Do you know the correct answer?

[Peerie Maa]

Wrong. When you made the guess it was 33%. It still remains the same. Look at the game of 100 doors. You guess one, it has a 99% chance of being right. Monty has 99 doors, of which 98 are for sure goats. He shows you 98 doors of goats. Just because he ended with two doors, does not mean that you guess is now 50-50. You could have selected any of those doors he opened. There is one move, and it is to switch, and you when 99% of the time.
ets try this way. No, the game has not restarted, the game has ended. There is no game, if one move will lead to a guaranteed outcome. And you said, that if you were wrong on the first pick, you are guaranteed to win by switching. So there is no game left. He ended the game by showing you which door has the car of the 99 you did not pick. Monty only had a choice which 98 cars to show you 1% of the time? correct. 99% of the time, he has no choice in the matter. So assume the game is over and switch.

That is garbled. Perhaps you should review it and do a rewrite.

Go get a friend, play the game with all 52 cards, trying to guess the queen of spades. You be Monty. He will bet you $100 ahead of time he will get the queen of spades, will you take the bet?

If I am Monty I can see all the cards. So what is the point? The odds of picking the Queen are I/52 or 1.9%, so of course Monty will take the bet.
Try running this past yourself

Look at it this way:
Door 1 33%
door 2 33%
door 3 33%.
OK so far?
We now progress through the game.
Open door 2, removing it from the game.

Door 1 33%
Door 2 no longer exists
Door 3 33%. Yes?

The odds are now even.

[rgthom]

But there are two possible choices of doors for him to open when the player has chosen correctly, and both choices should be tabulated.

The host is not making random choices with equal probabilities.
When the car is behind the door chosen by the contestant, the rule could be that a specific door number is chosen (say car in door 1 means host always reveals door 2. A computer might play that way). The result would be the same.

[Decourcy]

The host is not making random choices with equal probabilities.
When the car is behind the door chosen by the contestant, the rule could be that a specific door number is chosen (say car in door 1 means host always reveals door 2. A computer might play that way). The result would be the same.

I never said it was random. I said the table is ignoring 3 losses by combining 6 possible choices of the player, which happen to be losses, into 3.


Since the table purports to tabulate all possible choices and placements, they should all have a line

[peb]

If I am Monty I can see all the cards. So what is the point? The odds of me winning are I/52 or 1.9%, so of course Monty will take the bet.
Try running this past yourself

You will take the bet with only a 1.9% chance of winning?

[Tom Montgomery]

.
The following table of outcomes can be found here: The Monty Hall Problem: A Statistical Illusion

The article also explains why the answer is so profoundly counterintuitive and why our brains tend to scream, "The probability is 50/50!" The 50/50 solution is illusory. This is because our common experience tends to lead us to assumptions about the problem that are incorrect.

There are only nine different combinations of choices and outcomes.

You Pick	Prize Door	Don’t Switch	Switch
1	1	Win	Lose
1	2	Lose	Win
1	3	Lose	Win
2	1	Lose	Win
2	2	Win	Lose
2	3	Lose	Win
3	1	Lose	Win
3	2	Lose	Win
3	3	Win	Lose
		3 Wins (33%)	6 Wins (66%)

[Boatbum]

Why. How can a choice between only two unknowns have a skewed probability.
Look at it this way:
Door 1 33%
door 2 33%
door 3 33%.
OK so far?
Yes? We now progress.
Open door 2, removing it from the game.

Door 1 33%
Door 2 no longer exists
Door 3 33%. Yes?

The odds are now even.

Not quite...

Monty possess two chances of being right. 33%+33%=66% So there is a 66% chance that ONE of the two chances is a car. The fact that one is not the car is a certainty so when he reveals a goat he still has has 66% chance.

[Peerie Maa]

You will take the bet with only a 1.9% chance of winning?

I was in the middle of an edit when you posted this. Please go back and read the corrected post.

[Peerie Maa]

Not quite...

Monty possess two chances of being right. 33%+33%=66% So there is a 66% chance that ONE of the two chances is a car. The fact that one is not the car is a certainty so when he reveals a goat he still has has 66% chance.

Monty has a 100% chance of being right, he knows where the car is, that is why he always opens a goat door. So do forget about Monty and concentrate on the probabilities available to the punter.
The punter ends up with a choice between TWO doors. Two not thee. So he has a 50-50 chance of being correct.

[rgthom]

I never said it was random. I said the table is ignoring 3 losses by combining 6 possible choices of the player, which happen to be losses, into 3.


Since the table purports to tabulate all possible choices and placements, they should all have a line

The choice of one of the two doors available, if the first choice was the car, is not up to the contestant. It is determined by the action of the host.

[Peerie Maa]

The choice of one of the two doors available, if the first choice was the car, is not up to the contestant. It is determined by the action of the host.

What? How?
The host ensures that there is a door with a car, and a door with a goat. That is all he does. The choice is still a free pick by the punter. One from two. 50-50.

[CWSmith]

.
The following table of outcomes can be found here: The Monty Hall Problem: A Statistical Illusion

The article also explains why the answer is so profoundly counterintuitive and why our brains tend to scream, "The probability is 50/50!" The 50/50 solution is illusory. This is because our common experience tends to lead us to assumptions about the problem that are incorrect.

This is just a rewrite of your original table. You are confusing the outcome of the game with the realizations of the actions during the game. Monty opening either of 2 doors to reveal a goat are two different realizations. They should not be combined.

[Tom Montgomery]

.
Monty has a 66% chance of possessing the car. He knows if he possesses the car or not. He always opens a goat door because he does not want to reveal the location of the car should he have it. His choices are not random, so do pay attention to Monty and remember the initial odds that the contestant's door has a 33% chance of a car while Monty's two doors have a combined 66% chance of containing the car. The contestant's choice comes down to whether or not to switch out of his 33% chance of winning the car.

[CWSmith]

I never said it was random. I said the table is ignoring 3 losses by combining 6 possible choices of the player, which happen to be losses, into 3.


Since the table purports to tabulate all possible choices and placements, they should all have a line

How do you see so clearly and others not at all?

[rgthom]

What? How?
The host ensures that there is a door with a car, and a door with a goat. That is all he does. The choice is still a free pick by the punter. One from two. 50-50.

I mean the choice of the two remaining doors without the car, which of the two goats will be revealed. Then the contestant can make his choice to stay or change.
The choice of one of the two goats is not up to the contestant, and should not be in the table to calculate probability.

[leikec]

This has been a revealing thread.

Your initial probability of winning is 33% when you commit to a choice of a door.

Now the host reveals one of the three doors, and it doesn’t have the car.

Once this happens certain things change, and how things change depends upon different factors.

The host always knows where the car is. His probability of knowing is 100%—his probability of not knowing is zero.

You, the contestant, had an initial probability of being right which was 1 out of three, which is 33%. If you stay with your initial pick, that probability doesn’t change—there were three choices, you picked one, the probability of it being right is 33%.

The host offered you three choices for a chance to win the car. You could only pick one, so the probability that the show Would Not give you the car was 66%.

Now the host reveals one of the losing doors, and tells you that you may either keep your original choice (which still retains the 33% probability of being right), or choosing the other unopened door.

In effect, the host is giving you a chance to swap probabilities with him—he is offering to transfer the 66% probability to you, or you can keep your initial 33% probability.

That is your real world choice—while it is true that with only two choices remaining the abstract probability jumps to a 50-50 choice, it doesn’t affect you. The only way you can change your probability of winning is to switch doors.

If some new person entered the game with no prior knowledge of what had already happened, they would be picking one of two possibilities, and their probability of being right (being given no other information) would be 50%. Their probability wouldn’t be affected if the host offered them a chance to switch after they made their initial choice—they had a 1 out of two chance of being right, and switching doesn’t make a difference for them.

Jeff C

[peb]

I was in the middle of an edit when you posted this. Please go back and read the corrected post.

Ok, but you had the odds correct, and you were just wrong on taking the bet. The fact is Monty will lose 51 out of 52 times, as long as the other person switches. You should never take the bet. Get a friend, go try. Report back

[Decourcy]

This has been a revealing thread.

Your initial probability of winning is 33% when you commit to a choice of a door.

Now the host reveals one of the three doors, and it doesn’t have the car.

Once this happens certain things change, and how things change depends upon different factors.

The host always knows where the car is. His probability of knowing is 100%—his probability of not knowing is zero.

You, the contestant, had an initial probability of being right which was 1 out of three, which is 33%. If you stay with your initial pick, that probability doesn’t change—there were three choices, you picked one, the probability of it being right is 33%.

The host offered you three choices for a chance to win the car. You could only pick one, so the probability that the show Would Not give you the car was 66%.

Now the host reveals one of the losing doors, and tells you that you may either keep your original choice (which still retains the 33% probability of being right), or choosing the other unopened door.

In effect, the host is giving you a chance to swap probabilities with him—he is offering to transfer the 66% probability to you, or you can keep your initial 33% probability.

That is your real world choice—while it is true that with only two choices remaining the abstract probability jumps to a 50-50 choice, it doesn’t affect you. The only way you can change your probability of winning is to switch doors.

If some new person entered the game with no prior knowledge of what had already happened, they would be picking one of two possibilities, and their probability of being right (being given no other information) would be 50%. Their probability wouldn’t be affected if the host offered them a chance to switch after they made their initial choice—they had a 1 out of two chance of being right, and switching doesn’t make a difference for them.

Jeff C

Why do you ignore that when the host opens a door, the door you have chosen Initially, whether right or wrong, goes from 33% to 50%? Both doors have unknown results to you, but known to the host. The host will only remove an incorrect choice, increasing the odds on both remaining. Of course the host will always open one of the doors you haven’t chosen. One will always be a goat. If both are goats, he has two choices/events to be tabulated

[Boatbum]

Monty has a 100% chance of being right, he knows where the car is, that is why he always opens a goat door. So do forget about Monty and concentrate on the probabilities available to the punter.
The punter ends up with a choice between TWO doors. Two not thee. So he has a 50-50 chance of being correct.

We are just going to have to disagree on this. It doesn't matter what Monty knows because he is not the one that has to choose. It would only be a 50/50 chance if the location of the car was randomly reassigned to one of the remaining two doors after Monty reveals the goat behind one of the doors. Since there was no reassignment, there remains a 66% chance that Monti has the car and since one of the doors has been revealed it could only be behind the door he is offering to the contestant if he has it at all. The contestant still has a 33% chance that it is behind their door. Seriously, there are web sites that have the process automated so you can do as many trials as you want. The more you do it the more you will see that picking Monty's door will lead to a car much more than keeping your original choice. I think you need to actually put it in action and see for yourself before you are willing to understand the logic behind it.

[Tom Montgomery]

.
CWSmith: You are saying there are more outcomes in the game than the table in post #333 shows. That is incorrect.

[Tom Montgomery]

Why do you ignore that when the host opens a door, the door you have chosen Initially, whether right or wrong, goes from 33% to 50%?

Because that is incorrect and based on incorrect assumptions. The door the contestant chose initially remains with a 33% chance of containing the car regardless of what Monty says or does. His choice in the end is to either stay with his 33% chance of winning or switch out of it.

Do follow the link I posted in #333 above for an explanation why this is so.

[Decourcy]

^
You are saying there are more outcomes in the game than the table shows. That is incorrect.

There are 12 possible combinations of choices not 9

[Peerie Maa]

Ok, but you had the odds correct, and you were just wrong on taking the bet. The fact is Monty will lose 51 out of 52 times, as long as the other person switches. You should never take the bet. Get a friend, go try. Report back

No, the odds are 1.9% that the punter will select correctly. Monty will win 51 out of 52 times.
If Monty reveals one card the odd reduce to 1/51, which is as close to 2% as makes no difference, Monty is still on a safe bet.

[Decourcy]

We are just going to have to disagree on this. It doesn't matter what Monty knows because he is not the one that has to choose. It would only be a 50/50 chance if the location of the car was randomly reassigned to one of the remaining two doors after Monty reveals the goat behind one of the doors. Since there was no reassignment, there remains a 66% chance that Monti has the car and since one of the doors has been revealed it could only be behind the door he is offering to the contestant if he has it at all. The contestant still has a 33% chance that it is behind their door. Seriously, there are web sites that have the process automated so you can do as many trials as you want. The more you do it the more you will see that picking Monty's door will lead to a car much more than keeping your original choice. I think you need to actually put it in action and see for yourself before you are willing to understand the logic behind it.

It is always a two door puzzle. The third door is always going to be a goat, and always revealed.

[rgthom]

There are 12 possible combinations of choices not 9

No, there are 9. If the contestant chooses the car first pick then the host determines which door is still available. The contestant does not have a choice.

[Peerie Maa]

Because that is incorrect and based on incorrect assumptions. The door the contestant chose initially remains with a 33% chance of containing the car regardless of what Monty says or does. His choice in the end is to either stay with his 33% chance of winning or switch out of it.

Do follow the link I posted above for an explanation why this is so.

No. The contestant now has a choice between only two doors. Stay or switch is still between only two doors. So tell us how a choice between two is not a 50-50 bet?

[Peerie Maa]

No, there are 9. If the contestant chooses the car first pick then the host determines which TWO door is still available. The contestant does not have a choice.

Looks like a choice to me.

[rgthom]

Looks like a choice to me.

The host chooses, not random. The contestant does not.

[Boatbum]

Why do you ignore that when the host opens a door,

Because the host still has a 2 out of 3 chance that he has chosen the car (well more accurately that the car was chosen for him). He has 2 chances of winning but we know there is only one car so at least one door has to be a goat - no surprise when he shows it.

We can agree that in the beginning when the contestant chooses they have a 33% chance of having the car right? There are no further choices until after Monty revels the goat and offers to trade - nothing has really changed - we always knew he had at least one goat. But since he's got two chances to the contestants one, there remains a greater probability that his door has the car because he had more chances to win than the contestant.

[Ron Williamson]

Since Monty will NEVER open a door with a car behind it,the third door( the one that he opens)is a distraction.
A two door choice .
50/50
R

[peb]

No, the odds are 1.9% that the punter will select correctly. Monty will win 51 out of 52 times.
If Monty reveals one card the odd reduce to 1/51, which is as close to 2% as makes no difference, Monty is still on a safe bet.

OK, get a deck of cards and try it with a friend. I will do the same, lets compare notes tomorrow.

[Tom Montgomery]

.
I show you three cards: the King of Hearts and two Jokers. I shuffle them. You select one card and do not look at it. What are the odds that you hold the King of Hearts?

What in the world do you think I can do with any of the two cards I hold to change your initial odds of holding the King of Hearts?

You think that if I show you I am holding one Joker your odds of holding the King of Hearts increases to 50/50? It does not.
.

[CWSmith]

You, the contestant, had an initial probability of being right which was 1 out of three, which is 33%. If you stay with your initial pick, that probability doesn’t change—there were three choices, you picked one, the probability of it being right is 33%.

Jeff C

Probability is the ratio of desirable realizations to total realizations.

Initially, the contestant's chances of winning is 1 over 3.

The, one choice is removed. If the contestant's chances of winning are still 1 in 3, then that means there is a chance that the car is behind the door that was already opened.

Do you see that?

[Tom Montgomery]

Probability is the ratio of desirable realizations to total realizations.

Initially, the contestant's chances of winning is 1 over 3.

The, one choice is removed. If the contestant's chances of winning are still 1 in 3, then that means there is a chance that the car is behind the door that was already opened.

Do you see that?

No. See #359 above.

[CWSmith]

No, there are 9. If the contestant chooses the car first pick then the host determines which door is still available. The contestant does not have a choice.

The contestant has every choice. He chooses a door at first and then the game is redefined from 1 in 3 to 1 in 2. Then he has a choice to stick or switch.

[leikec]

Wow…

Jeff C