Monty Hall Problem: Win by Switching

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SUMMARY

The Monty Hall problem demonstrates that switching doors after one is revealed significantly increases the probability of winning the car from 1/3 to 2/3. The host's knowledge of the door contents is crucial; he will always reveal a goat, thus providing additional information that skews the probabilities. When a contestant initially picks a door, they have a 2/3 chance of selecting a goat. Therefore, switching doors after one goat is revealed is the optimal strategy. This conclusion is supported by exhaustive probability analysis and simulations.

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lowerlowerhk
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http://en.wikipedia.org/wiki/Monty_Hall_problem

This is a game about probability:
Say there is three doors. Two goats and a car are hidden behind.
Wanting to pick the car, you randomly picked a door.
Then the game host open one of the unchoosen door which happened to contain a goat.
And the host offer you a chance to switch door - to switch to the remaining unchoosen door.
The question (not my question though) is, should you switch door?

The correct answer is, switching is always better because the only way you could lose the game by switching is to pick the correct door at the first place, which is unlikely.

But I don't understand how the probabilities transfer during the host's action. Let's say the host is going to open a door which he knows a goat is inside. To me, before the door is opened it contain some probabilities to be a car. After the door is opened, that probability is transferred away. But why the car's probability only goes to the unchoosen door instead of distributed evenly between all unopened doors?
 
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Did you look at the diagram at the bottom?

You pick goat 1, Host shows Goat two, switch->you win
You pick goat 2, Host shows Goat one, switch->you win
You pick car, Host shows either goats, switch-> you lose

2 wins out of 3


In the beginning you have a 2/3 chance of picking a goat
 
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Doing an exhaustive list does yield the result, but I am not asking for that. I am asking why the probability transfers to one door only.
 
It said the probability of the unchoosen door increased, but it didn't tell why doesn't the probability of the choosen door increase accordingly. May be I missed something?
 
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lowerlowerhk said:
It said the probability of the unchoosen door increased, but it didn't tell why doesn't the probability of the choosen door increase accordingly. May be I missed something?
The probability of the original chosen door is 1/3. Opening a door when the host knows is not the car (and at least one of the two remaining has a goat) does not give you any more information about the door you chose. However, the fact that the host opened one of the other doors revealing a goat does give you more information about the remaining door.
 
The crucial thing is (which makes the probability "change") is that the host knows which door the goats are behind, and will only open a door with a goat behind. So you get more information about the game after he opens a door. He would never open a door with a car behind. If we randomly chose a door, it wouldn't matter if you chose the other closed one.
 
disregardthat said:
The crucial thing is (which makes the probability "change") is that the host knows which door the goats are behind, and will only open a door with a goat behind. So you get more information about the game after he opens a door. He would never open a door with a car behind. If we randomly chose a door, it wouldn't matter if you chose the other closed one.

Exactly. If the host had chosen a door arbitrarily, where there could have been a car, then switching would not make a difference.

Note the considerable difference between this and Deal or No Deal, where the case is chosen at random, without the "secret knowledge" seen in the Monty Hall problem. So switching cases does not make a difference.
 
  • #10
Another thing which is crucial is knowing that the host would open a door and ask you whether you wanted to switch regardless of what door you chose. If he did that only when you chose the car, you would always lose.
 
  • #11
One thing I never understood about the phrasing of this problem is that it seems necessary to make the choice to switch before the offer is even made to obtain the "edge" in the solution. During the second round of the game he is effectively asking you to make a choice between two doors, changing the problem.
 
  • #12
daveyp225 said:
One thing I never understood about the phrasing of this problem is that it seems necessary to make the choice to switch before the offer is even made to obtain the "edge" in the solution. During the second round of the game he is effectively asking you to make a choice between two doors, changing the problem.

How can you switch before the offer is made? Suppose you choose the first door -- you can't really "switch" since both the other two doors are both still closed. The advantage of switching comes from that you know that Monty knows which door hides which item, and so he was deliberate in his choice of which doors contents to reveal when making the offer to switch.

The second round of the game contains more information than (often) considered at first. So it's not exactly the same as randomly choosing a door from two in hopes that you win. You also know that (1) if you picked a goat, he would reveal the (only other) remaining goat; (2) if you picked the car, he would arbitrarily choose (either of) the remaining goats.
 
  • #13
Anonymous217 said:
Exactly. If the host had chosen a door arbitrarily, where there could have been a car, then switching would not make a difference.

Note the considerable difference between this and Deal or No Deal, where the case is chosen at random, without the "secret knowledge" seen in the Monty Hall problem. So switching cases does not make a difference.
I don't understand this...

Suppose Monty chose a door random. Before he opens the door, it may or may not have the car behind it. However, once he opens the door, we will know whether the car is behind it or not. If there isn't a car behind it, doesn't the situation revert back to the classic Monty Hall problem? How could it be different merely because of the method Monty used to choose a door? How does that express itself in the problem?
 
  • #14
Dr. Seafood said:
How can you switch before the offer is made? Suppose you choose the first door -- you can't really "switch" since both the other two doors are both still closed. The advantage of switching comes from that you know that Monty knows which door hides which item, and so he was deliberate in his choice of which doors contents to reveal when making the offer to switch.

The second round of the game contains more information than (often) considered at first. So it's not exactly the same as randomly choosing a door from two in hopes that you win. You also know that (1) if you picked a goat, he would reveal the (only other) remaining goat; (2) if you picked the car, he would arbitrarily choose (either of) the remaining goats.

I said that one must choose to switch before the offer is made, not switching before it is made. I understand the probability of making the choice to switch before he opens the door and being correct is 2/3. However, if I had suddenly incurred amnesia after the first step of the problem, and was presented with either keeping door 1 versus choosing door 2 over 1, how is the probability not now 1/2? Also, how is the Monty Hall problem any different than one with 4 doors, but one false door already open?
 
  • #15
daveyp225 said:
I said that one must choose to switch before the offer is made, not switching before it is made. I understand the probability of making the choice to switch before he opens the door and being correct is 2/3. However, if I had suddenly incurred amnesia after the first step of the problem, and was presented with either keeping door 1 versus choosing door 2 over 1, how is the probability not now 1/2?
Because in this scenariio, you have completely dropped the third door, reducing the problem to one out of two equally likely possibliities: probability of either, 1/2.

Also, how is the Monty Hall problem any different than one with 4 doors, but one false door already open?
Not at all different. That "fourth door" does not matter.
 
  • #16
Jocko Homo said:
I don't understand this...

Suppose Monty chose a door random. Before he opens the door, it may or may not have the car behind it. However, once he opens the door, we will know whether the car is behind it or not. If there isn't a car behind it, doesn't the situation revert back to the classic Monty Hall problem? How could it be different merely because of the method Monty used to choose a door? How does that express itself in the problem?

Not at all. The fact that there isn't a car behind it doesn't translate to Monty having the "secret knowledge" necessary to affect the probability. Given you chose a goat, it was merely 50% chance that he picked out the other goat. However, with the secret knowledge, it's 100% chance he picks out the other goat.
It's hard to elaborate conceptually since it's more or less intuitive. If all else fails, try an exhaustive list again. In terms of this list, note that the 50% causes more possibilities to arise, and these extra possibilities tie into make the 50% overall mark of switching.
 
  • #17
HallsofIvy said:
Because in this scenariio, you have completely dropped the third door, reducing the problem to one out of two equally likely possibliities: probability of either, 1/2.
I still don't get it. Monty opening one of the doors has changed the probability of both the door you didn't pick and the door you did pick.

So how does switching increase the likelihood of success?

Bah, I'm just going to brute force this problem and prove one way or 'tother once and for all...
 
  • #18
Monty Hall
A: goat B: goat C: car

1 You choose A
1.1 Monty opens B
1.1.1 You stay with A, You lose
1.1.2 You switch to C, You win

2 You choose B
2.1 Monty opens A
2.1.1 You stay with B, You lose
2.1.2 You switch to C, You win

3 You choose C
3.1 Monty opens A
3.1.1 You stay with C, You win
3.1.2 You switch to B, You lose
or
3.2 Monty opens B
3.2.1 You stay with C, You win
3.2.2 You switch to A, You loseConclusion: It makes no difference whether you switch or not. If you switch, you have a 50/50 chance of winning the car. If you do not switch, you have a 50/50 chance of winning the car. The fact that your first choice had a 1/3 chance of winning is a red herring, and is what keeps tripping people up.
 
  • #19
And yet...

...it's wrong.


Here's a simulator (only works in IE)
http://www.grand-illusions.com/simulator/montysim.htm

It shows manifestly that switching increases your wins from 1-in-3 to 2-in-3.

Logic has let me down. I am going to go join a church. Goodbye PF.
 
  • #20
Ok, suppose there are ten billion doors. You open one, and then monty reveals 9,999,999,998 goats. Still fifty-fifty? If so, please come to my new casino.
 
  • #21
Dave, your error is assuming that those outcomes are equally likely. They're not.
 
  • #22
zhentil said:
Ok, suppose there are ten billion doors. You open one, and then monty reveals 9,999,999,998 goats. Still fifty-fifty? If so, please come to my new casino.
No one suggested that.
 
  • #23
Better yet, I have a random number generator. you guess a number, then I tell you another number, which is the correct answer unless you guessed right. Want to switch?
 
  • #24
zhentil said:
Dave, your error is assuming that those outcomes are equally likely. They're not.
Show me the flaw in my simulation. I don't see it.
 
  • #25
zhentil said:
Better yet, I have a random number generator. you guess a number, then I tell you another number, which is the correct answer unless you guessed right. Want to switch?

Your analogies are lousy.
 
  • #26
DaveC426913 said:
No one suggested that.
Apply your enumerative proof to it. You get fifty fifty.
 
  • #27
DaveC426913 said:
Show me the flaw in my simulation. I don't see it.
Your simulation is flawless. Just tell us how you got fifty fifty out of it. That's the incorrect part.
 
  • #28
zhentil said:
Your simulation is flawless. Just tell us how you got fifty fifty out of it. That's the incorrect part.
It's a binary tree of actions. As many lead to wins as losses.

I concede (after running the simulation I linked to) that, not only is my conclusion flawed, but even my logic is flawed, since the simulation shows that in no case are the odds 50/50 - they are either 1-in-3 or 2-in-3.
 
  • #29
DaveC426913 said:
It's a binary tree of actions. As many lead to wins as losses.
So the probability you picked door c is one half?
 
  • #30
zhentil said:
So the probability you picked door c is one half?

There is no such thing as a probability of an event that has already occurred. Care to rephrase? :wink:
 

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