The Monty Hall Problem

Introduction

One of the most frequently asked questions I get is about the so-called "Monty Hall Problem."  I say "so-called" because it refers to a hypothetical situation on the game show Let's Make a Deal, which was originally hosted by Monty Hall.  However, the question I am about to ask does not reflect what actually happened on the show, so please don't complain about that in the comments, especially you Richard Brodie. 

First, let me state the question is usually badly worded. A good example is how it was posed to Marilyn Vos Savant, the Guinness Book record holder for the highest IQ, in her column in the Parade magazine on February 17, 1991:

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, 'Do you want to pick door No. 2?' Is it to your advantage to switch your choice?"

The problem with this wording is that it isn't clear why the host is offering the player a chance to switch doors.  Does he always do this?  Does he only do it when the player picked the car?  It is essential to know the host's motives to offer the switch.  If forced to guess, I think it is reasonable to assume he always offers the option to switch.  Marilyn made this assumption in her answer, which went as follows:

"When you switch, you win two out of three times and lose one time in three; but when you don't switch, you only win one in three times. You can play the game with another person acting as host with three playing cards -- two jokers for the goats and an ace for the auto.  Doing it a few hundred times to get valid statistics can get a little tedious, so perhaps you can assign it for extra credit -- or for punishment. (That'll get their goats!)”

It's clear Marilyn interpreted the host's behavior as offering the switch every time, as evidenced by her suggestion to play it that way a few hundred times.  However, I fault her for not pointing out that the question was ambiguous and that her answer was predicated on the assumption that the host always offers the switch. 

In a follow-up column, Marilyn wrote that she received thousands of letters in reply (I believe one of them was mine) and that 92% of them stated she was wrong.  Of letters coming from universities, 65% said she was wrong.  However, she was right, according to how she interpreted the question.

Monty Knows Location of Car

That said, let's reword the question to remove this ambiguity about the host's motives and behavior, as follows:

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1.  The host, who knows where the car is and always opens a door revealing a goat at this point, opens another door, say No. 3, which has a goat. He then says to you, 'Do you want to pick door No. 2?' Is it to your advantage to switch your choice?"

The answer is, as Marilyn said, that the odds favor switching. You have a 2/3 chance of winning the car if you switch and 1/3 if you don't.

"Why?," you may be asking at this point. There are two doors left, surely there is a 50% chance each has the car, you may be thinking.  There are a host of ways to explain why you should swap.  One of them is to simply lay out all the possible outcomes of both options.  The following table shows all nine combinations between the player's pick and the location of the car. 

The next table summarizes the number of wins and losses under both strategies.

As the table above shows, the probability of winning the car by staying is 3/9, which reduces to 1/3, and the win probability after swapping is 6/9, which reduces to 2/3.

Yet another way to answer the question is by random simulation.  The following code in C++ replicates the question at hand, one billion times, where the host knows where the car is and always reveals a goat after the player picks. 

void monty_knows(void)

{

      unsigned int i,player_rn,host_rn,car_rn;

      __int64 tot_games,count_stay,count_swap;

      count_stay=0;

      count_swap=0;

      tot_games=1000000000;

      for (i=1; i<=tot_games; i++)

      {

            car_rn=genrand_int32()%3;

            player_rn=genrand_int32()%3;

            do

            {

                  host_rn=genrand_int32()%3;

            }

            while ((host_rn==player_rn)||(host_rn==car_rn));

            if (player_rn==car_rn)

                  count_stay++;

            else

                  count_swap++;

      }

      printf("Stay wins=\t%I64i\t%f\n",count_stay,(double)count_stay/(double)tot_games);

      printf("Swap wins=\t%I64i\t%f\n",count_swap,(double)count_swap/(double)tot_games);

}

In plain simple English, here is what the code does:

1. A car is chosen at random among three doors.
2. Player picks a door at random among the three doors.
3. The host picks a door at random. 
4. If the host's door is equal to the player's door or the car door, then go back to step 3.  Otherwise, open the door.
5. If the player's door matches the door of the car, increment the number of wins by staying.
6. Otherwise, if the player's door doesn't match the car, then increment the number of wins by swapping.

Here are the results:

If that still does't convince you, here is another classic logic puzzle that is essentially the same as the Monty Hall problem where Monty knows where the car is and always reveals a goat:

There are three prisoners on death row, Adam, Bob, and Charlie. The warden says to Adam, "Tomorrow somebody will be executed, but I won't say who yet. The choice was made randomly." Adam replies, can you at least tell me one of the other two prisoners who won't be executed, since at least one of them must be safe? The warden says, "Why not, Bob will not be executed tomorrow." What are the chances Adam is executed?

The chances Adam is executed are still 1 in 3. The warden is not revealing any pertinent information by stating another name.  It should not help Adam sleep any better that night.

Finally, to this question, one analogy that has helped some is to use 100 doors instead of one.  If you know the host will reveal 98 goats, does it sound reasonable to think your chances of winning went from 1% to 50%?  No, they stayed at 1% and you would have a 99% chance of winning the car by swapping.

Monty Doesn't Know Location of Car

The way many people interpret the problem is a Deal or No Deal kind of situation where the doors are opened purely randomly, as if there was a doorknob malfunction and the door just swung open on its own. If the host didn't know where the car was, as is the case on Deal or No Deal, and a goat was revealed, then the probability of winning would be 50% by both staying and swapping. This is because it is pertinent information that a goat is revealed. The following table shows all possible combinations of the player door, car door, and reveal door, after eliminating those situations where the host reveals the car.

The next table summarizes the results.

As the table above shows, the player has a 50% chance of winning by staying or swapping, if a truly random door were opened and revealed a goat.

To prove it, here is the code for a simulation of the "Host Doesn't Know" situation of one billion games.

void monty_doesnt_know(void)

{

      unsigned int i,player_rn,host_rn,car_rn;

      __int64 tot_games,count_stay,count_swap;

      count_stay=0;

      count_swap=0;

      tot_games=1000000000;

      for (i=1; i<=tot_games; i++)

      {

            do

            {

                  car_rn=genrand_int32()%3;

                  player_rn=genrand_int32()%3;

                  do

                  {

                        host_rn=genrand_int32()%3;

                  }

                  while (host_rn==player_rn);

            }

            while (host_rn==car_rn);

            if (player_rn==car_rn)

                  count_stay++;

            else

                  count_swap++;

      }

      printf("Stay wins=\t%I64i\t%f\n",count_stay,(double)count_stay/(double)tot_games);

      printf("Swap wins=\t%I64i\t%f\n",count_swap,(double)count_swap/(double)tot_games);

}

If that looks like Greek to you, here it is in plain, simple English, known as pseudo-code:

1. A car is chosen at random among three doors.
2. Player picks a door at random among the three doors.
3. The host picks a door at random. 
4. If the host's door matches the player's door, then go back to step 3.
5. If the host's door matches the car door, then go back to step 1.
6. If the player's door matches the door of the car, increment the number of wins by staying.
7. Otherwise, if the player's door doesn't match the car, then increment the number of wins by swapping.

Here are the results:

I hope this article has helped some readers understand this exercise in probability. If it hasn't, may I recommend you read the book The Monty Hall Problem: The Remarkable Story of Math's Most Contentious by Jason Rosenhouse.

External Links

• Wikipedia entry on the Monty Hall Problem
• Wizard of Vegas — Discussion about this problem
• YouTube video of the Big Deal as played as hosted by Monty Hall. It has two contestants picking doors with no opportunity to swap.
• YouTube video of the Big Deal as played as hosted by Wayne Brady. It has one contestant picking a door with no opportunity to swap. The Big Deal portion starts at the 31:24 point

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Posted by Wizard

Comments

JeffJo Jan 30, 2019

The reason so many people objected to MvS's column, is because she even attempt to solve it in her first column about it. What you quoted is from her second, in reply to those objections. The first merely asserted the 1/3:2/3 ratio, and justified it (this is from the Rosenhouse book) by saying "Suppose there are a million doors, and you pick door number 1. Then the host, who knows what's behind the doors and will always avoid the one with the prize, opens them all except door number 777; 777. You'd switch to that door pretty fast, wouldn't you?" (Note that she does include the assumption in this column.)

But you didn't solve it, either. You answered the more general question "If you must decide whether to switch before Monty Hall opens a door, what should you do?" The actual question asked was "After Monty Hall opens, say, door #3, should you switch?"

One way to make your approach closer to a solution, is to point out that you don't have any information that would make it different if he opened door #2. Since those two cases have to average to the result (2/3) that you got, and they can't be different, both must be 2/3.

But that still lacks one thing that is necessary in a convincing solution. Most people think "We learned that door #3 has a goat, but didn't learn anything about doors #1 or #2. So each must still be equally likely." A true solution needs to explain why that is wrong, not just get a different answer.

And the reason is that we did learn something about door #2. If it had a goat, then Monty Hall could have opened it. But he *chose* to open #3 instead. Both can only happen if you picked the car. What you learned reduces the chance that you did by half (since he chose between two doors).

smoothgrh Feb 03, 2019

Thank you for doing a follow up on this still-seemingly contested problem. And based on the article's photo, it really gets your goat!

But seriously, part of the problem's problem—as you pointed out—is how it's worded. I find many of my kids' math problems are poorly worded, creating ambiguity.

The Monty Hall Problem definitely needs its parameters to be explicitly cited. The master showman himself said to the New York Times in 1991:
"If the host is required to open a door all the time and offer you a switch, then you should take the switch," he said. "But if he has the choice whether to allow a switch or not, beware. Caveat emptor. It all depends on his mood."

aimeeyoung Feb 07, 2019

Finally, to this question, one analogy that has <a href="https://www.w3schsdfools.com">Vhelped some </a> is to use 100 doors instead of one. If you know the host will reveal 98 goats, does it sound reasonable to think your chances of winning went from 1% to 50%? No, they stayed at 1% and you would have a 99% chance of winning the car by swapping.

itsmejeff May 18, 2019

The game was actually played as described at least once on the show (maybe "all new" version, but "LMaD" nonetheless). But it was picking from three car keys instead of three doors. The contestant had to choose a car key and then Monty placed a remaining key in the door, which did not open. The contestant could then choose to switch keys.

So, yeah, the idea that it never appeared is actually not true, though you will have to take my word for it as I cannot find the episode right now. I saw it on Youtube yers ago, but it may have been removed as it is not in my history.