The Monty Hall problem

[hgc]

No, what have "they" being doing when trying to define the game? Falling over themselves with corrections, and then again.
Quarky says that this is like ddwfftw. That similarity may be because both use arguments around intuition - Monty Hall has been used to support the ddwfftw's intuition argument.

But this one is a bit different, and it's Vos Savant, who is the cult leader, and peddler of intuition.
Martin Gardner also posed the same puzzle (in a different form), and wrote a large number of puzzle books, yet there are no cases of academics calling him an idiot, nor any "Martin is Wrong" sites.

There have been many papers written about this problem, and they all refer at one time or another, to Vos Savant. Why?

Intuition has nothing to do with it. Never did. Vos Savant is often referred to because she is the popularizer of the problem. What link Gardner has with your argument, whatever is is, is mysterious. I cannot intuit your intent.

 

[humber]

Intuition has nothing to do with it. Never did. Vos Savant is often referred to because she is the popularizer of the problem. What link Gardner has with your argument, whatever is is, is mysterious. I cannot intuit your intent.

The idea of intuition has been used in support of the reason why it the chance becomes 2/3 for the other door, and then the 100 door in support of that.
There is no need to invoke probability, apart from the initial choice. The rest is a direct consequence of the process embedded within the question.
Savant uses an intuitive probability argument to justify her solution. When pushed, she makes errors because of that. Not Martin Gardner.

ETA:
Vos Savant said that if Monty knows, switch, if he doesn't, don't. I have shown that even Monty he tells the contestant all he knows, it makes no difference to the odds.

 

[Rolfe]

I need my head examined, getting back into this....

This reasoning assumes Neutral Monty, because Biassed Monty is simply an exercise in second-guessing, not a true puzzle.

One third of the time Dave has chosen right, and should not swap.
Two thirds of the time, Dave has chosen wrong, and should swap.

So, if Monty always offers him the swap, then he will win twice as often if he swaps. Fine. We know this.

However, this is only true of Monty always offers the swap, that is, if he is deliberately avoiding opening the door with the car. If he's not deliberately avoiding the car, but is opening either of the doors Dave hasn't chosen, at random, it works differently.

One third of the time Dave has chosen right and should not swap.
One third of the time Dave has chosen wrong and is shown a goat, and should swap.
One third of the time Dave has chosen wrong but gets no opportunity to swap because Monty has opened the door with the car.

So, half the time Monty offers him the swap, he had already chosen the car and should not swap, half the time Monty offers him the swap he had not chosen the car and should swap. These occasions however only comprise two-thirds of the total number of iterations - the remaining third of the time he doesn't get the chance to swap. If he gets the chance to swap, it won't make any difference whether he does or not.

So in the end the answer is swap anyway, because it can't harm his chances, and if Monty is playing the first version of the game, it will double them.

Bored now.

Rolfe.

 

[Dr.Sid]

I like this idea (from wikipedia):

Switching loses if and only if the player initially picks the car, which happens with probability 1/3, so switching must win with probability 2/3.

 

[Rolfe]

Would that it were that simple!

Rolfe.

 

[Jack by the hedge]

The falsehood of the Vos Savant argument can be seen if you think about what must actually happen. If it is true that 2/3 of the time, the car is behind the "other" door, it must actually be there, 2/3 of the time.
That is also why the 100 door argument is false.

I'm still curious to know what the above means.

In vos Savant's 3 door game, the car really is behind the unchosen, unopened door 2/3 of the time. In the 100 door variant, the car really is behind the unchosen, unopened door 99/100 of the time.

So which argument's false?


Edit to add: In vos Savant's game, the contestant indicates an initial choice, then the host must open all but one of the other doors (all of which are losing choices) and then offer the player the chance to switch. Her analysis of that game is correct (whether it has 3 doors or 100).

Humber, have you tried the 52 "door" version, with a pack of cards? 51 times out of 52, Jeeves has the ace of spades, not you.

 

[humber]

I need my head examined, getting back into this....

No, it's interesting.

This reasoning assumes Neutral Monty, because Biassed Monty is simply an exercise in second-guessing, not a true puzzle.
One third of the time Dave has chosen right, and should not swap.
Two thirds of the time, Dave has chosen wrong, and should swap.
So, if Monty always offers him the swap, then he will win twice as often if he swaps. Fine. We know this.

Not "twice as often". If you say "2/3 of the time" you are testing a deterministic logic circuit to see if it works more than once, and if tried for all the doors, to see if it is position sensitive. Not much point to that.

Monty is a machine that responds to input, and can be tested by assertion.
The rules are in the original question. He shows a goat from one of the two remaining doors - if you have chosen the car or not. Should you swap? That is how the "2/3" is arrived at.

I object that it is a change in probability.
Dave uses logic, and so does Monty. Dave can "run" Monty.

You choose Door 1
(1) If Dave asserts the car is behind Door1, and that is TRUE, then Monty will show Dave there is a goat behind door 2 or door 3. A losing strategy.

(2) If Dave asserts the car is not behind Door1, and that is TRUE, then Dave is are asserting the car is behind Door 2 or Door3. Monty will show Dave a goat is behind Door 2 or Door3, and that allows him to conclude where the car is. A winning strategy.

That's it. That works every time. If Dave fully employs (2), he will win 2 out of 3 games.
Moving to 100 doors means Monty has to open 97 doors, until he gets down to three, and his logic is useful to Dave.

However, this is only true of Monty always offers the swap, that is, if he is deliberately avoiding opening the door with the car. If he's not deliberately avoiding the car, but is opening either of the doors Dave hasn't chosen, at random, it works differently.
One third of the time Dave has chosen right and should not swap.
So, half the time Monty offers him the swap, he had already chosen the car and should not swap, half the time Monty offers him the swap he had not chosen the car and should swap. These occasions however only comprise two-thirds of the total number of iterations - the remaining third of the time he doesn't get the chance to swap. If he gets the chance to swap, it won't make any difference whether he does or not
One third of the time Dave has chosen wrong and is shown a goat, and should swap.
One third of the time Dave has chosen wrong but gets no opportunity to swap because Monty has opened the door with the car.

So, half the time Monty offers him the swap, he had already chosen the car and should not swap, half the time Monty offers him the swap he had not chosen the car and should swap. These occasions however only comprise two-thirds of the total number of iterations - the remaining third of the time he doesn't get the chance to swap. If he gets the chance to swap, it won't make any difference whether he does or not

That has nothing to do with Monty's intent, but a change to his programme which introduces a "denial of choice".
The question asks for actions. If shown a goat, should you swap?
Always.
Deviations from the original question add nothing to how it works, but introduce more variables into the process.
Monty can't "deliberately" hide the car, unless you at least sometimes equate winning under one condition, with losing under another. If Dave swaps to the closed door, and the car isn't there, he loses. If it's there, he wins.
But, in the above version, he loses if the car is there, but the door is open.

So in the end the answer is swap anyway, because it can't harm his chances, and if Monty is playing the first version of the game, it will double them.Bored now.
Rolfe.

If Dave sticks, he will get the outcome of a 1/3 random trial. That is his worst case, and Monty can do nothing to change that. Dave will win 1/3 games. That means that Monty must lose 2/3 if Dave doesn't stick.
Dave wins 2/3 times if he does not stick.

 

[humber]

I'm still curious to know what the above means.

In vos Savant's 3 door game, the car really is behind the unchosen, unopened door 2/3 of the time. In the 100 door variant, the car really is behind the unchosen, unopened door 99/100 of the time.

So which argument's false?

Edit to add: In vos Savant's game, the contestant indicates an initial choice, then the host must open all but one of the other doors (all of which are losing choices) and then offer the player the chance to switch. Her analysis of that game is correct (whether it has 3 doors or 100).

Adding more doors says nothing, because it assumes its conclusion.

Humber, have you tried the 52 "door" version, with a pack of cards? 51 times out of 52, Jeeves has the ace of spades, not you.

There is an element of voodoo about the probability argument. Monty opens a door, and the result is a change from 1/3 to 2/3 for one specific door, which seems to depend upon which door has been "chosen". I suspect that was Erdos' objection. He was a number theorist, so saw things from a more philosophical point of view than a number-cruncher may. He is right. It can't happen.

When I said that the car must actually be there for that 2/3 result to appear, there must be a mechanism to ensure that it does.

It is possible to analyze the problem using Bayes, but I think the appropriate method is Boole. Write Monty as a set of logical operators, with the only input being the initial state of one door. Call that the "chosen" door if you like.

I won't post the Boole, but may if anyone is interested, but it's like this.

In a way, the probability argument confuses position and probability.

You can say "the probability that Dave's first guess is correct, is 1/3"

But also " the car is behind"

[car_1] or [car_2] or [car_3]

[car_1] or [car_2 or car_3]

Monty's logic of showing Dave a goat, when the car is behind 2 or behind 3 [NOT Car_2] etc, makes them as if one door, so the right hand becomes,

[car_1] or [car_2 and car_3]

Although at first choice, Dave may say the unconditional probability is 1/3, it is not, because Monty will operate. From the get go, the probability is 1/3, 2/3, when [NOT car_1] is input.

Only that initial state matters. Assert the car as the initial state [car_1] and you will be right 1/3 of the time. Assert the car isn't there [NOT car_1] as the initial state, and you will be right 2/3 of the time.

It can be seem as a truth table
http://en.wikipedia.org/wiki/Truth_table

 

[zooterkin]

There is an element of voodoo about the probability argument.

Only if you don't understand probability.

Monty opens a door, and the result is a change from 1/3 to 2/3 for one specific door, which seems to depend upon which door has been "chosen".

There isn't really a change; Monty opening all the remaining doors except one means that you are effectively offered the choice to swap from choosing one door out of N to choosing N-1 doors.

 

[humber]

Only if you don't understand probability.

There isn't really a change; Monty opening all the remaining doors except one means that you are effectively offered the choice to swap from choosing one door out of N to choosing N-1 doors.

Dave goes to the three doors and says "I have a 1/3 chance with door 1, so 2/3 of the time, the car is behind one of the other doors, so I have a 1/2 chance with each" Well, no.

But Monty makes that choice logically possible, if he shows Dave a goat behind Door 1, and certain, if he shows Dave a goat behind Door 2 or Door 3. ( When the initial condition in each is that the car is not behind Door 1.)

Do something else by all means, but Monty will be of no use to you at all.

It is a fixed and deterministic process. Applying probability iteration, is to count how many times it works (p=1) versus the number of times you are willing to ignore the in-built logic, so accept chance alone. (p =1/3) Hence, 2/3.

That is to say, when the initial condition is that the car is not behind Door 1 ( or any other) you are certain to win the car if you swap (p =1) But 1/3 of the time, the car will be there, so the outcome of swapping will be 1 - 1/3 = 2/3. Law of total Probability.

The game can be modified, but then there will be cases where the same outcome means both winning and not winning the car.

ETA: See De Morgan's Law to see how a 2 input NOR gate ( Car_not there) becomes the same as a 2 input AND gate ( Car is there). That is what Monty is.
http://en.wikipedia.org/wiki/De_Morgan's_laws

Some may take me to task for the above, but that's the risk of using natural language to talk about formal logic. Reading the latter often makes Wittgenstein look like a novel.
But as Niel Innes wrote:
Neil Innes – Protest Song (‘I’ve suffered for my music, now it’s your turn.’)

 

[quarky]

I suppose its been mentioned ad nauseum, but one needs to be sure of the rules to work the data advantageously. For instance, Monty wants to keep the car.
If Monty is flexible on that, odds change.

 

[Rolfe]

If Monty can manipulate the game at all, that is try to steer the contestant towards or away from the car, the problem is pointless and nothing more than an exercise in second-guessing. Only if he is constrained to act without reference to whether or not the contestant has already chosen the car, do you actually have a logic puzzle capable of solution.

Rolfe.

 

[humber]

If Monty can manipulate the game at all, that is try to steer the contestant towards or away from the car, the problem is pointless and nothing more than an exercise in second-guessing. Only if he is constrained to act without reference to whether or not the contestant has already chosen the car, do you actually have a logic puzzle capable of solution.

Rolfe.

How does he "steer the contestant away from the car"?
If he shows the contestant the car, he immediately loses, or there are inconsistencies in the term "win".
If he "avoids the car" that allows the contestant to conclude where it is.

The "Monty Fall" version allows for Monty to accidentally kick open the door, so revealing the car. When that is included, arithmetically, the chance for the contestant is 1/2, though that does not make that much sense within the context of the game.

But there is another error, and that is to confuse outcomes, with permutations.
If you are looking for your keys in a jacket hanging in the hallway, then you can say you have a 1/2 chance of finding them on the first attempt.
The probability of finding them in the first attempt, over two sequential trials is 1/2*1/2 = 1/4
But, if you do find them, you don't bother to see if they are not there in the other pocket. Two or three actions, can yield two desired outcomes.
Exhaustively it is 1/4 for the keys, and 1/2 for the Monty Fall, but the game does not really allow for that, unless showing the car is both losing and winning at different stages of the game.

 

[quarky]

geepers.

Am I confused now?

Or am I Monty?

 

[humber]

If Monty can manipulate the game at all, that is try to steer the contestant towards or away from the car, the problem is pointless and nothing more than an exercise in second-guessing. Only if he is constrained to act without reference to whether or not the contestant has already chosen the car, do you actually have a logic puzzle capable of solution.

Rolfe.

In the standard game, Monty shows Dave a goat after Dave has chosen. It is then said that Dave can increase his chances of winning to 2/3 by swapping.
Yes, but the reason is simple. It's only the initial chance that counts.
That's where the probability is.

If Dave sticks, he wins the car p =1/3.
If he swaps he wins p = 2/3
What is the assumption about the initial probabilty of a win? One in three chance p_initial =1/3

1) Keep Door 1 as Dave's door. All else remains the same, but the car is now behind that door p = 1/2
If Dave sticks, he wins p = 1/2
If he swaps, he wins p = 1/2.

2) Keep Door 1 as Dave's door. All else remains the same, but the car is now behind that door p = 1/10
If Dave sticks, he wins p = 1/10
If he swaps, he wins p = 9/10

If Dave sticks he wins p = p_initial
If Dave swaps he wind p = 1 - p_initial

Clockwork. If there is a car, and it's not behind Dave's door it is behind the other two. Monty tells which of those two it is, no matter what the distribution to each door is. Always swap.

That is why the 100 door argument is bogus. Start with 100 doors:
Initial chance 1/100, show a goat 1/99...etc.
Keep going until there are three remaining. That is the optimum number. Two is too few.

It is odd that so much has been devoted to a problem that is a gimmick.
The assumed initial probability is the inverse of the number of doors ! ( doh)
The rest is a fixed process.

 

[zooterkin]

In the standard game, Monty shows Dave a goat after Dave has chosen. It is then said that Dave can increase his chances of winning to 2/3 by swapping.
Yes, but the reason is simple. It's only the initial chance that counts.
That's where the probability is.

If Dave sticks, he wins the car p =1/3.
If he swaps he wins p = 2/3
What is the assumption about the initial probabilty of a win? One in three chance p_initial =1/3

Yes.

1) Keep Door 1 as Dave's door. All else remains the same, but the car is now behind that door p = 1/2
If Dave sticks, he wins p = 1/2
If he swaps, he wins p = 1/2.

2) Keep Door 1 as Dave's door. All else remains the same, but the car is now behind that door p = 1/10
If Dave sticks, he wins p = 1/10
If he swaps, he wins p = 9/10

What do 1) and 2) refer to? Is this something you're quoting from somewhere else with some context missing?

If Dave sticks he wins p = p_initial
If Dave swaps he wind p = 1 - p_initial

Clockwork. If there is a car, and it's not behind Dave's door it is behind the other two. Monty tells which of those two it is, no matter what the distribution to each door is. Always swap.

Yes.

That is why the 100 door argument is bogus. Start with 100 doors:
Initial chance 1/100, show a goat 1/99...etc.
Keep going until there are three remaining. That is the optimum number. Two is too few.

No. Monty opens all doors except one, not counting the one you have chosen.

It is odd that so much has been devoted to a problem that is a gimmick.
The assumed initial probability is the inverse of the number of doors ! ( doh)
The rest is a fixed process.

It is odd that you present the correct answer and gibberish in the same post.

 

[Rasmus]

That is why the 100 door argument is bogus. Start with 100 doors:
Initial chance 1/100, show a goat 1/99...etc.

Wait, what...?

Keep going until there are three remaining. That is the optimum number. Two is too few.

No, the best that can happen to the contestant is Monty opening all doors but the one he's chosen initially and one other door. (Okay, it would be better to just be given the car, but that would defeat the purpose of the puzzle.)

Above, you are just rambling on and actually skipping over the important points.

Assume 100 doors. As soon as the contestant picks one and Monty opens another door to reveal a goat - what exactly do you think is now 1/99?

The chances of the contestant having picked the car are *still* 1/100 assuming that Monty did not open a door randomly but was restricted to revealing a goat. If Monty then reveals a second goat the chances that the contestant picked the car are *still* 1/100.

... continue until Monty has opened 98 doors to reveal 98 goats: The chances that the contestant has initially chosen the car are still at 1/100.

Should he swap for the last other remaining door? Yes, yes he should! Only one time out of a hundred games will Monty have picked a door at random that he will leave closed to the very end. 99 times he will not have had a choice about which door to leave closed.

Interestingly enough, I think that if Monty offers to switch at any time it would be advisable to switch, since every revealed goat makes "all the other doors" look slightly better than "all the doors" you were initially choosing from. But I haven't done the math for that.

Now, it will get really interesting if Monty keeps offering you to switch your doors every time he reveals another goat. Especially once you reach the point where you can no longer chose a door that you haven't played with already! (Did I just come up with a new twist to the problem?)

 

[Jack by the hedge]

1) Keep Door 1 as Dave's door. All else remains the same, but the car is now behind that door p = 1/2
If Dave sticks, he wins p = 1/2
If he swaps, he wins p = 1/2.

No. That's wrong. You are either making a mistake or quoting something which is describing a different game.

Regarding changing probabilities, consider the 52 card version again for a moment:

If you pick a card (but don't look yet) you have a 1/52 chance of having the ace of spades.

If Jeeves then looks through the remaining pack and, one by one, begins discarding cards which are not the ace of spades, what happens to your probability of having the ace? It remains 1/52. Jeeves keeps going until he has one card left. What's the probability you have the ace now? Still 1/52. Nothing about your initial choice has changed. Nothing about Jeeves' actions gives you any useful information about your choice.

You can try this for yourself, and you don't even need Jeeves' help. Just pick a card. If it's not the ace of spades, you can infer that Jeeves would have whittled down his pack to the ace. You only win with probability 1/52.

 

[Dave Rogers]

How did this get to be all about me?

It seems to me that, if there is a strictly deterministic process, in which the only choice that's ever available to Monty after I've chosen a door is which goat to reveal if I happen to have chosen the car, then switching is always a better strategy. If Monty has the freedom to choose a course of action after I've made my initial choice, then the problem becomes one of game theory rather than one of probability, and I need to understand Monty's aims in order to determine the best strategy. So what it all boils down to is the rather obvious conclusion that, in order to analyse a game, it's necessary first to know what game you're playing.

Dave

 

[Dave Rogers]

Interestingly enough, I think that if Monty offers to switch at any time it would be advisable to switch, since every revealed goat makes "all the other doors" look slightly better than "all the doors" you were initially choosing from. But I haven't done the math for that.

Provided he always opens a door and offers a switch, then it's always better to accept. Suppose there are 100 doors initially, then my initial choice has a 1% chance of being correct. Once Monty reveals a goat, the remaining 98 doors have a cumulative 99% probability of being correct, so the probability of any one of them being correct is 0.99/98, or about 1.01%. The general case is that the initial door chosen from n has a 1/n chance, whereas the switch after eliminating one door has a chance of (n-1)/{n(n-2)}. This is the initial probability multiplied by (n-1)/(n-2), which must always be greater than 1.

Dave