Monty Hall Problem... For Newbies

[UKBoy1977]

I pick a door. Whichever door I pick, one of the other doors has a goat behind it. I'm confident of that. 100%. Isn't this just the one about the change and where did the other £2 go? This 'spare' 1/3 probability should surely be shared equally between the remaining doors, which are indistinguishable to the blind eye of chance.

But to give some of this 1/3 probability to your original choice of door would be to say that being shown a goat has increased the chance of your original choice being right - but you yourself have said that being shown a goat tells you NOTHING about the chance of your original door being correct, as you would be shown a goat whatever you chose.

Think of it this way:

When you choose your door (A, for example), you accept that at this stage the probabilities are 1/3 the car is in A, 2/3 that it is in one of the other 2 doors (B andC). Now, when Monty comes along and shows you a goat, it is as if he is saying the following - "IF the car is behind one of these 2 doors, it's behind THIS one" At that moment the 2/3 probability all goes into the door he didn't open. It's as if he is combining the 2 doors, and making them into one big door (the B AND C door), which contains the combined probability of them both.

 

[phunk]

I wrote a quick python script to simulate the game 10000 times. Here are the final results of three runs:

Contestant A always stuck and won 3292 out of 10000 games (32.92%)
Contestant B always switched and won 6708 out of 10000 games (67.08%)

Contestant A always stuck and won 3370 out of 10000 games (33.70%)
Contestant B always switched and won 6630 out of 10000 games (66.30%)

Contestant A always stuck and won 3446 out of 10000 games (34.46%)
Contestant B always switched and won 6554 out of 10000 games (65.54%)

 

[marting]

I wrote a quick python script to simulate the game 10000 times. Here are the final results of three runs:

Contestant A always stuck and won 3292 out of 10000 games (32.92%)
Contestant B always switched and won 6708 out of 10000 games (67.08%)

Contestant A always stuck and won 3370 out of 10000 games (33.70%)
Contestant B always switched and won 6630 out of 10000 games (66.30%)

Contestant A always stuck and won 3446 out of 10000 games (34.46%)
Contestant B always switched and won 6554 out of 10000 games (65.54%)

Running a Monte Carlo simulation is a very effective way to teach people about this. My gf, a middle school math teacher, would use this problem with a class. First people would write down what they would guess then the class would run lots of simulations. It made them think and is especially useful because it is apparently counterintuitive to most people. Usually, the next day some of the kids googled it at home and it made for a follow up conversation.

EtoA:
As an aside, the large majority of kids in the class had no idea who Monty Hall, being a rather old show they hadn't seen. Yet they still preferred not to switch and had a hard time believing the stated problem's solution. I suspect this is really an intuitive reaction to standard negotiating which says the other side is always trying to get an advantage and wouldn't have offered a chance to switch unless it was to their advantage. Never mind the controlling statement in the problem that a door with a goat would always be opened. It's a great example of how good analysis requires one to confront initial biases and expectations.

 

[Prometheus]

I think the 'solution' to this problem only makes sense from Monty's point of view, not the contestants.

There are essentially 6 possible outcomes:

Assume the car is behind door B.

1) I choose A, Monty shows me the goat behind door C. I stick with A, and I lose.
2) I choose A, Monty shows me the goat behind door C. I switch to B, and I win.
3) I choose B, Monty shows me the goat behind either A or C. I stick with B, and I win.
4) I choose B, Monty shows me the goat behind either A or C. I switch doors, and I lose.
5) I choose C, Monty shows me the goat behind door A. I stick with C, and I lose.
6) I choose C, Monty shows me the goat behind door A. I switch to B, and I win.

these boil down to:

1) stick, lose
2) switch, win
3) stick, win
4) switch, lose
5) stick, lose
6) switch, win

So, over a large number of trials, a strict policy of always switching will win two thirds of the time. Great. But this truth can only matter from the point of view of the host. Over many shows, it is obviously in Monty's interest to attempt to dissuade contestants from switching (more on that later), because the more people who switch, the more cars he'll have to give away.

But from an individual contestant's point of view there's only one trial.

Go back to the initial list of 6 possibilities. Suppose I choose door A at first.

I have now eliminated outcomes 4 thru 6 above. Since I only get to play once, I only have one shot at deciding whether to switch or not. From my point of view, the game doesn't even start until after I've made my initial choice. Monty's opening a safe door doesn't give me any new information, and it doesn't affect the reality of whether my first choice was right or wrong. In the end, I'm still left with a 50/50 choice (between outcomes 1 and 2 above--the other outcomes are eliminated) as to whether to switch or stick.

Now, back to my comment about Monty's interest in dissuading contestants from switching. My memory may be off, but IIRC Monty didn't always just open a safe door for every contestant. Sometimes, he would first offer to "buy" the the contestant's first choice: "You've chosen door A. So whatever is behind door A belongs to you. Before I open it, though, I will offer to buy whatever is behind door A for $100." If the contestant says no, he sometimes then upped his offer to $500 or $1000. Then after a contestant turned down cash he would open a safe door and offer the chance to switch.

So, sometimes, before being offered the chance to switch, a contestant had to first choose to stick to his/her first choice one or two times in the face of guaranteed cash. Most people, having made a decision, will more often choose to double down rather than reverse themselves later. Monty could even further reinforce his chances by sometimes NOT offering cash but just going directly to a safe door. By mixing it up a little, he could cause some people to assume that he only offered cash to people who chose the car on their first try, so that those who got offered cash falsely believed it was better for them to turn it down.

This was how Monty tried to ensure that a greater number of contestants would choose to stick rather than switch, so that, overall, the house won more often.

 

[Dr. Keith]

Now, back to my comment about Monty's interest in dissuading contestants from switching. My memory may be off, but IIRC Monty didn't always just open a safe door for every contestant. Sometimes, he would first offer to "buy" the the contestant's first choice: "You've chosen door A. So whatever is behind door A belongs to you. Before I open it, though, I will offer to buy whatever is behind door A for $100." If the contestant says no, he sometimes then upped his offer to $500 or $1000. Then after a contestant turned down cash he would open a safe door and offer the chance to switch.

So, sometimes, before being offered the chance to switch, a contestant had to first choose to stick to his/her first choice one or two times in the face of guaranteed cash. Most people, having made a decision, will more often choose to double down rather than reverse themselves later. Monty could even further reinforce his chances by sometimes NOT offering cash but just going directly to a safe door. By mixing it up a little, he could cause some people to assume that he only offered cash to people who chose the car on their first try, so that those who got offered cash falsely believed it was better for them to turn it down.

This was how Monty tried to ensure that a greater number of contestants would choose to stick rather than switch, so that, overall, the house won more often.

Don't worry about your memory, the problem is not an accurate description of how the show was played. It is only loosely based on Let's Make a Deal.

It is a problem made up by geeks to explore the ways that non-geeks don't really "get" probability. The funny thing was that so many geeks were thrown for a loop as well.

 

[timhau]

My young nephews struggled with the 3-door Monty Hall problem, but it seemed to make a little more sense to them when I explained the scenario with 100 doors instead:

Imagine there are 100 doors, 99 with goats and 1 with a car. You select a door at random; it could be a goat, or it could be a car. Monty Hall then opens 98 doors at random, showing 98 goats. Only two doors remain: do you switch? Hell yes you do.

The way I made sense of it was this:

Premise: Monty knows which door has a car behind it.

1. After you've made your initial pick, Monty gives you an option to choose between the door you picked and both the other doors. Do you switch? Of course you do.

2. After you've made your initial pick, Monty tells you that there's at least one goat behind the doors you didn't pick, and gives you an option to switch your single door for the two other doors. Do you switch? Of course you do, since he didn't give you any new information.

3. After you've made your initial pick, instead of telling you that there's at least one goat behind the doors you didn't pick, Monty shows you a door that has a goat behind it. Do you switch? Yup, because as per 2. above, he didn't really give you any new information here either, because you knew all along that the other two doors do have at least one goat behind them (and Monty, knowing which door the car is behind, can always open a door to show you a goat).

 

[barehl]

Indeed, as I said, I know I've had it explained in the past, got it (or took it on faith...), but still, I see the problem afresh every time. How does the act of confirming something we already knew alter the likelihood that you just picked the car?

In order for the probability to be the same, it would have to be the case that sometimes Monty would turn to you and say, "The door you picked has a goat." They open the one you picked and reveal the goat. "So, now choose one of the other two." If this were the procedure then it would have equal probability and there would be no statistical difference between keeping your first choice and changing.

So, let's look at the possibilities. There are only three:

You: goat; option1: goat; option2: car
You: goat; option1: car; option2: goat
You: car; option1: goat; option2: goat

This should be fairly clear. There are only three doors so only three possibilities. The car is behind your door or one of the other two options. Now, let's see what happens after Monty makes an elimination:

You: goat; option1: removed; option2: car
You: goat; option1: car; option2: removed
You: car; option1: goat; option2: goat--either option1 or option2 removed.

You can see by this that out of the three possibilities, two are that the car is behind the remaining option. So, you only have a 33% chance of winning if you keep your first choice.

But let's add in the possibility that Monty removes your pick.

You: goat; option1: goat; option2: car--either yours or option1 removed.
You: goat; option1: car; option2: goat--either yours or option2 removed.
You: car; option1: goat; option2: goat--either option1 or option2 removed.

Notice that with this arrangement there is no advantage in changing. All choices have a 50% chance of being right.

 

[ravdin]

I once got into a very long argument about this at work. One of my coworkers suggested that if you're still not convinced that you should switch, that he has a fun game to play and you should meet him in the alley after work with a roll of 20 dollar bills.

 

[Mark6]

I can think of three reasons people have such hard time with Monty Haul problem.

1. Conditional probability is something most people never learn. If you understand conditional probability, it is much easier.

2. A lot of people, including mathematically educated ones, tend to think of "probability of an event" as some objective quantity of said event, whereas in reality it is just a measure of one's ignorance. A tossed coin has "traditionally" 50% of landing on either side. But if you knew precisely the coin's velocity, rotation, aerodynamic properties, air density, and air currents, you would be able to tell "heads" or "tails" with 100% certainty. If you had only some of that information, you could say "80% chance of heads" (for example). As your knowledge of a situation changes, "probability of event" changes. And in Monty Haul you receive additional information when the host opens a door and shows you the goat.

3. The situation is incredibly contrived. Where else, outside a game show, you have to make a decision based on a partial information, from a knowledgeable source, who is deliberately withholding part of information? It is just not a common occurrence.

 

[Dr. Keith]

I once got into a very long argument about this at work. One of my coworkers suggested that if you're still not convinced that you should switch, that he has a fun game to play and you should meet him in the alley after work with a roll of 20 dollar bills.

Gambling is a very efficient effective profitable way to teach probability.

 

[ravdin]

By the way, it's very important that the presenter knows the correct door. If not then it really doesn't matter which door you should pick (because there is a 1/3 chance that he will accidentally reveal the prize without you having to make any decisions!)

 

[ddt]

Don't worry about your memory, the problem is not an accurate description of how the show was played. It is only loosely based on Let's Make a Deal.

It is a problem made up by geeks to explore the ways that non-geeks don't really "get" probability. The funny thing was that so many geeks were thrown for a loop as well.

Indeed. Just read the number of letters Marilyn vos Savant received that disagreed with her explanation.

 

[ddt]

3. The situation is incredibly contrived. Where else, outside a game show, you have to make a decision based on a partial information, from a knowledgeable source, who is deliberately withholding part of information? It is just not a common occurrence.

As a juror in a (criminal) lawsuit?

 

[ddt]

By the way, it's very important that the presenter knows the correct door. If not then it really doesn't matter which door you should pick (because there is a 1/3 chance that he will accidentally reveal the prize without you having to make any decisions!)

That still means that switching (but now to the revealed car door) is the best strategy. It doesn't change the probabilities. It's only bad TV.

ETA: and this is the third time I've mentioned that in this thread.

 

[IXP]

But after your original pick (indeed, even before it), we know that one of the doors you haven't picked hides a goat. Seeing the goat changes nothing. Your first pick is 1 in 3, and yes, I get that that means it is more likely that the car is behind a door you didn't pick. What I don't see is why the presence of a guaranteed goat shifts the odds on one of the remaining doors and not the other. What distinguishes the two remaining doors, that one of them should now be more likely to hide a car than before an event that must be possible and in no way comes as a surprise. One of the doors you didn't choose hides a goat - we always knew that. Why does it then make only one of the other doors more likely to hide a car? Why doesn't the 2/3 shift to the door you picked? Your choice cannot have affected the odds...

And for the record, the hypothetical 'you could open B and C!' hasn't helped (it isn't part of the problem - we know there's a goat, we see a goat, we merely confirm what we already knew. I do get that a choice to open two doors improves your chances, but seeing a safe goat isn't the same as opening both of them - the odds you'll see two goats remain the same) Imagining 100 doors doesn't help either. It doesn't matter (to me, at present) how many doors you start with, the question is why the odds shift to one of the doors and not the other?

You can go on and argue this way forever but you will still be wrong. Here is how you can convince yourself that switching is right.

Try all possible cases.

There are only 18 of them. 3 doors that the car is behind * 3 doors that you pick * 2 doors that Monty opens one of. Write out all these 18 cases, and determine in each case whether you win the car by switching and by not switching. You will find that you win in 6 of the cases by not switching in 12 cases by switching. Then maybe you can begin to understand why.

IXP

 

[catsmate]

But I swear the bridge equivalent is more complicated and causes even more arguments (among those with an interest, anyway).

I've found bridge to be the most anti-social card game for similar reasons.

Yeah that's the explanation I go with when people don't get it. Or imagine if Monty didn't reveal the goat, and after you picked door A, he said "Okay, you can stick with door A, or you can open BOTH door B and door C."

Good summary.

 

[UKBoy1977]

Actually, it doesn't matter - for the probabilities - if Monty Hall knows where the car is or not. The probabilities stay the same if he randomly opens another door.

I disagree. If Monty RANDOMLY picks his door and shows a goat then it increases the likelihood of your original choice being correct in the sense that he COULD have revealed the car, but didn't. This is more likely to have happened if he was choosing from two goats (i.e.you chose the car) than if he were picking from one goat and one car. Applying Bayes theorem would show it now becomes 50/50.

 

[Squeegee Beckenheim]

Write out all these 18 cases, and determine in each case whether you win the car by switching and by not switching.

Actually, don't even write it out. Physically do it. Get an ace of spades and 2 red cards and lay them face up. The ace is the car. Pretend you don't know where the car is and pick "door" 1. Remove a red card that is not the one you've picked (if "door" 1 is a red card), then see what the result would be if you stuck or switched. Do the same for "door" 2, then "door" 3.

After that (which should take approximately 30 seconds), nobody should question whether or not it's right, because it's obviously true that switching gets you the car 2 times out of every 3.

And, if not playing cards, then any 3 objects where one is distinctive enough to be a stand-in for the car.

 

[GlennB]

I've found bridge to be the most anti-social card game for similar reasons.

Yep. It takes 4 very easy-going people to play bridge without a major dispute breaking out.

 

[p0lka]

I thought it was this,

If you maintain your original choice, door 1 (1:3 that its the car) then its 2:3 that one of the other two doors is the car,

once you have seen the goat behind one of the other two doors, then its still 2:3 that one of the other two doors is the car, but there's only one door left to swap to.

so by swapping, you are swapping from a position of 1:3 that its the car, to a position of 2:3 that its a car, making it more likely to get the car by swapping than not swapping.

makes sense to me.