The Monty Hall Problem
You are a contestant on a game show. The host shows you 3 doors. He tells you that the prize behind one door is $1,000,000 and behind each of the other doors is a goat. He instructs you to choose a door; you will win whatever is behind it. You choose a door. “But wait,” he says. “Let me show you what’s behind one of the doors you didn’t choose.” He opens one of those doors to reveal a goat. “Would you like to switch your choice, or stick with what you have?” What should you do, stick or switch?
I put forth this problem on December 6, a month after the end of our Middle School Math Circle on Paradoxes in Probability, to make up our Hurricane Sandy class. Immediately, questions, conjectures, and assumptions started flying around the room:
- Can we assume that Monty Hall knows where the money is?
- Does he always reveal a goat?
- Should we assume that the money is more desirable than the goat?
- Are there situations in which the goat would be more desirable?
- (conjecture) Monty may be trying to trick the contestants.
- Using intuition might answer the question.
- There may be patterns in which door contains the money.
At this point, A mentioned that he had seen this problem addressed on the TV show Mythbusters. He said that the hosts ran experimental trials and determined that one particular strategy paid off more in the long run. I just happened to have with me some paper cups, paper goats, and paper million-dollar bills, so our group decided to run trials and collect data. The kids paired off, put the prizes under the cups, and played “Let’s Make a Deal.” C suggested that each pair adopt a different strategy (stick or switch) and record wins by strategy. After 10 minutes of this, the data seemed to clearly suggest that always switching resulted in more wins. “That’s all and good,” I whined, “but this is a Math Circle. Are the Mythbusters mathematicians or scientists?” The kids agreed that the Mythbusters are scientists and that in replicating their work, we weren’t exactly doing math. So we moved to the other side of the room to undertake a more mathematical approach.
We attempted to apply a bit of basic probability theory to the problem. It seems that each strategy should yield a 50% chance of success, right? So why do empirical results suggest an unequal chance? This problem, known as The Monty Hall Problem, is considered a paradox in probability. Our group discussed and questioned until the conversation found its way to randomness. The kids all realized that the choice of door is not really random since the host knows what’s behind each door and opens one according to his knowledge. This must be the reason for the paradox, but how does that work out mathematically? No one was sure. Discussion continued. Finally, A, who is always listening and thinking but does not speak up often, raised her hand high and said “I get it!” Everyone looked at her in surprise and excitement. She explained her reasoning. She had synthesized everyone else’s ideas into a solution, but not everyone understood. Once she explained it again, everyone understood. I’m not going to explain her reasoning here because the students left class excited to pose this question at home. If you haven’t already, give your kids a chance to pose the question to you. Since it’s been a while since this Math Circle took place, you may need additional explanatory help. You can find a concise clear explanation in Jim Tanton’s book Solve This, and a clear but lengthier explanation in his youtube video.
This Math Circle has spent six weeks discussing paradoxes and misrepresentations in probability and statistics. I think these students would find a recent study, seemingly on music and age, interesting. Researchers Simmons, Nelson, and Simonsohn demonstrated how easy it is to use statistics to support false hypotheses. The Penn Gazette explains in layperson’s terms how the researchers demonstrated that listening to The Beatles can make a person younger. Also, the older students in this group will probably enjoy our spring Math Circle for teens on Proofs. In it, we’ll continue this year’s Math Circle theme of how one can know for sure that something is true or false. It’s been a great six weeks! Wishing you all a Happy New Year.
Monty Hall paradox – why is it not 50:50 for two doors?
I understand why nearly everyone has this problem, even after the real probabilities are explained. I had this trouble, because simple maths says 1 prize behind 2 doors gives a 1/2 probability for each door, because the probabilities add up to 1. But this is only true if we assume the probabilities are equal for each door, to begin with. If we can show the probabilities are not equal, then the simple maths is not valid. Obviously, at the start of a horse race, 10 horses running does not give each horse a 1/10 chance of winning. Each horse has different probabilities. The maths involves adding the horses probabilities in such a way that the total =1.
This is the simplest Monty Hall explanation I know. If the contestant chooses door 1, he has a 1/3 probability of winning. And there is a 2/3 probability of the prize being behind one of doors 2 or 3. It doesn’t matter which door is left, as long as the loosing door is revealed. And the door left has the 2/3 probability of winning, double that of door 1. In fact, it doesn’t even matter if the loosing door was chosen by luck at random, or was chosen deliberately. A bit of prior knowledge on what has happened can improve your chances.
We need to challenge assumptions all the time. I’m looking for them all the time, as paradoxes often arise due to invalid ones. Hope I’ve shed some more light on teaching this problem. It’s easy to prove the result. But its dealing with the invalid assumptions that lead to a full understanding.
One thing that’s really interesting about the Monty Problem is that the question isn’t exactly “What’s the probability that the money is behind a certain door?” but instead “What is the better course of action after the host opens one door: keeping your original choice (“sticking”) or switching?” To quote mathematician Sandy Lemberg, a common issue with this problem is “a tacit and unmentioned transition from the single case presented in the problem to a series of cases.” This tacit transition makes the solution appear to be a paradox. Another issue with this problem that I did not mention in my blog posts is that it is technically irrelevant whether Monty Hall knows what’s behind the door. All that matters is which door he opens. Who cares why he opens it? (Thanks to Sandy for reminding me of these details.)
Hi Rodi et al,
I agree that how the second door was opened is irrelevant – and actually did mention it in my earlier post. Where I appear to differ, is that I think the paradox is fundamentally due to people not realising that the simple maths of 1 prize between 2 doors giving equal probabilities is invalid in this case. There would be no paradox if people accepted the door probabilities may be different in the first place. So proving it so would cause no surprise, and the best course of action is to go for the highest probability – which in this case is to swap.
Of course, as you referred to, every situation or case will have it’s own set of probabilities and best course of action. That’s why I like my horse racing analogy – every horse has it’s own probability of winning, and every race has different odds. I’ll bet bookies will have no trouble with the Monty Hall problem. They are doing these calculations every day. Thanks for the blogging opportunity on a problem I love. Best wishes, Brian
Brian you say “it doesn’t even matter if the loosing door was chosen by luck at random, or was chosen deliberately”. You do surely realise that this statement is totally incorrect.
Rodi compounds the error by agreeing ” it is technically irrelevant whether Monty Hall knows what’s behind the door. All that matters is which door he opens. Who cares why he opens it?”
Dear me, and (at least ) one of you is a maths teacher.
I’ll leave it to the two of you to work out the difference between Monty opening a losing at random and opening a losing deliberately.
To clarify (I hope)… The idea that Monty might not have to know what’s behind the door is not logically equivalent to the idea that he is opening at random. What matters mathematically in the way I presented the problem to my group (but maybe did not explain so clearly in the blog about it) is that we stated the assumption that the door that gets opened is always a door with a goat and not the money. It is entirely possible (of course, not likely) that Monty doesn’t actually know what’s behind each door but has an assistant off-screen holding up a sign just telling him which door to open. So his hypothetical lack of knowledge doesn’t definitely imply randomness. Certainly the answer to the question would be different were Monty to open the door at random. The answer to that, in fact, is the question I sent my group home with that day. Thanks for commenting on this and getting me to reflect further. Nothing like some mathematical debate to shake loose assumptions, etc.
BTW, my background sources for the treatment of this problem in my Math Circle were Jim Tanton’s Solve This and Ellen and Michael Kaplan’s Chances Are: Adventures in Probability.
Well the statement ““it doesn’t even matter if the loosing door was chosen by luck at random, or was chosen deliberately” is pretty clear and unambiguous, and is patently incorrect.
Rodi, you appeared to agree with this statement of Brian’s in your earlier post: you certainly didn’t correct it, hence my earlier comment.
Yes, Palmer, you are right that I didn’t correct it earlier. That’s the problem with doing so many of these blog posts after my bedtime! It’s time to find time during the day for this.
Monty Hall Problem – why is it a paradox at all?
Apologies for duplicating previous comments. This additional post is to give you some maths that helps in understanding why the paradox should not exist. For the moment, let’s take it as proven that swapping doubles the odds of winning – and deal instead with why people reject it. This is the real issue and has caused serious argument, myself included.
The paradox is fundamentally due to people assuming that 1 prize between 2 doors must give probabilities of 1/2 per door. But this is only true if we assume the probabilities are equal for each door, in the first place. If we can show that the probabilities are not equal, then the assumptions are not valid. Obviously, at the start of a horse race, 10 horses running does not give each horse a 1/10 chance of winning. Each horse has different probabilities. The maths involves adding the horses probabilities in such a way that the total =1.
This is the simplest Monty Hall explanation I know. If the contestant chooses door 1, he has a 1/3 probability of winning. And there is a 2/3 probability of the prize being behind one of doors 2 or 3. It doesn’t matter which door is left, as long as a loosing door is revealed. And the door left has the 2/3 probability of winning, because the sum for both doors is 2/3 and the loosing door is zero. In fact, it doesn’t even matter if the loosing door was chosen randomly or deliberately. All that matters is the result that the 2 doors have different probabilities, and one is twice the other. This is all that is needed to calculate the actual probabilities, as shown below.
Assign a probability p to one door and estimate the other. Sum the probabilities to 1, and solve for the value of p.
Door 1 Door 2
Relative prob p 2p
The sum is 3p, which =1 and means that p = 1/3
Absolute prob 1/3 2/3 ,
as expected, and no third door is involved in this calculation.
But if you think that’s invalid because it looks like I’m just verifying what I already knew, consider a more complex problem. For example, a 3 horse race with knowledge of their wins for their last 5 races.
Horse 1 Horse 2 Horse 3
Wins 2/5 3/5 5/5
Relative prob 0.4p 0.6p 1.0p
The sum is 2p, which = 1. Therefore p = 0.5
Absolute Prob 0.2 0.3 0.5
Betting odds 5:1 10:3 2:1
So there you are – the paradox should not exist at all! People only need to realize that the probabilities of the last two doors don’t have to be equal. And the horse racing odds calculation shows how simple it is to calculate the probabilities. If the bookies probabilities don’t add up to 1, the difference is their profit. I’ll bet they don’t have a problem with Monty Hall.
Have fun, Brian Farley
Hi Rodi and Palmer,
I actually agree with Palmer, who said in the previous post that the host opening his door at random is a different situation. Of course, he only has a 2/3 chance of getting a losing door first time – but he must reveal a losing door to satisfy the rules of the game. So, if he got a winning door, he would have to chose again to get a loosing door. But the audience mustn’t be present to see this – not very practical.
So, as far as the audience is concerned, the host has to know in advance which door to choose. How he finds this out doesn’t matter, and that’s really what I meant. I’m beginning to understand why Rodi said this problem is really just about taking the best course of action – swap or not swap. Life is so much easier if we don’t try to calculate the probabilities at every turn of events.
Best wishes, Brian Farley, retired electronic engineer
It might be helpful to take this game to an extreme: 100 doors, only one of which contains a prize. You place your hand on doorknob, about to open it, and Monty yells “Wait!” and opens 98 losing doors. He now asks: “Do you want to switch to the one remaining closed door?”
1. If you decided it was reasonable to assume that Monty knew beforehand where the prize was (after all, they would ruin the game if he opened the winning door by mistake or by chance) what should you do?
2. But what if your overhead the stagehand whisper “Phew! That was lucky! Monty didn’t know where the prize was and we sure lucked out he happened to open 98 losing doors!” what has changed?
OR take the game another way: 100 doors, one prize and 99 losing doors. But Monty opens to reveal only ONE losing door. You have the option to stick or switch to one of the remaining 98 closed doors.
2. What if you decide it is reasonable to assume Monty knows where the prize is and he opened a door with this knowledge in mind? Better to switch?
3. What if you decide this show seems to be run so sloppily that you doubt they even thought beforehand to tell Monty where the prize is. He’s just guessing too!
4. What if you overhear that he does know where the prize but hates to walk, so he’ll always open the losing door that is closest to the left side of the stage where he stands? (Suppose he opens door 2: What does that tell you?)
How do these two 100-door scenarios affect your thinking about the 3-door scenario?
[By the way: When you acted this out with the kids, I assume part of the instruction to the child playing host was to know which is the winning door beforehand and be sure to reveal a losing one. What if you repeated the experiment with the kids, have the host not know where the prize is, and just reject all those runs where the host happens to reveal the prize? Will your class now see a difference to switching and sticking to all the examples that were not rejected?]
1. Switch for a 99% chance (assuming the host knows) of winning
2. 50/50 No benefit in switching
2. It’s better to switch marginally – 1.01% chance of winning
3 No benefit in switching’
4. If he opens Door 3 then it’s 100% chance of being in Door 2
Hi James, I don’t think it matters whether Monty avoids the winning door deliberately or by luck. What matters is whether he opens the losing doors randomly or with a strategy in mind. Lets assume randomly. So,
1. Switch for 99% chance of winning
2. No change, still 99%
But if Monty only opens one door, then
2. Switch for 99/98 = 1.01% chance
3. Still 1.01%, as long as Monty avoids the winning door.
4. I think it’s still a 1.01% chance for switching. Because – If Monty chooses door 2 he is not revealing any information about where the prize may be, apart from the fact that it’s not door 2.
But to see it from the show organisers point of view, lets consider the probabilities for an infinite number of games. Switching doubles the overall chance of contestants winning to 2.02%, because there are now two ways of them winning. They have a 1.01% chance of choosing the prize themselves, plus Monty indicating that the winning door is 2 if he opens 3.
You ask “How has my thinking about the 3 door game been affected?” Well – if Monty applied strategy 4 above, I initially thought the chances would be 50% for either sticking or switching, . (See my separate post). But I now think I’ve fallen into the original trap of assuming that two remaining options means equal probabilities – most likely not true. Still thinking about it.
If Monty opens doors randomly then there is no benefit in switching, whether it’s 3 doors or 100 doors, because the possibility exists (even if it doesn’t occur) that Monty could open the winning door
I said that what matters is that he only chooses between the losing doors randomly, not all of them. Surely, what restrictions are placed on which doors he can choose from, are what affect the probabilities.
Have a good weekend, Brian
Sorry misread you post. yesi f he doesn’t open losing doors randomly then you need to know what his ‘bias’ is
I’ll be surprised if there was any difference between luck or prior knowledge in the host choosing the losing doors. And obviously, the odds change every time a losing door is revealed – just as in a horse race when a horse doesn’t leave the starting gate. We don’t know which horse is going to fail to run in advance of the race – only that it does when it happens.
I’m assuming that switching improves chances, purely because the contestants original door stays at it’s original probability. And the other doors have improving probabilities as we reduce the numbers of them, however many we start with. But if the host is restricted from choosing as many as 4 doors out of 10, then all those 4 doors have a fixed probability – a different situation. I suppose we just have to know the restrictions/rules placed on the host or game.
I like your the idea of a trial where the host choses randomly. Then comparing the results before and after eliminating the hosts disallowed choice of a winning door. Of course, he is also disallowed from choosing the contestant’s door! A while ago, I thought of manually compiling a table of theoretical results on just this, but it would be easier just running a computer simulation.
By the way, Rodi is the teacher with the kids. I’m just a retired research engineer at home in England. It’s nice to have a quality blog. Best wishes all, Brian Farley.
Your situation (4) is very interesting. Let’s assume contestant chooses door 1. If the host chooses door 3, then the prize is 100% behind door 2. But if he chooses door 2, there are only 2 equal options left: the prize is either behind door 1 or 3. So the probabilities are 50% for doors 1 or 3. So switching doesn’t make any difference to the contestant’s chance of winning. I like it ! But why does the contestant’s door not remain at 1/3 probability? Well, it’s due to the knowledge that it definitely doesn’t have the prize when the host chooses door 3.
If I can, I like using logic to solve these puzzles. It seems easier than making tables of all the possible outcomes.
Best wishes again, Brian Farley
Whoops! James, I assumed that scenario 4 was for the 3 door game show and not the 100 doors you considered. But it’s still an interesting problem.
There is no advantage in the switch
(Probability calculation by using the coins):
After the host throws open the empty door (let’s say C), which he always knows beforehand as empty (due to the placing of the prize behind a certain door and the player choosing a door for himself), the choice involving two remaining doors (A or B) for the prize and by the player in terms of two sides of coin can be considered as the head H (corresponding to door A) and the tail T (corresponding to door B). These choices are mutually exclusive in terms of H and T, implying that the player or the prize can’t be both at door A and door B.
Consider also that the prize and the player are represented now by two coins, coin 1 and coin 2, respectively.
We can then represent all the possible choices involving prize, player and the two remaining doors A and B (after the host has opened the empty door with the prior knowledge about the door being empty) in terms of coins 1 and 2 and their heads and tails.
From the above, prize at door A will be designated as 1H (head in the case of coin 1) and the player at door A as 2H. Similarly, prize at door B will be 1T and the player at door B as 2T.
The conditions for switch to work (leading to a win), i.e. either the prize is at door A (1H) and the player at door B (2T), or the prize at door B (1T) and the player at door A (2H): which in terms of probability, the probability for the switch to work,
P(s) = P(1H)*P(2T) + P(1T)*P(2H) ………….. (Eq. 1)
Similarly, the conditions for the switch to not work (i.e. failing to win), either the prize is at door A (1H) and the player at door A (2H), or the prize at door B (1T) and the player at door B (2T), which in terms of probability for switch not producing a win,
P(f) = P(1H)*P(2H) + P(1T)*P(2T) ……………(Eq. 2).
Since the probability for head or tail for any coin is 1/2, i.e. P(1H) = P(1T) = P(2H) = P(2T) = 1/2.
The probability therefore for the switch leading to a prize (from Eq. 1), P(s) = 1/2;
And the probability switch leading to no win (using Eq. 2), P(f) = 1/2
In other words, there is no advantage in the player making a switch.
What you are basically saying in your argument is :
1) Let the probability that the prize is behind Door A = the probability that a coin lands heads, and the probability that the prize is behind Door B = the probability that a coin lands tails
2) The probability that a coin lands heads = probability that a coin lands tails = 1/2
3) Therefore the probability that the prize is behind Door A = 1/2, and the probability that the prize is behind Door B = 1/2
This a common logical fallacy known as ‘Circular Reasoning’. There is no basis for the initial proposition you make in statement 1 and consequently the conclusion you draw in statement 3 is invalid.