Monty Hall Problem... For Newbies

[jt512]

The advantage of switching in the Monty Game can be quickly and easily demonstrated in practice using two black and one red playing cards.

Or by writing a few lines of code.

 

[ynot]

Or by writing a few lines of code.

Some people might not trust that evil code stuff.

“Monty” doesn’t even need to know the positions of the cards or reveal where a black card is after the first choice. A person that doesn’t switch is always staying with their first choice so all you have to do is see how often their first choice is correct. You will find (within the laws of average) the first choice will be correct one time in three which means switching would have been correct two times in three. Pretty sure that makes always switching a more winning strategy.

 

[ynot]

Here’s a coded version of the game for those that don’t think math is evil - http://betterexplained.com/articles/understanding-the-monty-hall-problem/

Keep clicking "Your Door" and see how well you do.

Cool thing is you can increase the amount of doors to better highlight the always switch advantage.

 

[Roboramma]

Largely unargued. But... Monty doesn't offer you both doors, and I can only speak from my conceptual framework, but claiming he's offering you both the goat and the (2/3 car)-door doesn't wash.

Before addressing this, let me first try to explain the "he's offering you both doors" argument in a way that I think should make it pretty clear:

Let's say you start out playing a different game, in which you choose your door, and then Monty doesn't show you a goat, but instead offers you to either stay with the door you choose or take whatever is behind both other doors. Let's further simplify things by getting rid of the goats entirely: there are three doors, one of which has a car, and the others have nothing behind them.

You play this way for a while, and, pretty obviously you find it a good idea to switch from your original choice to the other two doors.

After a while Monty has some fun with you: after you switch your choice, when it's time for the reveal, he opens an empty door first, then, opens the other door. If it's a car, great, if not he opens an empty door again (obviously). Either way there will always be an empty door to open.

At this point it's clear the game hasn't changed. You come to realise that every time you play he's going to open an irrelevant empty door before opening the door that matters.

Finally he switches it up again and opens that irrelevant door before you switch. But you already knew he was going to open it anyway. In what way has he changed the game?

He's also, at the same time and equally ('50/50') offering you your door plus the goat door. You can change to the other door and see a goat, or keep your door and see a goat. 50/50, right? Wrong - so, with the best will in the world, the explanation must be wrong. Right?

Your original choice of one door in three is clearly unaffected by Monty showing you a goat, so he's not offering you the original door plus a goat.

Which of the two remaining doors you choose from, though, is affected by him showing you the goat. So "equally" is wrong.

 

[gnome]

Of course, everyone has their own way of making this click.

Try this one on for size:

You pick a door initially. Monty offers you the choice, with no assistance, to pick one of the other two doors instead. This doesn't help you--since you don't know any more than before. If it is clear that there is exactly one car and two goats, your odds of improving your situation is 50-50 no matter which door you pick.

Having wrapped your head around that one, suppose before you make your decision, Monty offers to eliminate one of the goat doors for you among the remaining two. Like removing bad answers in "Who wants to be a millionaire"... would you ask him to do so?

Switching after he does that is only a bad choice if both of the remaining doors are goats--which is true exactly 1/3 of the time.

 

[jt512]

Of course, everyone has their own way of making this click.

Try this one on for size:

You pick a door initially. Monty offers you the choice, with no assistance, to pick one of the other two doors instead. This doesn't help you--since you don't know any more than before. If it is clear that there is exactly one car and two goats, your odds of improving your situation is 50-50 no matter which door you pick.

No. By blind switching, your odds of improving, given that you have a goat, are 50:50; that is, a probability of 1/2. But you don't know if you have a goat; you might have a car. Therefore, by blind switching, your probability of improving (Goat → Car) is only 1/3. In addition, your probability of no change (Goat → Goat) is 1/3, and your probability of worsening (Car → Goat) is 1/3.

 

[HumptyDumpty]

I think everyone agrees that the probability of guessing the right door in the beginning are 1/3.

Yes

That choice was made without knowing which door Monty is going to reveal, so that information does not alter the odds of winning if you stay.Therefore the odds of winning if you stay remain 1/3 and the odds if you switch have to be 1-1/3 = 2/3.

Not necessarily.

Odds of 50/50 would require that the information obtained by revealing the door with the goat is used when selecting the door in the first place. That can only happen if the goat is revealed prior to selecting the door in the first place - but it isn't, therefore that information cannot alter the original 1/3 odds.

The original probability of 1/3 that the door you picked contains the car can change, it depends on the 'algorithim' Monty uses for deciding which goat door to open when he has a choice. It only stays at 1/3 if Monty picks at random when he has a choice - which isn't stated in the OP, although it's a reasonable assumption to make.

 

[lionking]

Yes



Not necessarily.



The original probability of 1/3 that the door you picked contains the car can change, it depends on the 'algorithim' Monty uses for deciding which goat door to open when he has a choice. It only stays at 1/3 if Monty picks at random when he has a choice - which isn't stated in the OP, although it's a reasonable assumption to make.

The MH puzzle requires him to know what is behind each door. It's some other, rather trivial, problem without this condition.

 

[HumptyDumpty]

The HM puzzle requires him to know what is behind each door. It's some other, rather trivial, problem without this condition.

I am assuming Monty knows what's behind each door and that he is guaranteed to open a goat door - my point still stands.

 

[lionking]

I am assuming Monty knows what's behind each door and that he is guaranteed to open a goat door - my point still stands.

Well then, if you switch the probability improves to 2/3. Necessarily.

 

[HumptyDumpty]

Well then, if you switch the probability improves to 2/3. Necessarily.

Say Monty has a preference for opening the middle door (providing it contains a goat, and you didn't pick it), and say you pick Door1.

After Monty opens his (preferred)middle door the probability Door1 contains the car is 1/2, as is the probability Door3 contains the car. So no, not necessarily, but only under the assumption that given a choice of 2 goat doors to open he picks one at random - which as I said isn't stated in the OP.

 

[Prometheus]

Yes



Not necessarily.



The original probability of 1/3 that the door you picked contains the car can change, it depends on the 'algorithim' Monty uses for deciding which goat door to open when he has a choice. It only stays at 1/3 if Monty picks at random when he has a choice - which isn't stated in the OP, although it's a reasonable assumption to make.

Monty only has a choice if you've picked the door with the car on your first try. If you guessed wrong at first, then the car is behind one of the two unopened doors, thereby constraining Monty to open the other. So, if Monty has a choice, then no matter which algorithm he uses, the probability of the first door you picked hiding a car is necessarily 1.

 

[bruto]

I am assuming Monty knows what's behind each door and that he is guaranteed to open a goat door - my point still stands.

I think that's correct, and I made a poorly stated and poorly conceived argument heading in that direction earlier. If Monty has a choice of doors to open, it must be random. If, say, Monty always opens door 3 unless you've chosen it then we can presume that door 3 contains no car. You'd have to observe numerous instances, I think, in order to determine this, though. If door 3 is a dummy door and never contains a car, then if you choose door 3, you'll never win if you stick, and always win if you change.

 

[HumptyDumpty]

Monty only has a choice if you've picked the door with the car on your first try. If you guessed wrong at first, then the car is behind one of the two unopened doors, thereby constraining Monty to open the other. So, if Monty has a choice, then no matter which algorithm he uses, the probability of the first door you picked hiding a car is necessarily 1.

And your point is?
The example I gave is accurate - Doors 1 and 3 are equi-probable of containing the car.

 

[Roboramma]

The original probability of 1/3 that the door you picked contains the car can change, it depends on the 'algorithim' Monty uses for deciding which goat door to open when he has a choice. It only stays at 1/3 if Monty picks at random when he has a choice.

What makes you think this? Please explain.

 

[ynot]

What makes you think this? Please explain.

I would also like an explanation rather than just an assertion.

What magical process can change the odds of a choice when that choice doesn’t change in any way?

 

[ynot]

And your point is?
The example I gave is accurate - Doors 1 and 3 are equi-probable of containing the car.

Prove to yourself you are correct with this online version of the game - http://betterexplained.com/articles/understanding-the-monty-hall-problem/

Say Monty (the code) prefers to open the middle door. As the contestant playing along with Monty's preference you only ever choose door one or two. Obviously being merely code Monty doesn't know it's meant to always choose the middle door and half the time will choose an available end door. But you can fix that error by only using the times Monty chooses the middle door and ignore the times when it doesn't. Do this and stick with your original choice every time and I guaranteed you will wrong more than you are right. Not right half the time as you seem to be claiming.

 

[Modified]

I would also like an explanation rather than just an assertion.

What magical process can change the odds of a choice when that choice doesn’t change in any way?

Your relevant knowledge can change, as explained above, but only if you are aware that Monty is not choosing randomly and know his preference.

 

[HumptyDumpty]

Your relevant knowledge can change, as explained above, but only if you are aware that Monty is not choosing randomly and know his preference.

Very true.

 

[ynot]

I just played the Monty's Preference Game myself and out of 10 games where Monty revealed the goat in the preferred middle door my original choice was wrong 7 times (as expected).