Is Switching the Best Option in the Door Game Show Dilemma?

[jp79]

Some people can see the simulation and still not understand. Seeing why
and knowing why are two different things.

 

[Gib Z]

My preferred explanation: You pick a door, say Door 1. Note everything is still closed. Door 1 has 1/3 chance. Hence; Doors 2 and 3 have 2/3 chance together. So at this point you agree that if you could somehow pick the option of "Both door 2 and door 3", that would give you 2/3 chance. All opening the door does is allow you to pick the "Both door 2 and door 3" option because Monty opened one of them, now if you open the other, together you opened both. And we agreed before, that this has a 2/3 chance.

 

[Mensanator]

Some people can see the simulation and still not understand. Seeing why
and knowing why are two different things.

Sure, I can see that you still may not understand why, but certainly you must understand that if mathematical reality does not match your premise then your premise is false. No argument, no matter how clever, can change mathematics.

 

[elibj123]

Since you know definitely know what's behind door 3 ( a goat) the following reasoning occurs:

The chance of winning by switching=the chance you've picked a goat in the first place=66%

The chance of losing by switching=the chance you've actually picked the car=33%

 

[Mensanator]

Since you know definitely know what's behind door 3 ( a goat) the following reasoning occurs:

The chance of winning by switching=the chance you've picked a goat in the first place=66%

The chance of losing by switching=the chance you've actually picked the car=33%

Yet people STILL think the probability is 50/50.

 

[vector03]

As I understand probability, if I can "count" the TOTAL possible ways the game can play out and I can "count" the TOTAL number of ways to WIN (or lose), If I divide the number of ways to win by the total number of ways the game can be played then I would be able to calculate exactly the theoretical probability of winning.

Making a table of the 24 possible ways the game can be played out and then counting the 12 ways in which the game can be won, I would suggest the probability of a random selection of winning is 0.50.

 

[Indefinite]

As I understand probability, if I can "count" the TOTAL possible ways the game can play out and I can "count" the TOTAL number of ways to WIN (or lose), If I divide the number of ways to win by the total number of ways the game can be played then I would be able to calculate exactly the theoretical probability of winning.

Making a table of the 24 possible ways the game can be played out and then counting the 12 ways in which the game can be won, I would suggest the probability of a random selection of winning is 0.50.

You are suggesting that if one selects a single door at random from three doors, one of which contains a car, that the theoretical probability of selecting the car is 0.5.

The fact that the host eventually opens one of the doors containing a goat does not alter the fact that there is a 1/3 probability that you will select the door that hides the car.

Surely, you can agree that if the host does not open one of the doors and provide the opportunity for switching, there is a 1/3 probability that you have selected the winning door.

If you accept this, but still assert that there is a .5 probability that you have selected the correct door, you are asserting that if the host opens a door that he knows to contain a goat after you have made your selection, the probability that you initially made the winning selection increases.

If we accept your assertion that after the goat has been revealed, the probability that your door is the winning door increases to .5, then we would also expect that in a very large number of trials, you would likely win more often if the host opens a door after you have made your selection than if he does not, even if you stay with your selection after he reveals one of the goats.

This result is clearly absurd. If you stay with your initial selection, the fact that the host shows one of the goats will not change the frequency with which you make the correct selection.

While the "paradox" may appear initially counter-intuitive, the notion that if the host opens a door that he knows to contain a goat, the probability that you made the correct selection will increase to .5 should seem even more absurd.

We know that if the host never opens the door after you make your selection, and you have no opportunity to switch, your chances of winning are 1/3. How could the event of the host opening a door he knows not to be the winning door after you make your selection possibly increase your chances of winning? Would you really expect to win more often if you make your selection, and choose to stay with that selection after the host opens the door hiding a goat, simply because the host opened the door?

Suppose you roll a die, but do not look, and I look to see what it is. Suppose you guess what number was rolled, and I tell you which number it is not, and ask you if you wish to switch your guess. Suppose you choose not to switch. Will the fact that I told you which number it is not in any way increase the probability that you initially guessed right? Will the probability that you initially guessed right increase from 1/6 to 1/5 simply because I told you something I know after you guessed? Will the event of me telling you something I know after you guess cause you to guess correctly more often if you do not ever change your guess?

 

[pallidin]

I like that explanation. Very clear.

 

[vector03]

You are suggesting that if one selects a single door at random from three doors, one of which contains a car, that the theoretical probability of selecting the car is 0.5.

Not actually suggesting what you have stated above.
Suppose the player makes his door choice and the "host" opens a losing door. The player decides to switch doors as recommended.

Now suppose that a new player walks in and knows nothing about what has transpired between the host and original player. The new player simply sees two doors. The host asks the new player to make a choice of doors as well.

According to previous assertions, the original player has 2/3 chance of winning while the new player has a 1/2 chance of winning. What I'm suggesting is that one player cannot have a 2/3 chance of winning at the same time another player has a 1/2 chance of winning given exactly the same set of circumstances.

This implies that a "new" game ensues after the host opens a door and the chance of winning is simply changed from 1/3 to 1/2.

 

[trambolin]

That's not correct, you have caused the door open by the host, since you made a choice. That's why we need the premise of having the host aware of the things behind the doors.

The newcomer will have a new game, but not your game since he did not cause the open door to open.

Think it with the 1000 doors or lottery tickets nationwide. If I show you all the numbers except the last one. it might happen that all of them are the same but you don't know about the last one. Because you did not know you will miss it by one number initially. Your initial chance to win does not change. You still hold the ticket that you bought with a very small chance of winning.

Now I say, I am going to remove all the the dummy tickets and leave you with one. You are standing there with your ticket and slim chance. But the one I have left has a big chance because all the possibilities of the dummy tickets are stacked on that ticket. Why? Because you might as well picked up a dummy one which is very likely to happen...

 

[vector03]

If I understand, Your suggesting it is possible for 2 different players at a single point in time given an equal set cirumstances to have 2 different probabilities of winning?

 

[trambolin]

No, I am saying that one of them is cheating since there were 3 doors for one and afterwards 2 doors for the other.

 

[vector03]

Considering that point in time, what are the probabilities for both the original player and the new player?

 

[trambolin]

That's crux of this question!

The newcomer has %50.

The original player has 1/3 as he started but given a chance to modify the odds or switch sides. Moreover, they reduced the options down to one. So if I had 1/3, and I have one more option to choose, then that option has to have the probability 2/3.

That's why it is neither a multiplication of probabilities nor resetting to %50 after he made one decision. Hence, a "dice has no memory" argument does not apply here. Because the event is not repeated it is altered.

 

[vector03]

We're almost there... :-)
The new player has 1/2 chance of winning, it seems we agree.
What is the probability of winning for the original player at the point in time under consideration?

 

[trambolin]

Do you understand what I wrote there?

 

[jgens]

If we accept your assertion that after the goat has been revealed, the probability that your door is the winning door increases to .5, then we would also expect that in a very large number of trials, you would likely win more often if the host opens a door after you have made your selection than if he does not, even if you stay with your selection after he reveals one of the goats.

vector03, I suggest that you read this part very carefully.

This implies that a "new" game ensues after the host opens a door and the chance of winning is simply changed from 1/3 to 1/2.

This isn't true. Just because the host opened a door to reveal a goat doesn't mean that we can ignore the circumstances before it.

 

[vector03]

Do you understand what I wrote there?

Yes. I understand all the explanations presented and I understand the mathematics behind the calculation. Each one who has attempted an explanation has presented what I would consider a reasonably good explanation (for whatever that's worth). My big "hang-up", and I'm not sure I agree, at least yet, that it's possible for a player to have a 1/2 chance of winning while another player at the same point in time with exactly the same set of conditions can have a 2/3 chance of winning.

Either something wrong with the theory or there is something wrong with the application that doesn't consider (or account for) the "hypothetical" new player (3rd observer).

I think to achieve 2/3 chance of wiinning requires and depends on an event that has a 100%chance of occurring --> "mechanically" requiring the original player to "switch" everytime which takes away from, in my opinion, some of the "randomness". An event which must occur 100% of the time is, in my opinion, not random.

So bottom line, yes... I've understood your explanations and appreciate them yet I'm just having a hard time "wrapping" my thoughts 100% around them.

 

[DrGreg]

My big "hang-up", and I'm not sure I agree, at least yet, that it's possible for a player to have a 1/2 chance of winning while another player at the same point in time with exactly the same set of conditions can have a 2/3 chance of winning.

When you calculate a probability, your answer depends on how much information you have. The second player is lacking some information, so the probability they calculate is 1/2. The original player has more information and calculates the probability as 2/3. The game show host has even more information and will calculate the probability to be either 1 or 0.

The probability represents how often you would win if you repeated the experiment many times using the information you have and assuming all the information you don't have is random.

 

[vector03]

Personally, I would vote for the host's chances as 0 since the host is generally not allowed to play.

I note the qualifications of "long run" or "repeated many times" and respectfully submit that the theory is based on the "long run" assumption. In this particular case, that assumption is not met. This experiment is setup as a one time chance of winning. If the player had hundreds of chances, in the long run, his chances would approach a limit of 2/3. However, the player only get's 1 chance and that invalidates any use of the "repeated many times" assumption. The player only has one chance.

Applying any theory that is based on certian assumptions being met to the solution of a problem where those assumptions are not being met does not seem cosistent

 

[dennynuke]

I've read all these comments and theories and I understand what's been said, but to me it boils down to this:

When you make your initial choice, you would of course have a 33% chance of winning. But, no matter which door you choose, the host is going to open one of the losing doors, then present you with a new choice. So based on these facts, your chances of winning are 50% from the start.

 

[pallidin]

I've read all these comments and theories and I understand what's been said, but to me it boils down to this:

When you make your initial choice, you would of course have a 33% chance of winning. But, no matter which door you choose, the host is going to open one of the losing doors, then present you with a new choice. So based on these facts, your chances of winning are 50% from the start.

Ok, you almost got it! Not quite, but your on the right track, sort of.

Consider this. If you intially choose a door, you have a 1/3 chance. Right?
Right.

If you do nothing further with that choice(that is, you KEEP your door) it DOES NOT change the fact that your chance is 1/3, because 1/3 what you started with. Right? (as long as you keep your initial selection)
Right.

But, if you switch to the remaining door, your odds change to 50/50.
Not if you STAY, ONLY IF YOU SWITCH!
That was the flaw in your logic. You think that by staying your odds somehow magically convert from 1/3 to 1/2 by your doing nothing.
That's NOT true.
But, if you switch, it IS true! And does indeed become 50/50

That's why switching increases your odds.

 

[pallidin]

Here's another, rather brilliant way of looking at this that finally conviced me (I was a sceptic too!)

You have a standard deck of playing cards and are told that the Ace of Hearts is the winning card.
You choose one "blindly"
What's your chance of having the Ace of hearts? 1/52 Very poor odds. Agreed?!

Now for the "magic":
Take that card, DO NOT LOOK at it, and lock it into a safe.
1/52 odds, agreed?
Nothing can change that at all as long as it is kept, unkown, in the safe. Agreed?
1/52 odds, agreed?

The magician takes the remaining 51 cards and looks through them. You are not allowed to do this.
The magician knows that the winning card is the Ace of hearts. His task is to select 1 card from the remaining 51 and discard the rest.
If the magician finds the ACE of HEARTS, he selects that card, but doesn't tell you.
If the magician does not find the ACE of HEARTS(because you have it), he selects a different card but doesn't tell you.

In either case, either you or the magician has the Ace of Hearts at this point, agreed?
Read that over again if it doesn't make sense. Sip on a beer and agree with this to this point...

OK, now, you OR the magician HAS the Ace of Hearts. This much IS CERTAIN, because the magician would select it if you did not have it, or select a different card if you did.

At this point, what is going on?
Nothing.
Your card in the safe still has a 1/52 chance of being the Ace of Hearts.
If you do NOTHING, and keep your card, it's 1/52. Nothing has changed.

But NOW, the magician poses this question: Keep your card or switch it for mine.
You NOW know that either you OR the magician has the ACE of Hearts.
It most certainly at that point is a 50/50

To keep your card is 1/52
To switch it, the odds are now 1/2.
Those odds WILL NOT GO FROM 1/52 to 1/2 if you keep your card. YOU MUST SWITCH.

You could have the winning card if you do not switch. THIS IS TRUE! Could happen!
But it will be 1/52.
If you switch the odds go to 1/2. You COULD still lose, but the odds are much better for a win.

 

[DrGreg]

To keep your card is 1/52
To switch it, the odds are now 1/2.

This doesn't make sense. As there's a 1/52 chance you have the winning card, the chances the other card is the winning card are 51/52, so if you switch, the odds of winning are 51/52. (Assuming the magician is telling the truth and has done what you described.)

And, similarly in the original car/goat game, the odds are 1/3 if you don't switch, and 2/3 if you do.

When you consider all possible outcomes, the odds have to add up to 1.

 

[dennynuke]

OK, here is the same situation presented to us in different words. The crux of the matter (and the most important) is that in each case we are presented with a new problem in the end, with a new set of variables.

In the case of the three doors, your first choice is 1 out of three, then you are asked to make your next choice with only two options. It does not matter how you got to this point. The past has nothing to do with the future odds, so your chances are 50/50.

The same is true with the cards. your first choice was 1/52. Then, no matter how you look at it, once the rest of the cards are elimnated then your new choise is between 2 cards, neither of which you know. There is no other answer than 50/50.

Again, the past has nothing to do with future odds. No matter how you end up at a choice, if that choice is between two unknowns, then your odds are always 50/50.

 

[pallidin]

OK, here is the same situation presented to us in different words. The crux of the matter (and the most important) is that in each case we are presented with a new problem in the end, with a new set of variables.

In the case of the three doors, your first choice is 1 out of three, then you are asked to make your next choice with only two options. It does not matter how you got to this point. The past has nothing to do with the future odds, so your chances are 50/50.

The same is true with the cards. your first choice was 1/52. Then, no matter how you look at it, once the rest of the cards are elimnated then your new choise is between 2 cards, neither of which you know. There is no other answer than 50/50.

Again, the past has nothing to do with future odds. No matter how you end up at a choice, if that choice is between two unknowns, then your odds are always 50/50.

No, no, no!
Denny, if you leave the card in the safe(don't switch) the 1/52 odds CANNOT be violated no matter what happens to the other 51.
It's only when you DO switch that the odds change dramatically in your favor.
Ok 2/3, not 1/2, my bad(thanks DrGreg, I forgot).
Anyway, the IMPORTANT point is to grasp that NOT switching is NOT the same odds as switching.
Bizarre? I know, but it's true, and makes perfect sense once you think about it enough.

 

[dennynuke]

In the beginning the magician is allowed to select from and eliminate all but one card, thus giving him a 51/52 chance of winning and I have a 1/52 chance. You're saying that once he eliminates 50 of the cards that I still have a 1/52 chance of having the right card.
I submit that it is now a new problem with different variables. My new (and now my only) choice is to keep my card or choose a different card. It doesn't matter if we started with 1,000,000 cards, or what the odds were when we started. That was a different problem. The only choice NOW is between two cards that are unknown to me.

I can't see it any other way. I appreciate your examples, but each boils down to the same choice between two unknowns. 50/50

 

[DrGreg]

Just consider what would happen if you repeated this game 5200 times. You pick a card at random.

In about 100 games you pick the ace of hearts, your opponent picks some other card at random. If you swap, you lose.

In the other 5100 games you pick some other card (about 100 games the 2 of hearts, about 100 games the 3 of hearts, about 100 games the 4 of hearts, etc), and always your opponent picks the ace of hearts. If you swap, you win.

So if you swap, in about 100 games you lose, in about 5100 games you win. Is that a 50:50 chance?

 

[pallidin]

In the beginning the magician is allowed to select from and eliminate all but one card, thus giving him a 51/52 chance of winning and I have a 1/52 chance. You're saying that once he eliminates 50 of the cards that I still have a 1/52 chance of having the right card.
I submit that it is now a new problem with different variables. My new (and now my only) choice is to keep my card or choose a different card. It doesn't matter if we started with 1,000,000 cards, or what the odds were when we started. That was a different problem. The only choice NOW is between two cards that are unknown to me.

I can't see it any other way. I appreciate your examples, but each boils down to the same choice between two unknowns. 50/50

What you and others fail to understand(including myself at one point) is that those are NOT 2 entirely separate problems.
They are "entangled" if you will.
That is, the potential outcome of problem #2 is definitely influenced by problem#1
Hope that makes any sense.

 

[dennynuke]

Just consider what would happen if you repeated this game 5200 times. You pick a card at random.

In about 100 games you pick the ace of hearts, your opponent picks some other card at random. If you swap, you lose.

In the other 5100 games you pick some other card (about 100 games the 2 of hearts, about 100 games the 3 of hearts, about 100 games the 4 of hearts, etc), and always your opponent picks the ace of hearts. If you swap, you win.

So if you swap, in about 100 games you lose, in about 5100 games you win. Is that a 50:50 chance?

OK, there's a glimmer of understanding starting here... Because the magician is allowed to see all the cards, he has a 51/52 chance of finding the ace so he has skewed the probability in his favor. I only have a 1/52 chance based on my initial random selection. So at this point I'm offered a choice between two cards which are unknown to me, but the magician's card has a high probability of being the ace since he was able to eliminate all but my one card. Is that it?