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Monty Hall Problem

I may be repeating someone up above. If so, I apologize. A couple of points:
  • The problem does not necessarily have anything to do with the TV show.
  • The problem states that Monty opens a door he knows has a goat.

~~ Paul
 
Paul C. Anagnostopoulos said:
I may be repeating someone up above. If so, I apologize. A couple of points:
  • The problem does not necessarily have anything to do with the TV show.
  • The problem states that Monty opens a door he knows has a goat.

~~ Paul
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But does he have to offer a second choice, or can he not offer the second choice and stay with the first?
 
Robin said:
But does he have to offer a second choice, or can he not offer the second choice and stay with the first?
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I'm not sure why this matters, since he did offer the contestant the chance to switch. There is no suggestion in the problem statement that his offer was conditional.

~~ Paul
 
Paul C. Anagnostopoulos said:
I'm not sure why this matters, since he did offer the contestant the chance to switch. There is no suggestion in the problem statement that his offer was conditional.
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The problem is, there is no suggestion in the problem statement that his offer was unconditional, either.

I agree that his offer being unconditional is one interpretation of the problem--In this case, you do double your chances if you switch.

However, I also agree that the offer being conditional is another interpretation--In this case, the probabilities are indeterminate; they depend on that condition on which Monty chooses whether or not to open a door and offer a switch.

It's this ambiguity that's a flaw in the problem as it is often stated.

If we rephrased the problem, replacing Monty Hall with some con man running a sleazy carnival, would you still agree that it's safe to assume his offer to switch is unconditional?
 
Paul Anagnowhatchamacallit is right.

He makes the offer. So the problem (which I posted) states. Therefore there is no conditionality. The situation before you is:

Contestant chooses one of three doors.

Monty offers one of the remaining two doors as an alternative selection, after opening the other of the two remaining doors.

Contestant either switches or doesn't switch.
 
And then Cabbage whipped out his big ten inch ........ Crayola:

Cabbage said:
Okay, one more time (just on the offhand chance you actually do have a brain):

If you reread, you'll notice I didn't say "1/3 of 2/3"; I said "1/3 out of 2/3" (there's that reading comprehension I was referring to earlier). 1/3 out of 2/3 is 1/2. Just like 1 out 2 is 1/2. Or 3 out of 6 is 1/2. Or 1/4 out of 1/2 is 1/2. Or, in general, just like x out of 2x is 1/2 (if x is any nonzero real number).
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I did reread it, and my reading comprehension is quite good, thank you very much.

You have introduced a novel usage of the phrase "out of" by applying it to fractions. ARE YOU TRYING TO SAY THAT 1/3 DIVIDED BY 2/3 IS ONE HALF? If so, you are correct, but I must give you no marks at all for clarity of expression.

Plus, you're wrong about the problem itself.
 
Goodbye, TeaBag420; I have neither the time, patience, nor inclination to educate you.

To anyone else still interested, say you encounter someone on the street offering to play this game with you, where you're trying to locate a pea hidden under a shell, or something like that.

You pick one of the shells, he lifts another one to show you the pea isn't underneath it. You either switch or don't switch, and you either win or you don't win; I don't care, that's not the point.

Now, after playing the game you run in to another person. He says, "You look like a smart man, I'll bet you know that once he offers the switch, you've got a 2/3 chance of winning when you do switch, and only a 1/3 chance when you don't switch. Let's say the two of us hang out here for a while, watch this guy playing his game with other passersby, and make a bet. Everytime a person switches and ends up winning, I'll give you $50, but if he switches and loses, you'll owe me $50. You can't lose--You ought to win twice as many times as I do! Whaddayasay?"

Do you take the bet?
 
Cabbage said:
Goodbye, TeaBag420; I have neither the time, patience, nor inclination to educate you.
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You've already played that card.

To anyone else still interested, say you encounter someone on the street offering to play this game with you, where you're trying to locate a pea hidden under a shell, or something like that.

You pick one of the shells, he lifts another one to show you the pea isn't underneath it. You either switch or don't switch, and you either win or you don't win; I don't care, that's not the point.

Now, after playing the game you run in to another person. He says, "You look like a smart man, I'll bet you know that once he offers the switch, you've got a 2/3 chance of winning when you do switch, and only a 1/3 chance when you don't switch. Let's say the two of us hang out here for a while, watch this guy playing his game with other passersby, and make a bet. Everytime a person switches and ends up winning, I'll give you $50, but if he switches and loses, you'll owe me $50. You can't lose--You ought to win twice as many times as I do! Whaddayasay?"

Do you take the bet?
Click to expand...

No, because you will probably lose every time. The reason is that this is a con, not a game show. He's expecting you to be stupid enough not to know that there is no pea under any of the three shells at the time the player is given the choice to switch. Just like Cabbage is.
 
Shell game is exactly the same as the Monty Hall game. As long as the rules state that an offer to change must be made and as long as the rules are followed then it is advantageous to change your choice.

The following article contains a demonstration of this:

http://www.willamette.edu/cla/math/articles/marilyn.htm

So if you offer me 100 games and guarantee me that each time you will reveal one empty cup when I have put my finger on the choice and guarantee that I will be given the chance to switch, and I win the $50 on best of 100 then certainly I will take the bet. As long as you don't cheat it does not matter if you try to make me lose.
 
Monty Hall Problem - General Solution

A few people have suggested that this has been settled and I would be interested to see their solutions.

Here is my solution and it incorporates all the objections that have been raised. I have only used one unjustified assumption - that Monty never reveals the prize, mainly to simplify, but also because nobody has suggested that he does.

The problem is that as stated the problem might be read as a specific scenario and not a description of the game. This means that we cannot assume that Monty is required to open a door and offer a second guess. So we cannot assign a probability of 1 to this event and we cannot assign an even probability to the event because we do not know what basis he makes the decision on.

So clearly we need a little basic game theory and I have devised the following categories:

Contestant strategy:
1. Always switch when given the chance
2. Never switch when given the chance
3. Toss a coin if given the chance

Host behaviour:
1. Good Monty - always offers a second choice when initial guess was wrong
2. Bad Monty - always offers a second choice when initial guess was right
3. Indifferent Monty - tosses a coin about offering a second choice
4. No Choice Monty - is required by the rules of the game to offer a second choice

Method:
Simulate contestants in all possible combinations through 100,000 iterations each, create a grid of probabilities

Premises:
1. Host never reveals prize (unjustified)
2. The probability of picking the car is 0.33
3. The probability of picking a donkey is 0.67
4. If Monty offers a second choice AND I accept the offer AND I have initially picked a donkey THEN I will win the prize.
5. We do not have any knowledge about Monty's motivation
6. We do not know if Monty is obliged by the rules of the game to offer a second guess

Result:
A risk averse person will always stick to their original choice when given a chance to change - p min/p_mean/p_max=0.33/0.33/0.33
A risk neutral person will toss a coin to decide whether to change when given the chance - p_min/p_mean/p_max=0.17/0.44/0.67
A risk happy person will always change - p_min/p_mean/p_max=0.00/0.54/1.00

Summary:
<html>
<table border=1>
<tr><td rowspan=2 valign=center>Contestant Strategy</td><td align=center colspan=4>Host Behaviour</td></tr>
<tr><td>Good Monty</td><td>Bad Monty</td>
<td>Indifferent Monty</td><td>No Choice Monty</td></tr>
<tr><td>1. Always switch if offered</td><td>
100.00%
</td><td>
0.00%
</td><td>
49.98%
</td><td>
66.84%
</td></tr>
<tr><td>2. Never switch if offered</td><td>
33.16%
</td><td>
33.16%
</td><td>
33.18%
</td><td>
33.16%
</td></tr>
<tr><td>3. Toss a coin if offered a switch</td><td>
66.69%
</td><td>
16.77%
</td><td>
41.68%
</td><td>
50.02%
</td></tr>
</table>
</html>


That is all.
 
pgwenthold said:
But you do not know why, and knowing why is critical, because:

1) If Monty only reveals a door after the contestent choses the car, then it means you have selected the car and chosing means you will lose

2) If Monty only reveals a door after the contestant choses wrong, then it means you have selected the wrong door and should switch.

3) If Monty always reveals doors when someone picks one, or does it indiscriminately, then switching means you will win 2/3 of the time

If you have a single instance, and you are shown a door, then you don't know if it is because Monty knows you have chosen a car and goes by (1), Monty knows you have chosen a goat and goes by (2), or this is just the way he always does it, or this is the way it happens sometimes.
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Your logic is sound. The problem should be reworded to exclude the behavioral ambiguity -- for the simple version.
 
Robin said:
Shell game is exactly the same as the Monty Hall game. As long as the rules state that an offer to change must be made and as long as the rules are followed then it is advantageous to change your choice.

The following article contains a demonstration of this:

http://www.willamette.edu/cla/math/articles/marilyn.htm

So if you offer me 100 games and guarantee me that each time you will reveal one empty cup when I have put my finger on the choice and guarantee that I will be given the chance to switch, and I win the $50 on best of 100 then certainly I will take the bet. As long as you don't cheat it does not matter if you try to make me lose.
Click to expand...
I don't know if you were responding to me or TeaBag420 with this, but the point of my example with the shell game is that you don't know how the man running the shell game is playing (and he is playing it fair, so there is a pea somewhere, and the person playing the shell game can certainly win).

At the point you make the bet with the second man, you've only see the game played once--when you played. Should you take the bet without knowing how the shell game man plays from game to game?
 
Cabbage said:
I don't know if you were responding to me or TeaBag420 with this, but the point of my example with the shell game is that you don't know how the man running the shell game is playing (and he is playing it fair, so there is a pea somewhere, and the person playing the shell game can certainly win).

At the point you make the bet with the second man, you've only see the game played once--when you played. Should you take the bet without knowing how the shell game man plays from game to game?
Click to expand...

In practice no. In theory if he could guarantee beforehand that he would offer you the chance to switch whatever happened then you would still have a greater chance by switching.
 
By the way, Robin, I wasn't actually addressing that shell game example to you, since you already agree that Monty's motivation as to whether or not he offers the switch actually does make a difference in the original problem. I was addressing it to those who think otherwise.
 
Robin said:
Shell game is exactly the same as the Monty Hall game. As long as the rules state that an offer to change must be made and as long as the rules are followed then it is advantageous to change your choice.

The following article contains a demonstration of this:

http://www.willamette.edu/cla/math/articles/marilyn.htm

So if you offer me 100 games and guarantee me that each time you will reveal one empty cup when I have put my finger on the choice and guarantee that I will be given the chance to switch, and I win the $50 on best of 100 then certainly I will take the bet. As long as you don't cheat it does not matter if you try to make me lose.
Click to expand...

Let me explain to you how the shell game works.

There IS NO PEA by the time the shells stop moving. You will NEVER win.

The person who offers you the bet is working with the person manipulating the shells and the pea.

Again, IT IS A CON.

If you run into someone who tries to run it like a legitimate game, then once again, it is nothing like the Monty Hall puzzle because you have a legitimate chance of knowing exactly where the pea is.

You guyses are way overthinking this.
 
SkepticJ said:
What were the odds that this thread would have over 150 posts after one day?!
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Do you want to stay with that question or take what's behind door number 2?
 
JamesM said:
Just to make sure I've got this clear:

He didn't always offer a switch.

When he offered a switch, he always opened another door.

That door was NEVER the door that had the 'real' prize in.

Correct?
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Exactly so.
 
hgc said:
Your logic is sound. The problem should be reworded to exclude the behavioral ambiguity -- for the simple version.
Click to expand...
I told you these threads are always a blast.
 
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