Monty Hall Problem: Win by Switching

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In summary, the Monty Hall problem is a game of probability where the player must choose between three doors, two of which hold goats and one holds a car. The host then reveals one of the remaining doors to have a goat and offers the player a chance to switch to the other unchosen door. The correct answer is to always switch, as the only way to lose is by initially choosing the correct door. The probability of the unchosen door increases because the host knows which door contains the car and will not open it, giving the player more information. This is different from other similar games where the outcomes are chosen at random without any secret knowledge.
  • #1
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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  • #2
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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  • #3
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.
 
  • #5
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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  • #6
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.
 
  • #8
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.
 
  • #9
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. :grumpy:
 
  • #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:
 
  • #31
DaveC426913 said:
There is no such thing as a probability of an event that has already occurred. Care to rephrase? :wink:
Schrodinger's cat does not appreciate your jest.
 
  • #32
Dave, I have modified your list to include probabilities. Some of these are conditional probabilities; for "1.1 Monty opens B (1)", (1) means this would always happen given that "1 You choose A"
There is an equation for the probability of each outcome. Winning outcomes are in bold.

1 You choose A (1/3)
1.1 Monty opens B (1)
1.1.1 You stay with A (1), You lose, probability = 1 * 1 * (1/3) = 1/3
1.1.2 You switch to C (1), You win, probability = 1 * 1 * (1/3) = 1/3

2 You choose B (1/3)
2.1 Monty opens A (1)
2.1.1 You stay with B (1), You lose, probability = 1 * 1 * (1/3) = 1/3
2.1.2 You switch to C (1), You win, probability = 1 * 1 * (1/3) = 1/3

3 You choose C (1/3)
3.1 Monty opens A (1/2)
3.1.1 You stay with C (1), You win, probability = 1 * (1/2) * (1/3) = 1/6
3.1.2 You switch to B (1), You lose, win probability = 1 * (1/2) * (1/3) = 1/6
or
3.2 Monty opens B (1/2)
3.2.1 You stay with C (1), You win, probability = 1 * (1/2) * (1/3) = 1/6
3.2.2 You switch to A (1), You lose, probability = 1 * (1/2) * (1/3) = 1/6

So, the total probability of winning with the strategy of staying is 1/6 + 1/6 = 1/3.
The total probability of winning with the switch strategy is 1/3 + 1/3 = 2/3.
 
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  • #33
^ Read the second post at the top of this page and was going to post exactly what you did. Thank you.
 

1. What is the Monty Hall Problem?

The Monty Hall Problem is a famous probability puzzle named after the host of the game show "Let's Make a Deal". It involves a game where there are three doors, behind one of which is a prize. The player chooses one door, and then the host reveals one of the other doors that does not have the prize. The player is then given the option to switch their choice to the remaining unopened door. The question is whether or not the player should switch in order to increase their chances of winning the prize.

2. What are the three possible outcomes of the Monty Hall Problem?

The three possible outcomes are: 1) The player chooses the door with the prize on their first try, 2) The player chooses a door without the prize on their first try, and then switches to the door with the prize, and 3) The player chooses the door without the prize on their first try, and then switches to the other remaining door without the prize.

3. What is the probability of winning if you switch in the Monty Hall Problem?

The probability of winning if you switch is 2/3, or approximately 66.7%. This means that switching doors will increase your chances of winning the prize.

4. Why is the Monty Hall Problem considered counterintuitive?

The Monty Hall Problem is considered counterintuitive because it goes against our natural instincts and intuition. It may seem like there is an equal chance of winning regardless of whether you switch or not, but the math and logic behind the problem show that switching actually increases your chances of winning.

5. Is the Monty Hall Problem applicable in real life situations?

Yes, the Monty Hall Problem can be applied in various real-life situations such as decision-making, risk assessment, and game theory. It highlights the importance of understanding probability and making informed choices rather than relying on intuition or gut feelings.

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