[JTG]

This problem is named after the host and developer of the old game show "Let's Make a Deal"

The rules of the game are as follows:

• You are presented with three doors, one concealing a good prize
• You choose one door as yours
• The host then reveals a consolation prize behind one of the unchosen doors
• You then choose whether you want to keep your door or swap to the remaining other

Obviously without the glimpse of a consolation prize behind one of the other doors, your chance of winning the grand prize would always be 33%. However, does the revelation change your chances?

After the swap, is there strategy to be employed, or are you still shooting in the dark except now with a 50/50 chance?

[topquark]

If you swap, you have a 2/3 chance of winning. Here's why:

There are two possible scenarios:

1: you chose the right door 1st time. Probability = 1/3
2: you chose a consolation prize 1st time. Probability = 2/3

In Scenario 1, you win if you stick and lose if you swap.
In Scenario 2, the host opens another door to show you the other consolation prize. The real prize must be behind the unopened door. You lose if you stick and win if you swap.

At the time you make your decision, you don't know which scenario applies to you. However, you do know that Scenario 2 is twice as likely to occur as Scenario 1. Therefore, swapping is the best strategy.

[Photolysis]

It's always amused me how many people get this problem utterly wrong and think that at stage 2 the chances have changed to 50-50.

[Tocsin]

I thought the "Monty Hall Problem" was whether or not to grope the show girls and risk a civil lawsuit.
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[D4P]

Does the line of thinking represented in the thread so far assume that the contestant might choose to accept the revealed consolation prize, or does it assume that the contestant will only choose between the two remaining unopened doors?

[CallMeSpooky]

I love this problem. I use it often to explain to people why some mathematical understanding is useful, and to show how immediate intuitive judgments may be wrong, people usually view it as counter-intuitive that swapping your choice doubles your winning probability.

[JTG]

I was hoping most of the respondents would be people who hadn't seen it before
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	Originally Posted by D4P To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Does the line of thinking represented in the thread so far assume that the contestant might choose to accept the revealed consolation prize, or does it assume that the contestant will only choose between the two remaining unopened doors?

The two consolation prizes are equal to each other and a worst-case scenario. For example, imagine a new personal media system (home stereo, speakers, headphones, portable mp3 player, etc) behind the winning door and just a pair of headphones behind each of the other doors

[CrudeHypothesis]

You can see through this easier if there was 1,000,000 doors.
You pick one and the host opens 999,998 of them, leaving just your door and another door.
Your chances of picking it right first were 1 in 1,000,000
for some reason out of 999,999 doors, that other door wasn't opened.
Still feel like sticking?

[Uncle Mort]

	Originally Posted by JTG To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	I was hoping most of the respondents would be people who hadn't seen it before To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

There where two chances of that happening:

fat chance and no chance.

[D4P]

This is how I would intuitively approach the problem. I'd be curious to know if I'm making a mistake somewhere.

It seems to me that there are 4 possible scenarios:

Scenario 1. Choose the right door the first time, choose the right door the second time (probability equal to 1/3 * 1/2 = 1/6)

Scenario 2. Choose the right door the first time, choose the wrong door the second time (probability equal to 1/3 * 1/2 = 1/6)

Scenario 3. Choose the wrong door the first time, choose the right door the second time (probability equal 2/3 * 1/2 = 2/6)

Scenario 4. Choose the wrong door the first time, choose the wrong door the second time (probability equal 2/3 * 1/2 = 2/6)

In each scenario, the probability of making a correct second choice is equal to 1/2, and the overall probability of ending up with the right door will be equal to the sum of the probabilities for Scenarios 1 and 3, which equals 1/6 + 2/6 = 3/6 or 1/2. It seems to make no difference which door you choose first: you will then always have a 1/2 probability of choosing the right door on your second choice because one of the wrong doors will have been eliminated and only two doors will remain.

Put another way, before one of the doors is opened to reveal a consolation prize, there is a 1/3 probability that the door you chose first will contain the grand prize and a 2/3 probability that it won't. But once one of the other doors is opened, there becomes a 1/2 probability that your door contains the grand prize and a 1/2 probability that it doesn't.

Am I mistaken somewhere? Does this make me look stupid?

[babsa]

You guys are still attributing the original chosen door as a 33% chance. When the host removes one of the doors, that automatically raises the chance that your original door is a winning door to 50%.

[Ghostwheel]

	Originally Posted by JTG To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	The two consolation prizes are equal to each other and a worst-case scenario. For example, imagine a new personal media system (home stereo, speakers, headphones, portable mp3 player, etc) behind the winning door and just a pair of headphones behind each of the other doors

A consolation prize is not significantly exciting for me to try for it. Whether the odds are 50% or 66.67% really doesn't matter much.

[Photolysis]

	Originally Posted by babsa To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	You guys are still attributing the original chosen door as a 33% chance. When the host removes one of the doors, that automatically raises the chance that your original door is a winning door to 50%.

No it does not.

[Psychotropic]

	Originally Posted by babsa To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	You guys are still attributing the original chosen door as a 33% chance. When the host removes one of the doors, that automatically raises the chance that your original door is a winning door to 50%.

The reason that your second decision in the original scenario is NOT 50/50 is because you made your choice when the chance of picking the grand prize was 1/3. So there is a 67% chance that your initial selection is the wrong one and a 33% chance that it was the right one. So you switch because it is more likely that you initially chose the wrong door.

The probabilities do not change to 50/50 because your initial selection was made before you knew what was behind any of the doors. Knowing what is behind one of the other doors does not change the fact that your initial selection was made when you had 3 options.

EDIT: The second decision is 50/50 if you do not consider strategy because you will pick one door randomly and one consolation prize will be shown every time (i.e. the second decision is always one grand prize and one consolation irregardless of your first decision). If you flip a coin to decide which of the two remaining doors to open you will win 50% of the time. However, in the example problem your choice is not random; it is deliberate (i.e. do you win more if you hold or if you switch?)

[JTG]

	Originally Posted by Psychotropic To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	The probabilities do not change to 50/50 because your initial selection was made before you knew what was behind any of the doors. Knowing what is behind one of the other doors does not change the fact that your initial selection was made when you had 3 options.

But after one door is removed from play, if you do not swap, is that not like re-choosing the door now that there are only two options?

Regardless of the fact that you chose Door1 previously, there are now two doors, and you are tasked with choosing one of the two.

[Psychotropic]

	Originally Posted by JTG To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	But after one door is removed from play, if you do not swap, is that not like re-choosing the door now that there are only two options? Regardless of the fact that you chose Door1 previously, there are now two doors, and you are tasked with choosing one of the two.

No, because there is only a 1/3 chance that your initial selection was the correct one. Since the door you picked is not eliminated there is a still a 33% chance you picked the correct door.

Door 1: Grand Prize
Door 2: Consolation
Door 3: Consolation

Outcomes
Pick 1. Show 2 or 3. Hold. WIN (Switch. LOSE)
Pick 2. Show 3. Hold. LOSE (Switch. WIN)
Pick 3. Show 2. Hold. LOSE (Switch. WIN)

Hold = 33% win
Switch = 67% win

In all 3 instances you will be shown a consolation prize. Therefore, when you pick door 1 it only counts once because doors 2 & 3 are exactly the same.

---------- Post added 06-16-2012 at 10:51 PM ----------

People end up with 50% because they count ALL of the outcomes as a total. The question wants to know which strategy results in the best odds of winning so you consider the "Hold" outcomes and the "Switch" outcomes separately.

[Tocsin]

At the outset, you have a 1 in 3 chance of having made the right choice: which equates to 33% chance of success.

Regardless of the timing of when one of the"wrong" choices is revealed, it has the effect of reducing the outcome to either having made a correct choice, or the other wrong choice (50/50), but it only applies to two of the three possible original choices.

So what you are left with is a 50/50 choice between two of the original three doors.

50% * 66% = 33%

So, despite the revelation of a mistake, your odds of having made the right choice to begin with remain at 1 in 3, though, once one of the wrong choices has been revealed, the odds become 1 in 2 between the remaining two doors.

[INTJRyan]

Originally Posted by Psychotropic To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

No, because there is only a 1/3 chance that your initial selection was the correct one. Since the door you picked is not eliminated there is a still a 33% chance you picked the correct door. Door 1: Grand Prize Door 2: Consolation Door 3: Consolation Outcomes Pick 1. Show 2 or 3. Hold. WIN (Switch. LOSE) Pick 2. Show 3. Hold. LOSE (Switch. WIN) Pick 3. Show 2. Hold. LOSE (Switch. WIN) Hold = 33% win Switch = 67% win In all 3 instances you will be shown a consolation prize. Therefore, when you pick door 1 it only counts once because doors 2 & 3 are exactly the same. ---------- Post added 06-16-2012 at 10:51 PM ---------- People end up with 50% because they count ALL of the outcomes as a total. The question wants to know which strategy results in the best odds of winning so you consider the "Hold" outcomes and the "Switch" outcomes separately.

One door is eliminated so unless you consider picking the consolation prize, it's 50/50.

[Psychotropic]

	Originally Posted by INTJRyan To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	One door is eliminated so unless you consider picking the consolation prize, it's 50/50.

If you read my post I wrote out the possibilities and they come out to 1/3 and 2/3 when you eliminate the shown door.

[ktgrey]

You are missing an important detail in the statement of the question, which is that the host always opens an empty door, as opposed to the host sometimes opening an empty door and sometimes opening a door with the prize. In the first case, the chances are 33%/66% for the original door and the swap, and in the second case it is 50%/50% regardless of the number of original doors.

[Latro]

To those suggesting that the usual analysis is invalid, please install a Python interpreter if you don't have one and run the following:

( Show Spoiler )

import random
def monty_run(switch):

correct = random.randint(1,3)
choice = 1
if correct == 1: openable = [2,3]
elif correct == 2: openable = [3]
else: openable = [2]
opened = random.choice(openable)
if switch:

if opened == 2: choice = 3
elif opened == 3: choice = 2

if choice == correct: return 1
else: return 0

if __name__ == '__main__':

switch_str = raw_input('do you want to switch at each stage y/n?')
run_count_str = raw_input('how many times do you want to play?')
run_count = int(run_count_str)
if switch_str == 'y' or switch_str == 'Y': switch = True
else: switch = False
nsuccesses = 0
for i in xrange(run_count): nwins += monty_run(switch)
print float(nwins)/run_count
raw_input('press enter to exit')

If you need instructions for how to do this, ask me and mention what OS you are using.

I tested this just now and with 100,000 runs with switching I got about 0.665 vs. about 0.335 without switching. It should be fairly clear that this code simulates the problem correctly.

[Malle]

Image over possible scenarios as well as final probabilities:

( Show Spoiler )

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	Originally Posted by ktgrey To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	You are missing an important detail in the statement of the question, which is that the host always opens an empty door, as opposed to the host sometimes opening an empty door and sometimes opening a door with the prize. In the first case, the chances are 33%/66% for the original door and the swap, and in the second case it is 50%/50% regardless of the number of original doors.

That depends on which probability it is you are talking about now. Is it the probability to win the game from the start of the game or is it the probability to win the game once we're in the situation that Monty Hall has opened a door with no prize behind it? Because once we're in the position that Monty Hall revealed a losing door, whether or not he knew that there was a prize behind it does not matter.

If it is the previous:
If we pick the winning door first (P = 1/3) Monty will always reveal a losing door (P = 1) and sticking is the winning choice.

If we pick a losing door first (P = 2/3) then either
a) a losing door is revealed (P = 1/2) and switching is the winning choice or
b) the winning door is revealed (P = 1/2) and there is no winning choice

Total:
P(Win if Stick) = 1/3
P(Win if Switch) = 1/3


If it is the latter:
If we picked the winning door (P = 1/3) then sticking is the winning choice.
If we picked a losing door (P = 2/3) then switching is the winning choice.

Total:
P(Win if Stick) = 1/3
P(Win if Switch) = 2/3

[topquark]

	Originally Posted by INTJRyan To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	One door is eliminated so unless you consider picking the consolation prize, it's 50/50.

Actually, the elimination of one door by Monty gives you more information about the other doors that makes them no longer equivalent, so 50/50 doesn't apply.

[INTJRyan]

	Originally Posted by topquark To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Actually, the elimination of one door by Monty gives you more information about the other doors that makes them no longer equivalent, so 50/50 doesn't apply.

Please explain.

	Originally Posted by Psychotropic To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	If you read my post I wrote out the possibilities and they come out to 1/3 and 2/3 when you eliminate the shown door.

I did read your post. The analysis is incorrect because of this, so I discounted it:

Originally Posted by Malle To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

That depends on which probability it is you are talking about now. Is it the probability to win the game from the start of the game or is it the probability to win the game once we're in the situation that Monty Hall has opened a door with no prize behind it? Because once we're in the position that Monty Hall revealed a losing door, whether or not he knew that there was a prize behind it does not matter.

Of course the original odds are still 1/3. But once one is revealed, I'm not understanding how it's not a 50/50 guess if you do not consider the revealed door at that point. There are only two choices remaining at that point. Is not the game reduced to a coin flip? Perhaps I don't get the game.

[Latro]

	Originally Posted by INTJRyan To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Please explain.

Suppose you always choose 1, and the winning door is chosen randomly. Then with probability 1/3 you chose the correct door, in which case either of the remaining doors can be opened and switching to the other one results in losing while staying results in winning. With probability 2/3 you chose the incorrect door. In this case the other incorrect door (never the correct door) is opened, and switching guarantees that you win while staying guarantees you lose. Cf. my numerical simulation for concrete evidence.

You cannot look at the system after a door has been opened as a simple choice between two items, because the rules of the game make the state have intrinsic history.