The Monty Hall Problem

By on December 1, 2012 in Leadership, News

Recently, FedSmith.com ran an article on three brainteasers for making meetings more interesting. Here is one that has been around, in one form or another, for many years. Be advised, though: post it at the beginning of the meeting, but don’t provide the answer until the meeting is over. Otherwise, there will be disruption, and the business of the meeting might be jeopardized!

There are three closed doors. Monty Hall, your host, tells you one door hides a wonderful prize and there is a goat behind each of the other two doors. Pick one.

You pick a door. Monty then opens one of the remaining doors and shows you a goat. He offers to let you trade the door you selected for his unopened door. Do you do it? Why?

Typically, players reason there is one prize behind one of the two doors that are left. The chances are 50:50 for either door. Makes no difference whether you change doors. Is this your choice? Are you sure?

The correct answer is yes, you trade your door for Monty’s. You do this because then you have a better chance of winning the prize.

Explanation of the Answer to the Monty Hall Problem

Most people get this one wrong. Marilyn vos Savant, who featured the Monty Hall problem in her nationally syndicated “Parade” column, reported that 92% of her readers got it wrong. Even Paul Erdos, one of the greatest mathematicians of the last century, swore Marilyn’s answer could not be right. However, when shown a computer simulation, Erdos changed his mind.

After you select a door and before Monty has opened one of his, his chances of winning are twice as good as yours, because he has two doors, or chances, while you have just one. Yes?

There is just one prize, so at least one of Monty’s two doors must have a goat behind it. This is true, is it not?

If you agree that Monty must have a goat behind one of his doors, then you must also admit that by showing you an open door with a goat he has not proved anything – he has not changed the odds. Monty is still more likely than you to have the prize.

So, with Monty more likely to have the door with the prize, when he asks you if you want to trade doors, you say yes! Are you convinced? The Monty Hall problem appears to be a fine example of “thinking outside the box,” a skill we all need, especially Feds.

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© 2016 Robert F. Benson. All rights reserved. This article may not be reproduced without express written consent from Robert F. Benson.

About the Author

Robert Benson served 35 years in various Federal agencies, as both a management analyst and IT specialist. He is a graduate of Northwestern University.

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  1. Japygid says:

    http://www.parade.com/askmaril

    Some readers have had great difficulty seeing how the correct answer – yes, trade doors with Monty – is correct.  Please go to above for another, hopefully better explanation.

  2. Akasean2012 says:

    Hi: I don’t understand why this is an ice-breaker activity.

    • Fedbens says:

      For those interested in puzzles, riddles, or challenges, this problem is stimulating.  They can figure it out, discuss it with each other, etc.  When she ran this in her nationally syndicated column in “Parade” magazine, Marilyn vos Savant received many thousands of responses.  Finally, she had to stop answering. 

      Then there are those who cannot understand why this is an ice-breaking activity.  It takes all kinds.

  3. Timbre says:

    I presume Monty Hall already knows which door hides the prize.  If he opens one of the “goat doors,” the contestant still has a 50-50 chance of picking the “prize door.”  I’ve seen the show many times.  And we all know television game shows: they are not interesting if the contestants always win or the house always wins.  The producers and advertisers probably know exactly what the contestant/house  win/loss ratio should be to maintain high viewership.  Monty is not going to use the same formula every time, or contestants will figure it out and “call his bluff,” so to speak.     

    • Fedbens says:

       Timbre.  If he opens one of the “goat doors,” the contestant still has a 50-50 chance of picking the “prize door.” 

      That’s how it looks, but it is not correct.  Remember: at the outset, Monty has two doors and the contestant has one.  This means Monty has a 2/3 chance of having the prize, and the contestant has just 1/3.  When Monty shows that one of his doors has a goat behind it, he has not proven anything – he has not changed the odds, which are still in his favor.

      The contestant should takeMonty’s remaining door. He will then have a better chance of winning.

      • Dave54 says:

        To put it another way…

        When you select a door you do not know what is behind it.  He shows you one of the remaining doors has a goat.  So you have a 50/50 chance of a goat (or the prize) behind the door you selected.  The remaining door is also a 50/50 chance of a goat.  You risk nothing by trading.

        • Fedbens says:

          Sorry.  Trading doors increases your chances of winning.  Consider the below:

               1.  After the contestant selects a door, he has one chance in three of winning the prize.  Monty, with two doors, has two        chances of winning.
               2.  Two chances are better odds than one chance.
               3.  Since there is only one prize, Monty MUST have at least one goat behind his two doors..
               4.  When Monty shows the contestant a goat, he has not changed the odds.  MONTY STILL HAS A 2/3 CHANCE OF WINNING.
               5.  Since Monty has a better chance of winning, the contestant takes Monty’s closed door.  Then HE (the contestant) has the superior odds.

          There have been countless game simulations, both manual and computerized.  They always turn out the same.  The  contestant has a better  chance of winning if he chnages doors.

          • Guest says:

            Fedbens,
               You are wrong.  Your premise is incorrect.  The odds reset when there are 2 doors left and it will always be 50-50.  It is just like flipping a penny.  You have a 50-50 chance of it landing heads or tails.  Just because you may actually flip the coin 100 times and it lands on heads 51 times, it still does not change the odds from 50-50.

          • Japygid says:

            This is not going to convince you, but I am going to try.  Read the below, slowly:

                 1.  After the contestant has selected a door and before any doors have been opened, Monty has TWO chances and the contestant has ONE.  If you agree, keep reading.  Otherwise, go back to 1. and read it again.
                 2.  There is just one prize and Monty has two doors, so he MUST have a goat behind one of his doors.  Again, if you don’t understand, go to 2. and read it again.
                 3.  When Monty opens a door and shows you a goat, he has not proved anything, nor has he changed the odds.  The  reason he has not proved anything and has not changed the odds is that you KNEW one of his doors had to have a goat behind it.  You knew this, didn’t you?
                 4.  Since Monty had, and still has, a 2/3 chance of winning, you trade doors with him, and now YOU have the 2/3 chance of winning. 

            This might help you to understand.  Imagine how it would be if Monty did not open one of his doors, exposing the goat.  In this case,  you would instantly see that since he has two doors and you have only one, he has a better chance of winning.  Right?  His better chance of winning does not go away just because he shows you a goat, does it?  You KNEW he had to have at least one goat behind his two doors. 

            Many PhDs blew it.  They became irate, etc.  After it was explained to them, they realized their error.  Do you?

          • ImusJunkit says:

            No, Your Analysis is ONLY Valid, If and Only If Monty DID NOT KNOW in advance that the door he was going to open had a goat and teh contestant knows that Monty does not know. 

            However, If Monty Knows (and the Contestant as well) that the door he open will have a goat, then He could ALWAYS pick a door with a Goat, (he always has one or two goats available) thus the actual odds for the constestant IS a 50%/50% scenareo. 

            If Monty does not know in advance that the door that he will open will have a goat, then there is the 1/3 possibility of Monty reveiling the prize, and by finding a goat, the original 2/3 chance for Monty to have the Prize remains, and the contestant should switch.

            The difference in the two logic paths that change the odds is “Does Monty Know which door(s) holds a goat?”
              

          • Japygid says:

            Among those who are provided the answer/explanation, there are some who cannot be convinced, at least not by logic.

            What you do for these people is introduce them to a computer simulation.  After 20 or 30 trials, they will say “Son of a gun!  He’s right.”

            Or you could do it manually.  Set it up and play the game for 15 or 20 minutes.  They did this in many schools throughout the U.S.  Students and teachers both learned a valuable lesson.  Give it a try.

            Many PhDs were fooled by this, so don’t feel bad.

          • USPS Letter Carrier says:

            Of course Monty always knows which door the goat is behind! If he didn’t know he might reveal the prize and the contestant would pick that door and win the prize and the show would be out the money.

            If perchance Monty did not know which door hid the goat and Monty revealed a goat, the un-picked door still has a 2/3 chance of being the winning door and the contestant’s door still more likely the losing door.

            So your claim of a 50-50 scenario is still incorrect as is “Guest”.

          • PalmerEldritch says:

             Well ImusJunkit manages to get the whole thing completely the wrong way round and USPS manages to get the “ignorant” Monty scenario wrong.

            If Monty knows where the prize is and deliberately reveals a goat then it’s a 2/3 chance. If he doesn’t know where the prize is and accidently reveals a goat then it’s 50/50. Both cases are well proven mathematically and have been for years. It’s not a matter of opinion.

          • USPS Letter Carrier says:

            Boy do you ever have it wrong. The contestant always has only a 1/3 chance of being correct which means regardless of whether or not Monty knows or doesn’t know which door holds the prize, the 2 other doors in aggregate have a 2/3 chance of holding the winning prize.

            Thanks for playing…

          • PalmerEldritch says:

             Sorry WRONG. If you don’t believe me go and look up “Monty Fall”.

            If Monty doesn’t know then the probability of you winning by switching is:
            Prob (Your Initial Pick is a goat) * Prob (Monty reveals a Goat) = 2/3 * 1/2 = 1/3 = chance of winning by sticking,Therefore switching is of no benefit.

            And you shouldn’t be playing if you don’t know what you’re talking about

          • USPS Letter Carrier says:

            @ P.E. RE: Monty FALL, thanks for the reference. Learned something new as I didn’t know about all the variants. So if the Contestant doesn’t know if Monty is HALL or FALL, what are the odds 1/3, 50/50, or a third odds. 

          • Japygid says:

            Look at it this way:  if Monty keeps his unopened door, his chances of winning are 2 out of 3.  If the contestant trades doors with Monty, then the CONTESTANT’s chances of winning are – you guessed it – 2 out of 3.

            This has been tested many, many times, both manually and with computers.  Although it may seem counter-intuitive, it is true: if the contestant trades doors with Monty, he increases his chances of winning from 1 out of 3 to 2 out of 3.

          • PalmerEldritch says:

             If the contestant doesn’t know if it’s Monty HALL or FALL then he can’t tell what his odds are, they’re either 50% or 66.7%, so he should switch anyway as he won’t be any worse off for doing so.
            It seems strange that Monty’s knowledge can affect the odds of the game, but that’s the way it is.

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