flip Phone)We shall assume we have n doors with a prize, a smartphone behind one door and flip phones behind the rest, or n – 1 doors. The participant chooses a door, which has a 1n probability to have a smartphone behind it.The host then reveals b doors where 0 ? b ? n – 2 because the host must leave the door you chose, and another door for you to switch to, closed. The probability that the contestant will choose a flip phone given they have won is n-1n and the probability that the door they chose has the smartphone behind it is 1n-b-1, since b doors will be opened by the host and 1 door has already been chosen by the participant. We know from the Savant riddle that the chance the host will choose a flip phone will always be one hundred percent or 1.

0, therefore:P(Win | Flip Phone) =n-1n1n-b-1 = 1nn-1n-b-1 All Doors OpenedIf all possible doors are revealed by the host where b = n – 2 then,n-1n1n-(n-2)-1 = n-1nThe deduced equation n-1n is the probability that the contestant will win if they switch after the host reveals a flip phone.Graph of 1 ? n ? 20 At Least One Door is OpenedIf at least one door is revealed by the host where b > 0, then n-1n-b-1> 1 and n-1n-b-1> 1n.Graph of 1 ? n ? 20 with varying b valuesFrom these graphs, it is clear that when all doors are opened by the host, the probability of winning if the contestant switches increases sharply then flattens out at approximately n=5. On the other side, if only a few doors are opened, as the number of doors is increased, the probability of winning if they switch decreases very rapidly.ConclusionWhen looking back at how the Ph.D.’s solved the Monty Hall riddle, it may seem like they mistook the instructions and assumed that Monty Hall does not know where the hidden items are. In reality, they merely jumped to what seemed like an obvious conclusion and ignored the complexities of the riddle.

Through my calculations, I found that there is often a higher likelihood of winning a smartphone when the contestant switches. This can be applied directly to real life because we are more likely to get something out of our lives if we go for it, take risks, and don’t always take the easy route. This problem very much reflects this idea, because most contestants, even if they knew the math, would choose to stay because switching doors felt to them like taking a risk. Risks do not always pay off, but in the end, you are more likely to be successful if you push yourself beyond your upper limits and take new leaps into unfamiliar territory.