Monty Hall Problem... For Newbies

[Red Baron Farms]

I'm more fascinated by the interest in explaining this problem than I am by the problem itself. I definitely don't intend to sound snarky, but I am amused by all the 'this is how you do it' posts, especially as so many haven't worked for me. If only all learners were identical sponges, education would be a breeze...

Adding doors isn't helping with the conceptual blocks I stumble over. 100 doors, a milliard doors, it really doesn't matter to me. We know there's an inevitable (irrelevant) goat. 98 irrelevant goats are still irrelevant.

OK think about it this way. See if this helps. Make it two separate problems.

There are two studios, studio A has 3 doors, 2 goats and a car. Studio B has 2 doors, 1 goat and 1 car.

Which studio would you rather pick from? A or B? Everyone realizes the better odds are Studio B. But what if Monty "cheats" to help you? He will eliminate a goat after you choose and let you pick again to double your odds! Then studio A is best.

 

[jiggeryqua]

OK think about it this way. See if this helps. Make it two separate problems.

There are two studios, studio A has 3 doors, 2 goats and a car. Studio B has 2 doors, 1 goat and 1 car.

Which studio would you rather pick from? A or B? Everyone realizes the better odds are Studio B. But what if Monty "cheats" to help you? He will eliminate a goat after you choose and let you pick again to double your odds! Then studio A is best.

Thanks for playing Seriously, if more doors isn't helping, how would more studios?? I have clearly stated the conceptual blocks as I see them, yet nobody addresses them.

 

[jiggeryqua]

I reduced your table to the important part. You're counting that as two separate events, but really they're one in terms of probability.

Picked Door|Car Is Behind|Monty Opens|Change|Stick
X|X|Y or Z|LOSE|WIN

is how it should read. Half the times the setup is X-X Monty will open Y, and half Z.

And I had acknowledged in the body of the post that disagreement on that point was likely to be the issue. But you can't just repeat what I'm objecting to - that's like shouting your english to foreigners. It's no more understandle. Again, I see you get the result you 'want' by combining two of the possible situations. But they're two distinct situations, why do you get to combine them. My table shows all the possible combinations of elements, leading to 12 distinct scenarios. In 6, changing wins. In 6, changing loses. Certainly, if you count two of them as one you get a different total. I'm not a complete klutz, I'm quite confident when it comes to counting. But '1, 2, 3, 4, 4, 5' is still 6.

 

[Red Baron Farms]

Thanks for playing Seriously, if more doors isn't helping, how would more studios?? I have clearly stated the conceptual blocks as I see them, yet nobody addresses them.

OK try again:

but what distinguishes the two available doors??

When you choose originally there were 3 doors, so a 1/3rd chance. Those odds are locked in forever because WHEN you made it, the odds were 1/3rd you picked the car and 2/3rds it was under one of the other two doors.

Now Monty comes in and helps you by cheating to give you an edge. He knows where the goats are. He eliminates one door. Now you still have your 1/3rd chance from when you picked, but now you can shift to the 2/3rds side without fear of hitting that goat Monty eliminated.

 

[Andy_Ross]

I have read the whole thread. I still think it is 50/50

You choose a door he shows you a goat, he is going to show you a goat no matter what is behind your door.
Your next choice will always be 50/50
I think you are all over thinking this.

 

[jiggeryqua]

As clear and concise as I'm able, then, with no verbose bush-beating (too late... )

• I cannot believe anyone here is claiming your choice of door can affect the actual result. Your thoughts do not affect the material world. I hope we're all good on that.
• Monty opening a door to show a goat (an action that is always available regardless of your choice) cannot affect the result. Whether you picked right or wrong, seeing a goat is always going to happen. Something that is always going to happen cannot be affecting the probabilities?
• Conceded: the odds shift from 1 in 3 when there are only two doors left that might hide a car. The big stumbling block is the insistence that the 'spare' probability from the now open door should go to the door you choose (as oppposed to the door I chose) - that is, the claim appears to be that the probability becomes greater for one door but not the other depending on your thoughts. I thought we dismissed that notion in the first point...

 

[jsfisher]

And I had acknowledged in the body of the post that disagreement on that point was likely to be the issue. But you can't just repeat what I'm objecting to - that's like shouting your english to foreigners. It's no more understandle. Again, I see you get the result you 'want' by combining two of the possible situations. But they're two distinct situations, why do you get to combine them. My table shows all the possible combinations of elements, leading to 12 distinct scenarios. In 6, changing wins. In 6, changing loses. Certainly, if you count two of them as one you get a different total. I'm not a complete klutz, I'm quite confident when it comes to counting. But '1, 2, 3, 4, 4, 5' is still 6.

Maybe this will help:

For all the situations where you do not first pick the door with the car, Monty has one and only one possible door he can open. He will do that 100% of the time in each of those situations.

If you pick the door with the car, Monty can open either door. 50% of the time, he'll open one, and 50%, the other.

The rows in your table are not all equally likely. The probability of each occurring is the probability you'll pick that particular door (1/3) times the probability of the car being behind a certain door (1/3) times the probability Monty will open the specified door. It's that last probability the confuses things. For most rows, it is 1, but for the ones GlennB said to combine, it is 0.5 for each.

 

[jiggeryqua]

OK try again:



When you choose originally there were 3 doors, so a 1/3rd chance. Those odds are locked in forever because WHEN you made it, the odds were 1/3rd you picked the car and 2/3rds it was under one of the other two doors.

Now Monty comes in and helps you by cheating to give you an edge. He knows where the goats are. He eliminates one door. Now you still have your 1/3rd chance from when you picked, but now you can shift to the 2/3rds side without fear of hitting that goat Monty eliminated.

Now we're getting somewhere. My initial instinct was 'But if you picked the car, he opened part of your 2/3, the other door is still only 1/3.'

But...if you didn't pick the car (and 2 out of 3 times you didn't), he does appear to be collapsing the choice. 2 out of 3 times you'll have been wrong in your first pick, so you should change. The goat is still irrelevant...the car always starts out being more likely to be behind a different door than the one you chose. When you've chosen, it's still more likely to be behind a different door. We know the goat is irrelevant, so when we see an irrelevant goat, the car must still be more likely to behind a door other than the one you chose.

And that is how you explain the Monty Hall problem to me. Well done, RBF

 

[GlennB]

And I had acknowledged in the body of the post that disagreement on that point was likely to be the issue. But you can't just repeat what I'm objecting to - that's like shouting your english to foreigners. It's no more understandle. Again, I see you get the result you 'want' by combining two of the possible situations. But they're two distinct situations, why do you get to combine them.

Because it's correct - they're not 'distinct situations', they're different paths stemming from an individual situation. Taking the location of the car and the door you choose as 'the setup' then there are 9 different setups. Each has an outcome, win or lose. You've described 12 setup/outcome combinations.

A parallel case - I can set off down road A or road B to get to work and these are very different routes, but I like a little variety. Route A leaves me no further choices. Route B forks somewhere and I can take fork X or fork Y, both of which take the same time.

Route A takes 40 minutes, route B 50. I set off down A and B equally often. What's my average journey time? 45, obviously.

Route chosen|Fork chosen|Journey time
A|-|40
B|X|50
B|Y|50
Average| |46.66

Nuts, huh? This has happened because I've taken B-X and B-Y as separate events and given them equal weight to A-

 

[Red Baron Farms]

edit problem solved already. see above.

 

[bruto]

I have read the whole thread. I still think it is 50/50

You choose a door he shows you a goat, he is going to show you a goat no matter what is behind your door.
Your next choice will always be 50/50
I think you are all over thinking this.

And yet, this is not how it works. If you reverse your bet when offered the choice, your result is not to get the car 1/2 the time, it is to get the car 2/3 of the time. If you stick, you get the car 1/3 of the time. It doesn't take over-thinking. It takes thinking about what is actually happening, and that is the trick, and the reason this little puzzle keeps coming up and keeps fooling people.

 

[Rasmus]

And I had acknowledged in the body of the post that disagreement on that point was likely to be the issue. But you can't just repeat what I'm objecting to - that's like shouting your english to foreigners. It's no more understandle. Again, I see you get the result you 'want' by combining two of the possible situations. But they're two distinct situations, why do you get to combine them. My table shows all the possible combinations of elements, leading to 12 distinct scenarios. In 6, changing wins. In 6, changing loses. Certainly, if you count two of them as one you get a different total. I'm not a complete klutz, I'm quite confident when it comes to counting. But '1, 2, 3, 4, 4, 5' is still 6.

Are all these scenarios equally likely, though? Why, or why not?

Picked Door|Car Is Behind|Monty Opens|Change|Stick
X|X|Y|LOSE|WIN
X|X|Z|LOSE|WIN
X|Y|Z|WIN|LOSE
X|Z|Y|WIN|LOSE

So, these are all the possible situations in which the player choses door X. But are they equally likely?

No, they are not.

The car is not going to be behind door X twice as often as it is going to be behind all other doors! Why would it be? But your distribution here is XXYZ, rather than XYZ.

Let me add a few numbers to the table:

Picked Door|chance|Car Is Behind|chance|Monty Opens|Change|Stick
X|1/3|X|1/3|Y|LOSE|WIN
X|1/3|X|1/3|Z|LOSE|WIN
X|1/3|Y|1/3|Z|WIN|LOSE
X|1/3|Z|1/3|Y|WIN|LOSE

You do not have 4 opening scenarios, you have 3: You always pick door X, and the car is either behind x (1), y (2) or z (3).

These scenarios happen with the same chances or likelihood.

*then*, for each of those scenarios, you have a sub-distribution:

If the car is behind the door you picked (one third of the time), Monty will open either door y or door z. He'll open each door half the time he gets to make that choice (one sixth of the time.)

If the car is behind one of the doors you didn't chose (two thirds of the time), Monty will open the only door he can. (Two thirds of the time total, half these times (or 1/3 of the total time) it will be door Y, the other half of these times, door Z.

Let's look at this with some actual numbers. Let's play the game 300 times -just looking at games where the player choses door X.

So, 100 times, the car will be behind door X.
1oo times, the car will be behind door Y,
and 100 times, the car will be behind door Z.

Trials|Picked Door|Car is behind
100|X|X
100|X|Y
100|X|Z

What happens next?

Trials|Picked Door|Car is behind|Monty opens
100|X|X|Y or Z
100|X|Y|Z
100|X|Z|Y

In the first case, he will open Y 50 times, and Z 50 times, too. Each of the scenarios happens less often than the other two where Montys hand is forced.

 

[Andy_Ross]

Don't see it there are two doors, you have a 50/50 choice. It's a new game after he opens the first door.

 

[jiggeryqua]

Don't see it there are two doors, you have a 50/50 choice. It's a new game after he opens the first door.

May I? It is a new game. But we're also talking about your old pick, because it's the same old doors and the car hasn't moved for this 'new' game. You were probably wrong in the old game (1/3 probability). You should, therefore, probably change in the 'new' game. I don't know why people have such trouble with this one...

 

[wollery]

Okay, for all the 50/50 people out there, I'll try to explain why you're wrong by changing the rules slightly. We'll keep everything the same, except now you aren't allowed to change door, you have to stick with your original choice. If you're right and the chance is 50/50 then you should win half the time.

So let's try it.

The car is behind door A, and there are goats behind doors B and C.

Let's start with picking door A. I open one of the other doors, to reveal a goat. I then open the other door to reveal a goat. Congratulations, you won the car.

Now you pick door B. I open another door to reveal a goat. I open the other door to reveal the car. Bad luck, you lost.

Last turn, you pick door C. I open another door to reveal a goat. Long pause for effect, I slowly open the other door to reveal.... the car! Oh dear, you lost again.

Three doors, three choices, one car. If you stick (as you were forced to in this game) you lose 2 out of 3 times.

But if revealing the first goat makes the probability 50/50 how can you lose 2 out of 3 games?

The point is that the probability doesn't change, only your perception of it. At first you were given a choice of one out of three doors. Then one of the doors was opened to reveal a goat, leaving two doors. But you always knew that at least one of the doors hid a goat, the only difference is that you've had it confirmed that there was a goat. Big deal, your original choice was from 3 doors, hiding 1 car and 2 goats. Your probability of choosing right to start with is 1 in 3. Having it confirmed that one of those other doors is a goat is actually a red herring.

So let's change the rules again to really ram the point home.

This time you get to choose your first door, then I don't open one of the other doors, you have to choose whether to stick with your one door or change to having the other two doors. You win whatever is behind the door or doors that you choose, so if you change and get a goat and a car you drive away in a brand new car with a goat in the boot.

Okay, now what do you do? When offered that choice do you stick or change?

Let's suppose the car is behind door B this time.

You pick door A.
I don't open a door.
Do you stick or change? If you stick you win a goat. If you change you win the car and a goat.

Okay, now try door B. If you stick you win the car. If you change you win two goats.

Lastly, door C. Stick, win a goat. Change, win a goat and the car.


Sticking wins you a car 1 out of 3 times, changing wins you a car and a goat 2 out of 3 times.

The only thing opening one of the doors does is show you where one goat is. You knew one of those doors was a goat, it had to be, but by changing to having two doors you're doubling your chance to win the car. You are effectively getting two picks instead of one. How does knowing for certain that one of the doors hides a goat affect that probability?


Okay, last rule change, I promise.

This time you get to pick 2 doors right off the bat. Then I open one of the doors you picked. Then I offer you a swap to the last remaining door. What do you do this time?

Okay, for this one the car is behind door C

You pick doors A and B. I open one of the to reveal a goat.
Stick with your choice and you get two goats, change and you get the car.

Now you pick doors A and C. I open door A to reveal a goat.
Stick with your choice and you win a goat and the car, change and you just get a goat.

Lastly you pick doors B and C. This time I open door B to reveal a goat.
Stick with your choice and you win a goat and the car, change and again you just get a goat.

Again, you always knew you were going to get at least one goat playing by these rules, but going for 2 doors gets you the car 2 out of 3 times. Going for 1 door gets you the car just 1 out of 3 times.

I hope I've shown that it doesn't matter how you change the rules around, the revealing of the goat doesn't alter the probabilities one iota.

 

[Red Baron Farms]

Don't see it there are two doors, you have a 50/50 choice. It's a new game after he opens the first door.

Actually there are still 3 doors, 2 unknown and 1 known. Your original choice is still only 1/3rd chances being right because it is still unknown.

Try again for you. Lets say Monty hands you a "free spin" instead of opening a goat door.

You can now either stick with your original choice, or pick one of the other two. If you pick wrong you can use your "free spin" to take the other one.

So you are now down to 2 choices. Stay put with your 1/3rd chances or shift to the other side with your "free spin". Which do you do?

 

[phunk]

Let's try it this way:

You pick one door. Instead of eliminating any doors, monty says "you can have the door you picked, or every other door." Do you switch? Of course you do.

The only difference between this and the original problem is monty keeps the goat.

 

[bruto]

Let's try it this way:

You pick one door. Instead of eliminating any doors, monty says "you can have the door you picked, or every other door." Do you switch? Of course you do.

The only difference between this and the original problem is monty keeps the goat.

Several of us have tried it this way, or variants of it, but it doesn't always register, because of the way the deal is offered. Because there are only three doors in the game, and because Monty opens a goat door (a red herring, really) before offering the swap, it is not obvious to all that what is really being offered for the swap is "every other door." It is, but the opening of a door causes confusion. It looks as if it changes the odds in a new game, but the choice is still really just a swap of the original odds: if you chose a winner initially, now you lose. If you chose a loser initially, now you win.

 

[jiggeryqua]

Let's try it this way:

You pick one door. Instead of eliminating any doors, monty says "you can have the door you picked, or every other door." Do you switch? Of course you do.

The only difference between this and the original problem is monty keeps the goat.

Also very good, though RBF got there first (and I'd suggest appending "which will always include an irrelevant goat" to Monty's verbal offer). I like this one too...

May I? It is a new game. But we're also talking about your old pick, because it's the same old doors and the car hasn't moved for this 'new' game. You were probably wrong in the old game (1/3 probability). You should, therefore, probably change in the 'new' game. I don't know why people have such trouble with this one...

But if you should probably change, regardless of the goat, then...you should probably change regardless of Monty's intervention. That is to say, suppose the game was:

• Monty shows you three doors, behind one of which is a car.
• The other two hide goats.
• You may pick one.
• Now, if Monty doesn't open any of the doors, if there are No Goats Visible, should you change your choice? You're still as probably wrong as you used to be...

Not so tough, I know - you're not going to improve your probability by changing your choice. It's less a maths question than a philosophy question - now I understand the Monty Hall question I can only perceive my choices as 'probably wrong'

 

[pgwenthold]

OK, while you are arguing the numbers, there is something being almost completely ignored: that Monty is playing fair with you.

As has been noted, this problem can only be done if you know beforehand that Monty is going to offer you the option to switch, or at least that you know he does not bring with it any motives.

For example, how do you know that the only reason he is offering you the option to switch is because he knows that you have selected the car, and that you think you are smart and have solved the problem? In that case, the chance of being right if you switch is 0% and if you don't switch, it is 100%.

Thus, if ALL you know is that he knows where the car is, and has offered you the option to switch, you do not have enough information to answer the problem. You must know that the offer to switch is not dependent on your initial guess being right or wrong. IOW, he always does this (or, he does it randomly).

If that is the case, then the answer is switch, because you have a 2/3 chance of winning.