The Monty Hall problem

[humber]

You might find it to your advantage to observe what befell previous competitors.
Probability is not about frequency?
Jeeves is playing a different variant.
The probability of winning by switching with Evil Jeeves is zero, since Evil Jeeves only opens a door if you originally guess right.
All variants make assumptions. The 100 door version only makes the same assumptions as the familiar 3 door version with the 1/3 vs 2/3 probabilities. i.e. after you choose, Jeeves will open all but one of the remaining doors and will not reveal the prize.

Intuition helps people grasp this version as intuition tells them a) their first guess is almost certainly wrong and b) there's probably something special about the door Jeeves doesn't open. And 99 times out of 100 their intuition will be right.

All wrong. Try reading.

 

[Jack by the hedge]

"All wrong" is trivially disproven:

Evil Jeeves only opens a door if you originally guess right. Therefore if evil Jeeves opens a door and you switch, then you must now be wrong. QED

I presume "try reading" is your equivalent to saying you have no substantive response.

 

[AvalonXQ]

Evil Jeeves only opens a door if you originally guess right. Therefore if evil Jeeves opens a door and you switch, then you must now be wrong.

As has already been pointed out, this may not be the best strategy for Evil Jeeves. If, for instance, Evil Jeeves opens a door 1/3 of the time when you guess wrong and 2/3 of the time when you guess right, and this is known, it might provoke some players who guessed right to switch anyway.

 

[humber]

"All wrong" is trivially disproven:
Evil Jeeves only opens a door if you originally guess right.
Therefore if evil Jeeves opens a door and you switch, then you must now be wrong. QED

Oh, I also have another Evil Jeeves. That is my puzzle.

If Evil Monty has a bias to one of two doors ( he must have if he includes the contestants door, if the original choice is right), then the chance that the car is not behind the third door = 1/2. The contestant can't know that Monty knows he has selected the car, but it is still advantageous to swap, because the unconditional probability is 2/3. All conditional probabilities are at least 1/2.
All covered by game theory, and Bayes too.

Oh, I see that he opens the door, if the contestant guess right? Thanks Jeeves, there's the car I was looking for...

 

[Dave Rogers]

Oh, I also have another Evil Jeeves.

Irrelevant. You're claiming that you've established a rule to which there is no counter-example. This claim is refuted by the counter-example of the first order Evil Jeeves. Citing further supporting examples will not invalidate the counter-example.

Dave

 

[humber]

Irrelevant. You're claiming that you've established a rule to which there is no counter-example. This claim is refuted by the counter-example of the first order Evil Jeeves. Citing further supporting examples will not invalidate the counter-example.

Dave

It is not invalidated by Evil Jeeves.

The matter really only concerns the original problem,. But, as I explained the worst case for the contestant is 1/2. Evil Monty risks that, and the chance that the contestant will gain 2/3 by swapping, or retain 1/2 by swapping. If you want him to show the car to the contestant, then the contestant wins. Asked to swap? Refuse. You now need to say if the contestant wins only if the door is closed. That is a problem for Evil Monty, because he has to take part. If he doesn't the contestant still has a 1/3 chance.
As for Jeeves, I can make a variant where he can't win. Within two steps, and under all cases, I find the car 100% of the time.

 

[Rolfe]

That's because it is gibberish.

This is a poster that has posted tens of thousands of times - literally, no exaggeration - over the last two years or more on the topic of DDWFTTW, and still denies it's possible despite overwhelming evidence to the contrary (including a world record).

I don't know what his (?) deal is, but I advise you to ignore his posts unless they entertain you.

Not entertained, I have to say.

I find discussion of biassed Monty to be pointless. If he is allowing his knowledge of whether or not the correct door was selected first time to influence what he does next, it stops being a solvable puzzle and becomes an exercise in second-guessing.

I like the formulation of the discussion to focus on Monty's choice whether to offer the switch or not, rather than randomness in picking doors.

Rolfe.

 

[humber]

Not entertained, I have to say.
I find discussion of biassed Monty to be pointless. If he is allowing his knowledge of whether or not the correct door was selected first time to influence what he does next, it stops being a solvable puzzle and becomes an exercise in second-guessing.

I like the formulation of the discussion to focus on Monty's choice whether to offer the switch or not, rather than randomness in picking doors.

Rolfe.

I would not listen to Sol. He is bitter because he gets beaten.

All variants are all soluble.

If you think they are not, then you should accept that solutions based on multiple iterations cannot force an exact solution, and are largely irrelevant.

You mentioned Monty's motive, but that is again only ( perhaps) a consideration in the case of multiple iterations. The question does not concern that, but only should the contestant swap or not, using the given information.

The problem is generally presented like this:
A car is equally likely to be behind any one of three doors. You select one of the three doors (say, Door #1). The host then reveals one non-selected door (say, Door #3) which does not contain the car. At this point, you choose whether to stick with your original choice (i.e. Door #1), or switch to the remaining door (i.e. Door #2). What are the probabilities that you will win the car if you stick versus if you switch?

The usual solution is that there is an initial 1/3 chance that the contestant has picked the car. The goat is revealed, so the probability that the car is in remaining door is 2/3. (Why would you find that argument more convincing if there were 100 doors?)

Then there is this "accidental version"
In this variant, once you have selected one of the three doors, the host slips on a banana peel and accidentally pushes open another door, which just
happens not to contain the car. Now what are the probabilities that you will win the car if you stick with your original selection, versus if you switch to the remaining door?

No. It does not matter to the contestant how he learns that the car is not there.

But, it is possible to consider that case.
Suppose you select Door 1, and the host accidentally opens Door 3. The probabilities that Door 3 happens not to contain a car, if the car is behind Door 1, 2, 3, are respectively 1, 1, 0.

The general probabilities for the contestant (car is actually behind the door) are 1/2, 1/2, and 0.
Now there is no advantage in swapping. That is contrary solution that says that method is wrong.

 

[Jack by the hedge]

That appears to be simply restating the vos Savant explanation. And she's still right about the 100 door version.

 

[humber]

That appears to be simply restating the vos Savant explanation. And she's still right about the 100 door version.

Three doors. Contestant does not choose. Monty opens one door, reveals goat. Common reply 50/50 for the contestant.
What makes you think you can keep the same initial value - or a final value of 1/3 - for the contestant's choice, as the doors are opened one by one, unless you believe in magic?

 

[Jack by the hedge]

Three doors. Contestant does not choose. Monty opens one door, reveals goat.

That is a different game.

What makes you think you can keep the same initial value - or a final value of 1/3 - for the contestant's choice, as the doors are opened one by one, unless you believe in magic?

100 doors. You choose one. 99% likelihood you are going to be wrong.

Monty starts opening other doors, revealing goats. What makes you think you are not still 99% probably wrong, unless you believe in magic?

Monty ends up with your choice plus one door. Swap. 99 times out of 100 you win the car.


Another example you can try yourself:

Shuffle a pack of cards. You're trying to find the ace of spades. Pick one but don't look.

Now ask your butler to remove 50 cards from the pack which are not the ace of spades.

So now there are 2 cards left, one of which is the ace. Your card has 1/52 chance of winning and Jeeves' card has 51/52 chance. Do you want to swap?

This version can be simplified as follows: Pick a card. Is it the ace of spades? No? OK, then you know that the card Jeeves would have ended up with would have been the ace so you should have swapped.

Try again - shuffle then pick a card - is it the ace of spades?
Try again - shuffle then pick a card - is it the ace of spades?
Try again - shuffle then pick a card - is it the ace of spades?

Do you get it now?

 

[humber]

That is a different game.

Really?
Doors closed. Contestant chooses. Door opens. Contestant reconsiders probability.
Second time. A door is open, and then another. Contestant reconsiders probability.

 

[Jack by the hedge]

Yes really.

Game a) Contestant chooses first - contestant has 1/3 chance of being right.
Monty opens a losing door - contestant swaps and has 2/3 chance of winning.

Game b) Monty opens a losing door first - contestant has 1/2 chance of choosing right.

Different game.

So, do you understand the 100 door variant now?

 

[humber]

Yes really.

Game a) Contestant chooses first - contestant has 1/3 chance of being right.
Monty opens a losing door - contestant swaps and has 2/3 chance of winning.

Game b) Monty opens a losing door first - contestant has 1/2 chance of choosing right.

Different game.

So, do you understand the 100 door variant now?

Never mind.

 

[zooterkin]

Yes really.

Game a) Contestant chooses first - contestant has 1/3 chance of being right.
Monty opens a losing door - contestant swaps and has 2/3 chance of winning.

Game b) Monty opens a losing door first - contestant has 1/2 chance of choosing right.

Different game.

So, do you understand the 100 door variant now?

Never mind.

I think you'll find it's spelt, "No".

 

[Rolfe]

Bored now.

Rolfe.

 

[Rolfe]

Well, maybe not that bored.

I was just thinking about a different formulation of the problem.

You are on a game show, and Monty shows you three closed doors. Behind one there is the prize, a car, and behind the other two there are booby-prizes, in this case goats. You are invited to choose a door, and do so.

Monty has two basic options now. He can open a door so as to end the game then and there - that is, either by opening the door you chose and reveal whether you won or not, or opening one of the doors you didn't choose, to reveal the car (too bad, you lost). Alternatively, he can open one of the doors you didn't choose, and reveal a goat, and if he does this, he will offer you the opportunity to switch your choice to the other unopened door.

Monty is operating under only one constraint. He has to make up his mind whether or not he will offer you the chance to switch, before you choose your door.

Should you switch?

Discuss.

Rolfe.

 

[epepke]

I see your point, but that wasn't really the angle I was looking at. Monty may vary the frequency with which he offers the switch as much as he likes - the only stipulation is that he can't base his decision on whether or not the switch is offered to any given contestant on whether or not the contestant already picked the prize door.

So, if you aren't offered the switch, your chance of winning is 1/3rd. However, if you are offered the switch, and you switch, your chance of winning is 2/3rds.

Overall, going into the game, the chance that you will win depends on how often Monty is deciding to offer the switch.

Rolfe.

I'll now demonstrate why it is I no longer try to explain this by trying to explain it and watching it fail. It's been said many times in this and countless other discussions. It hasn't worked then, so the probability that it will work now seems to me pretty low. I am trying to make it as simple as possible, so you don't call me a troll. In my defense, I think this is something that skeptics should understand to protect themselves against cheaters and confidence tricksters, who take advantage of people's willingness to jump to conclusions.

If Monty's strategy is to reveal a goat when and only when the contestant originally picked the curtain with the car, then the strategy of always switching will always result in ]b]a probability of 0 of winning the car.[/b] That is, the strategy strongly implied by your statement, to wit: "However, if you are offered the switch, and you switch, your chance of winning is 2/3rds."

 

[Rolfe]

I don't think you read the post you quoted.

the only stipulation is that he can't base his decision on whether or not the switch is offered to any given contestant on whether or not the contestant already picked the prize door.

The explicit stipulation was made that Monty was not allowed to let his knowledge of whether or not the contestant had already picked the car to influence his decision on whether or not to offer the switch.

I was trying to make the point that the puzzle is only solvable if you make that stipulation. If the terms allow for Biassed Monty (either Helpful Monty or Evil Monty) you simply start down a never-ending chain of second-guessing, and it's pointless.

Rolfe.

 

[epepke]

I don't think you read the post you quoted.

Fair enough.