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#1 2006-06-17 15:56:20

Simon
Member
Registered: 2006-06-06
Posts: 41

Variation on the Monty Hall conundrum.

I hope you find this a challenging variation of the conundrum.

Before I begin, let's make sure we're all familiar with the old three door Monty Hall concept. Here is a brief refresher.

"Contestants are given three doors to choose from. Behind two doors are goats, and behind the other door is a car. The host knows what's behind each door. The contestant gets to pick one door and the host opens one of the doors the contestant did not pick to reveal a goat. The host then gives the contestant a chance to change his mind.

    The question is, is it more likely for the contestant to win the car if he sticks with his original door, or if he changes doors, or are the odds equal either way?"

Answer: as long as the host knows where the car is (and so never cancels a game by revealing it) the contestant's chance of winning if he sticks is 1/3 and if he swaps is 2/3.

For a full discussion of this solution which many find hard to beleive, see:

http://www.mathsisfun.com/forum/viewtopic.php?id=3106

And for those who still can't beleive the answer, here are a couple of excellent online simulators to test it.

http://www.curiouser.co.uk/monty/montygame.htm

http://math.ucsd.edu/~anistat/chi-an/MonteHallParadox.html

The Monty Hall debate has already established the following:

As long as the host knows where the car is - and so never cancels a game by revealing it - the contestant's chance of winning if he sticks is 1/3 and if he swaps is 2/3.

However, when the host doesn't know where the car is but reveals a goat randomly - the odds whether you stick or swap become 50/50.

The following variation is a simple, but perhaps more mysterious, version of the original. It is aimed at those who already agree on the above. Hope you're ready for this:

There are three doors. Behind one is a car. Behind the others are goats.  You select one door.

Enter Host 1, who does not know where the car is. From the two doors you didn’t pick, he randomly reveals a goat. He then closes that door and leaves, but you remember what's behind it. (If a car was revealed the game is void. Shuffle the doors and start again. Keep going until you get a game where the host has randomly revealed a goat).

Because the host's elimination was random, we're agreed at this point that your chances of finding the car are now 50/50 whether you stick to your original selection or swap to the door that wasn't opened.

However, before you make your decision:

Enter Host 2, who did not witness the above but knows where the car is and will not reveal it. As it happens, he opens the same door as Host 1. You are shown nothing new.

What are the odds now?

Are they the same as when Host 1 opened this door? Or should your estimate be different?

(If Host 2 opens a different door to Host 1, once again the game is void. As before, 'shuffle' the doors and keep going until you get a complete game where Host 1 randomly opens a door with a goat and Host 2 opens the same door.)

Simon

Last edited by Simon (2006-06-22 23:51:36)

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#2 2006-06-20 03:30:45

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Variation on the Monty Hall conundrum.

You say he picks a door by random, and that he must pick a door with a goat behind it.  But that's a contradiction.  He must either pick it by random, or he must pick the door with a goat behind it.  You can't have it both ways.

I guess technically you can.   But either way, you're giving me that information in the problem, specifically, that he is picking a door with a goat behind it.  So whether he knows it or not, I know that he "subconsciously" knows it.  And thus, it is no longer "random".

Edit: I should have noticed this before...


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2006-06-20 03:34:07

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Variation on the Monty Hall conundrum.

These are some good twists to a problem which is fairly difficult to grasp in the first place.  If you have more, keep 'em coming.  I'll try to see if I can come up with a few of my own.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#4 2006-06-20 03:53:23

Simon
Member
Registered: 2006-06-06
Posts: 41

Re: Variation on the Monty Hall conundrum.

Actually no Ricky. In the Monty Hall debate it's already established that the 2/3 advantage in swapping only occurs when the single Host knows where the car is and will avoid revealing it.

If you doubt this, try runnning a one host simulation where he may randomly open either door. Then discount all cames where the host prematurely revealed the car. These are obviously voided games because all has been revealed and you the player have no decision to make.

You will find that among the non-voided games, the odds are now 50/50 whether you stick or swap.

So strangely enough, when you get a single game where the host has revealed a goat but had no knowledge where the car was, the odds are 1/2 whatever you do.

Once you're convinced of that, see what you think about my amended version of the two host game.

Last edited by Simon (2006-06-20 04:01:24)

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#5 2006-06-20 04:39:32

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Variation on the Monty Hall conundrum.

Right on.  If I choose the door that does have the prize, the random host is always successful in choosing a non-prize door.  If I choose the door that doesn't have the prize, the whole process gets started over again if the random host chooses the prize door.  Thus, 50% of the time, I don't even get to my final choice if I'm not on the prize door.

But this is a phenomena which only occurs when I start the process all over again.  If the random host were to choose the prize door, and then I let the random host keep trying till he chooses a non-prize door, I'm back at 66%. 

But then it's not very random, now is it?  See what I mean by the contradiction?  You say he chooses a random door, then you define what that random door can and can't be.  But that isn't random.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#6 2006-06-20 08:19:46

Simon
Member
Registered: 2006-06-06
Posts: 41

Re: Variation on the Monty Hall conundrum.

OK, so before we proceed to the two host scenario, I need to convince you that in any game where a host doesn't know where the car is, but nevertheless avoids revealing it, the odds are no longer 1/3 that you originally picked the car. They are 1/2.

There is no contradiction. The best way to illustrate it is to show the six possible outcomes.
So let's look at six games where those outcomes get played out.

Imagine three doors. The door the player selects we call X. The one the host opens is Y. The remaining door we call Z. Behind each could be Goat 1, Goat 2 or the Car.


     X    Y   Z

  -------------

1.  G1  G2  C

2.  G2  G1  C

3.  G1  C   G2 voided

4.  G2  C   G1 voided

5.  C   G1  G2

6.  C   G2  G1


The above is simply the most probable average set of outcomes. You see what is most likely to happen? Outcomes 3 & 4 get eliminated.

In those six games you will most likely originally choose 4 goats and 2 cars. However, because the host doesn't know where the car was, he is most likely to nullify two out of the six games. On average, you will originally pick the car one third of the time. On average, the random host will prematurely reveal the car one third of the time. This cancels your advantage. A non-random host will never reveal the car. Therein lies the difference

So two games have been voided. Of the remaining non-voided games you picked 2 goats and 2 cars. If you swapped in all four of those games, you would still end up with 2 goats and 2 cars. The two games voided by the random host were the extra games you would have won by swapping if the host had known where the car was and avoided revealing it.

So in any single game where the host randomly opens a door, but nevertheless avoids revealing the car, the odds are 1/2 for the two doors left. Why? Because you know there was always a chance he might have revealed the prize, even though he didn't. That's why the host's state of knowledge is crucial.

Have tried this ready made simulator?

http://math.ucsd.edu/~anistat/chi-an/MonteHallParadox.html

Ok are we agreed on that yet?

If so we can move on to the two host game!

Last edited by Simon (2006-06-20 08:22:46)

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#7 2006-06-22 14:10:33

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Variation on the Monty Hall conundrum.

I agree with everything you said in the last two posts.  I still don't like calling it "random" though.

If the game were to start completely over everytime the random host picked the car, then yes, it would be 1/2.  However, if you just let the host pick again in the same game till he gets it right, the odds are again 2/3.

Now moving on to two hosts.  If the random host checks first, then the normal host, I believe the odds will be 2/3.  Your thoughts?  Oh, and if you haven't already, don't make a simulator just yet.  Lets debate this a bit first.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#8 2006-06-22 14:11:19

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Variation on the Monty Hall conundrum.

Originally posted by Simon.  Accidentally deleted and replaced by Ricky - Ricky

If the host doesn't know where the car is and reveals it, then naturally the game must start over. This goes without saying, because the player now knows where the car is 100% and there is no decision to make. There are no longer any odds at all.

The same applies to the Host 2. Although Host 2 will always avoid revealing the car, he could reveal a different goat to Host 1. If that happens, the player knows he picked the car 100% and once again he has no decision to make.

In the two host scenario, the only time the player gets to make a decision is if a game makes it through to both rounds.  A complete game is where Host 1 (who doesn't know where the car is) opens a door with a goat and Host 2 (who does know where the prize is) opens exactly the same door.

Now, assuming those rules have been followed, your answer is:

After Host 1 reveals a goat, you assess the odds to be 50/50. (I think we're now agreed on that.)

After Host 2 reveals the same goat, you assess the odds to be 2/3 in favour of swapping.

Is that your answer?


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#9 2006-06-22 14:11:39

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Variation on the Monty Hall conundrum.

Maybe not.  Let me rethink this.

There are two cases:

Player picks the door with the car.  The the random host must reveal one door.  Now there is a 50% chance that the other host will open the other door, letting the player know that he has picked the door with the car.  This happens 33% of the time, so there is a 1/2 * 33% chance here.

Case 2, the player picks a non-car door.  Then by the time the random host is done, there is a 50% chance the the player has picked the right door.  Now the other host must pick the exact same door.  So the 50% chance remains the same.  This happens 66% of the time, so there is a 1/2 * 66% chance here.

1/2*.33 + 1/2*.66 = 1/2.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#10 2006-06-22 14:11:48

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Variation on the Monty Hall conundrum.

If that's right, this is a great problem.  The more information you get, the less likely you are to know anything!


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#11 2006-06-22 14:13:32

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Variation on the Monty Hall conundrum.

Everything in this topic should be back to normal.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#12 2006-06-22 18:04:56

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,550

Re: Variation on the Monty Hall conundrum.

Ricky wrote:

Everything in this topic should be back to normal.

Yay!


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#13 2006-06-27 05:59:30

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Variation on the Monty Hall conundrum.

Did my post deleting episode scare Simon away?  I'm still waiting to hear if he agrees with me...


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#14 2006-06-27 09:35:50

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,550

Re: Variation on the Monty Hall conundrum.

I hope not ... this was good stuff.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#15 2006-06-28 15:01:16

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: Variation on the Monty Hall conundrum.

With the new scenario with two hosts, you should switch.
50% of the time it is voided.
1/6th of the time you should stay.
1/3rd of the time you should switch.
I'm not going to try to explain myself because it wouldn't help with my poor facility.


igloo myrtilles fourmis

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#16 2006-06-28 15:22:33

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Variation on the Monty Hall conundrum.

And the other 1/2 the time?  You should choose the third option wink

I would really like to hear how you came to that though.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#17 2006-07-01 02:33:46

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: Variation on the Monty Hall conundrum.

Okay, my notes are sketchy, as I see the old piece of paper still beside my computer, but first I
said that lower case g means goat not chosen.  I also put the car always in door # 1 on left.
Why bother to do all cases where car is in middle and right doors; waste of time, same thing over again.
So hence, capital letters mean they were chosen by contestant or host, if it is a capital C, then the
contestant chose it because hosts don't pick cars in non-voided plays.
So anyway, there are probably mistakes in this, but this is what my notes said...
CGg 1/12
CgG 1/12
cGG  2/(3x2) void
cGG  2/(3x2) continue
CGG voided  1/6
Looking back at these notes, I don't think they are right and I don't understand them again.
But I will do it again soon and continue to post later on this interesting subject.

Last edited by John E. Franklin (2006-07-01 02:34:20)


igloo myrtilles fourmis

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#18 2006-07-04 08:14:34

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: Variation on the Monty Hall conundrum.

Okay, I did it again.  Maybe I was right with the messy unexplainable try.
In this photo (click on photo to enlarge), the bold outlined letters refer to
what the contestant guesses.  The letters with the subscripts 1 and 2 refer
to what the host #1 and host #2 pick.  Voids and percentages are given.
I still get 1/3 switch, 1/6 don't switch, 50% of rounds get redone so it
repeats with the second round with same results: 1/3 switch, 1/6 don't switch, and
50% of 2nd rounds go to 3rd round, etc, etc...

View Image: monkey.JPG

igloo myrtilles fourmis

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#19 2006-07-04 12:14:29

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Variation on the Monty Hall conundrum.

I'm sorry John, I tried to read your post, but it is just too confusing.  I had the same problem when I wrote mine.  Everytime I tried to explain something, it quickly got complicated.

What I found worked well is to start at the beginning of the game, and set up different cases.  "If such and such were to happen, then blah blah blah and you only have X percent.  On the other hand, if such and such happened, then blah, and you have Y percent."  And I would recommend not using symbols, as trying to keep track of them can be a real bother.

Would ya give it another shot?


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#20 2006-07-05 02:31:02

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: Variation on the Monty Hall conundrum.

There are eight rows in my hand-written scan.
First row says the contestant picks the car, and host 1 picks a goat and host 2 picks the other goat.
Chance of this is 1/12, and it is voided because two goats are revealed.
Second row says the contestant picks the car, and host 1 picks a goat and host 2 picks the other goat, just like row1 except the hosts swap doors on each other: chance is 1/12 and is voided because two goats are revealed.
Third row says host picks car and hosts pick door #2 goat: chance 1/12
Forth row says host picks car and hosts pick door #3 goat: chance 1/12
Rows 1 to 4 must add to 1/3 of the time because car is chosen by contestant 1/3 of the time.
Row 5 says that contestant picks goat on door # 2 and host picks car:  this is a void.
Row 6 says that contestant picks goat and hosts 1 and 2 pick goat on door #3.
Row 7 says contestant pick goat on door 3 and host 1 voids it by picking car on door 1.
Row 8 says contestant picks goat on door 3 and host 1 and 2 both pick goat on door #2.
After all the possibilities are listed out, then you have to think about how the 100% breaks
down for each round.  (A round being that the contestant picks a door and the hosts pick doors)
So 1/3 of the time a contestant picks a car on door #1 because the car is always behind door #1 in
my scenarios.  And 1/3 of the time contestant picks goat behind door#2 and 1/3 of time contestant picks goat behind door #3.  Then you have to break up the four cars picked scenarios into percentages by logic.
Since the 2 goats are left, both hosts will pick randomly, so divide the 4 scenarios (rows 1,2,3,4) equally.
Hence 1/12 each of time.
Then if contestant picks goat on door #2, there are two possibilities: rows 5 and 6 above.
Host #2 picks randomly so it is half and half, that's how I got 1/6 and 1/6 for rows 5 and 6.
Row 7 and 8 are also divided equally after contestant picks goat on door#3, then host#1 picks randomly, so
divide 1/3 by 2 for each row 7 and 8.
My eight permutations only have the car on door #1 because if you do all 24 permutations then
it would come out the same because of symmetry.

Last edited by John E. Franklin (2006-07-05 02:53:16)


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#21 2006-07-20 23:10:55

Kurre
Member
Registered: 2006-07-18
Posts: 280

Re: Variation on the Monty Hall conundrum.

Simon wrote:

  X    Y   Z

  -------------

1.  G1  G2  C

2.  G2  G1  C

3.  G1  C   G2 voided

4.  G2  C   G1 voided

5.  C   G1  G2

6.  C   G2  G1

the car isnt randomly placed, it is "forced" to be placed behind the two doors that the first host doesnt chose, which gives you a 50% chance, so it doesnt matter if you switch or stay. When the 2nd host comes, he is "forced" to open the same door as the first host, so the 2nd host doesnt do any different, the 50% chance remains and it still doesnt matter if you switch or not...did i get that right?

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