Brain teaser.

[Timless Timber]

You are a contestant on a game show.
You are presented with three doors.
Behind one of the doors is a new car.
Behind the other two doors are goats.

You are allowed to choose one of the doors but it isn’t opened yet.
Instead, the host of the show opens one of the other doors, to reveal a goat.
Two doors remain closed.
Host then gives you a choice:
A ) stick with your original door
or
B ) switch to the other unopened door instead.

Should you switch?
Before we go any further, there are a few things that must be clarified.
The first is that the car is always placed randomly amongst the three doors.
Second, the host always knows where the car is.
The third is that the host will always open a door to reveal a goat.
The last important detail is that the contestant (that’s you) knows all of this information ahead of time.

Switch or not? :?

[doug3030]

I would stick with the original decision because I would rather have a goat than a car. :D

[_fly_]

You ALWAYS switch for you best chance.

If you stick you have a 1/3 chance.
If you swap you get 2/3 chance, double.

Didn't mythbusters do something on this?

Called the Monty Hall problem

[Timless Timber]

Indeed and often referred too as the Monty Hall Problem. :2tsup:

The popular reasoning goes like this:

after opening a door to reveal a goat, Host leaves you with two possibilities, so the odds must be 50/50 and therefore there is nothing to be gained by switching.
You could switch, but there is no reason to do so.
The popular reasoning is wrong.


As strange as it may seem, you should switch doors – it doubles your chances of winning!
Without switching, you would only win a third of the time.
By switching doors, you will win two thirds of the time.


How is that possible?

Perhaps the easiest way to see that switching is the proper strategy is to work through an example.

Let’s assume you choose Door 1.
If the car is behind Door 2, then Host (Monty) has no choice but to reveal Door 3 (a goat).

Switching from Door 1 to Door 2 wins for you.

Likewise, if the car is behind Door 3 then Host (Monty) is forced to reveal Door 2.

Switching from 1 to 3 wins for you.

The only time you lose by switching is if you happened to choose the right door from the outset.

But we know that only happens one third of the time.

The other two thirds of the time switching is the correct strategy.


If you’re still not convinced, imagine there were 100 doors.
Behind 99 of them are goats and behind one is the car.
You choose a door.
Host (Monty) then opens 98 of the doors with goats and leaves only your original door and one other.
He then gives you the option of switching.
What’s more likely, that you picked the right door at the very beginning or that Host (Monty) was avoiding the winning door when he was opening up all of the losers?
Of course it’s the latter.
In 99 times out of 100 Host Monty has to avoid the correct door.
He’s practically pointing it out to you!
In this extreme example, only one time out of 99 will a switching strategy lose for you.


The same logic applies to the standard three-door example.
Statistically speaking, two thirds of the time you’ve chosen the wrong door initially and Host (Monty) is literally showing you the other wrong door. The one remaining is the correct one and you should switch to it as soon as you can.

Simple huh? :D

[switt775]

Sorry to burst your bubble, but your math is wrong. You outline 3 scenarios, there are actually 4.

In the case you selected the correct door to begin with, Monty can do one of 2 things. He can open the first incorrect door, or he can open the second incorrect door. The outcome is the same, but these are still 2 different scenarios.

When you made your original choice, your odds were 1 in 3. Once he has opened one of the incorrect doors, you have two doors remaining. He has now improved your odds to 1 in 2. No amount of gyrating changes those odds to 2 in 3.

Same applies to the 100 doors by the way. There is 1 correct door, and you have made 1 choice. By eliminating 98 wrong doors, Monty has improved your odds to 1 in 2 in the final choice.

[doug3030]

Sorry Switt, the maths in TT's solution was right.

The person whose strategy is to switch will in the balance of probability win 2/3 of the time as opposed to 1/3 of the time if they dont switch.

Look at it this way. If you dont switch you only win the car if you pick the door with the car, which is 1/3 of time. In this case the host has two doors to pick from to reveal a goat.

If you do switch you win the car every time you initially pick a door with a goat which it 2/3 of the time. The host has the choice of only one door to open to reveal the other goat.

It takes a bit of thought to grasp it, but the odds do actually double for a player whose strategy is to switch every time.

Cheers

Doug

PS: I would still rather have the goat, they are cheaper to run and less trouble.

[Avery]

Don't know anything about the maths or the logic. I just know that I would always get the goat and when I opened the door it would probably spring back, hit the goat, startle it, cause it to charge at me breaking two ribs and an arm.

At least that's what happened last time.

[switt775]

I stand by my earlier post. Here's the proof.

To simply things I've only included the scenario where the car is behind door 1, and goats are behind doors 2 and 3. But if you work through D2 = Car and D3 = Car you will see they only extend the results with the same outcome.


<colgroup><col span="4"><col><col span="2"></colgroup><tbody>
D1
D2
D3
I Select
Monty Eliminates
Switch?
Result


Car
Goat
Goat
D1
D2
N
W


Car
Goat
Goat
D1
D2
Y
L


Car
Goat
Goat
D1
D3
N
W


Car
Goat
Goat
D1
D3
Y
L


Car
Goat
Goat
D2
D3
N
L


Car
Goat
Goat
D2
D3
Y
W


Car
Goat
Goat
D3
D2
N
L


Car
Goat
Goat
D3
D2
Y
W


</tbody>


You will see there are twice as many possible options where you have selected the correct door. That's because Monty has two different wrong doors to play with, and he can eliminate either one. So there are 4 possible paths, and two of them (the ones where you stick with your original choice) result in winning the car.

If you originally selected a wrong door (D2 or D3) Monty has only 1 option. He eliminates the remaining wrong door. So in this case you need to switch doors in order to win the car.

Bottom line, there are 8 possible scenarios.

In 2 of them, you stick with your original choice and win.
In 2 of them, you stick with your original choice and lose.
In 2 of them, you change your original choice and win.
In 2 of them, you change your original choice and lose.

I rest my case.

[_fly_]

OK, your a former geek.

try reading the proof here Monty Hall problem - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Monty_Hall_problem).

or look up the "monty hall problem" in your favourite search engine.

[switt775]

OK, your a former geek.

try reading the proof here Monty Hall problem - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Monty_Hall_problem).

or look up the "monty hall problem" in your favourite search engine.


Yep, read that. The analysis is flawed, because as shown in my proof above, it ignore the fact that there are 4 possible outcomes where you have initially selected the correct door, and only two possible outcomes for each wrong door selection.

Wikipedia is wrong (not the first time).

[doug3030]

Switt, there are only three scenarios to start out.

1, You pick the door with the car

2. You pick the door with goat 1

3. You pick the door with goat 2

If your strategy is to stick with your first pick, you win the car only when you pick the door with the car, statistically once in 3

If your strategy is to switch no matter what, you win the car every time you initially pick a goat, statistically 2 in three. If you switch and you have a goat in your chosen door, the other goat has been eliminated by Monty, so you must be switching to the car. This happens in 2 of the 3 initial selection possibilities.

If you initially chose a door with the car you would be switching to the remaining goat, but your initial chances of picking the car are only 1 in 3.

What is behind the door that Monty chooses not to open, when he has a choice, is not totally random because of information we already have.

I hope this helps.

Cheers

Doug

[doug3030]

I stand by my earlier post. Here's the proof...

You will see there are twice as many possible options where you have selected the correct door. That's because Monty has two different wrong doors to play with, and he can eliminate either one. So there are 4 possible paths, and two of them (the ones where you stick with your original choice) result in winning the car.

Right here is where you logic goes wrong Switt. You are looking at the possible options that Monty has to play with, not what the contestant is left to choose from.

Yes, if you have picked the door with the car Monty has a choice of two doors to open, but whichever one he opens, the contestant,and thats whose choices we are evaluating, still is left with only one door to switch to. If he does not have the car behind the door he initially chose then the car must be behind the only remaining door.

The puzzle is looking at the choices the contestant has left after Monty has opened a door, not how many doors Monty has a choice of opening.

Cheers

Doug

[Skew ChiDAMN!!]

Sorry Switt, your chart lays out the possible outcomes, but does NOT lay out the chances (probability) of each one.

In simple terms:

Your first pick has a 1 in 3 chance of winning. Which means the odds are 2:3 it's behind one of the other doors.

It is a certainty that Monty will open a "losing" door... so the remaining, unopened door still has 2:3 chance of being the winner.

1:3 for your first pick, compared to 2:3 for a switch.



Hell, even if I didn't believe the maths, in the same situation I'd tell myself - at the final choice - "I have a 50/50 here, so I'll switch simply because the savants (who're s'posed to understand these things) say it's the better choice."

A bit like my consistently choosing heads in a coin toss, which everyone "knows" is 50/50 but actually is not... it depends on how well-balanced the coin in question is. :rolleyes:

[switt775]

Switt, there are only three scenarios to start out.

1, You pick the door with the car

2. You pick the door with goat 1

3. You pick the door with goat 2

If your strategy is to stick with your first pick, you win the car only when you pick the door with the car, statistically once in 3

If your strategy is to switch no matter what, you win the car every time you initially pick a goat, statistically 2 in three. If you switch and you have a goat in your chosen door, the other goat has been eliminated by Monty, so you must be switching to the car. This happens in 2 of the 3 initial selection possibilities.

If you initially chose a door with the car you would be switching to the remaining goat, but your initial chances of picking the car are only 1 in 3.

What is behind the door that Monty chooses not to open, when he has a choice, is not totally random because of information we already have.

I hope this helps.

Cheers

Doug

Doug, I stand by my analysis because it is mathematically correct.

Initially there are 3 possible outcomes. But because Monty always eliminates a door with a goat, there are 4 outcomes, as shown in the matrix above. Monty is intentionally improving the odds of success to 50%. No more, no less.

There are many pitfalls in logic, but mathematics don't lie.

Here is another logic / math puzzle to ponder. You decide to flip a coin 10 times and record the results. The coin and your flipping technique have been carefully examined by Monty himself, and there is no bias in the coin, or your way of flipping it.

Q1. What are the odds that you will flip 10 consecutive Heads?

Q2. You have amazed yourself and your friends by flipping 9 consecutive Heads. What are the odds that the 10th flip will be Heads?

[Timless Timber]

Before we go any further, there are a few things that must be clarified.
The first is that the car is always placed randomly amongst the three doors.
Second, the host always knows where the car is.
The third is that the host will always open a door to reveal a goat.
The last important detail is that the contestant (that’s you) knows all of this information ahead of time.

Switt 775's 'mathematical response' ignores this "precondition". :wink:

Here

https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-prn2/1497450_688388197848413_53952321_n.jpg

Cheers! :p :D

[switt775]

Is that the goat from Jurassic Park? If so, you might want to get the kids out of the room..:C

[doug3030]

Q2. You have amazed yourself and your friends by flipping 9 consecutive Heads. What are the odds that the 10th flip will be Heads?

I wont bother with question 1 but the answer to question 2 is 50/50. It is not dependent of the throws that preceded it.

If you wish to maintain that your analysis is correct, when it is analyzing the choices that Monty has and not the choices that his actions leave the contestant with, I have three cups and a little ball. Bring your money and come down for a little game. :D

Cheers

Doug

[switt775]

I wont bother with question 1 but the answer to question 2 is 50/50. It is not dependent of the throws that preceded it.

If you wish to maintain that your analysis is correct, when it is analyzing the choices that Monty has and not the choices that his actions leave the contestant with, I have three cups and a little ball. Bring your money and come down for a little game. :D

Cheers

Doug

Happy to Doug. As long as you follow Monty's rules.:D

I included the second puzzle because it illustrates the trap of using logic rather than math. You are correct, the chances of getting heads on any individual throw is 1/2, so the fact you've just rolled 9 consecutive heads is irrelevant. A coin has no memory. Many reason because the odds were so small at the beginning ( 1 in 1024) they must still be very low.

On the 1st puzzle, we will have to agree to disagree.

[_fly_]

the difference is each throw is independent from all other throws.
The three doors are not independent, the prize is only behind 1 of the three.
And it makes no difference where the doors are in the puzzle, makes no difference if they open door 3 or door 2 to show a goat.
But I'll let you disagree anyway.

My rule explain it twice and if you still haven't got it, your on your own.

[doug3030]

Happy to Doug. As long as you follow Monty's rules.:D

Ok We will follow Monty's rules. I will be the contestant you can be Monty.You put up $50 as the prize for each game and I will pay you $50 for every game I lose.

To reiterate the rules from TT's initial post:

"You are presented with three doors.
Behind one of the doors is a new car.
Behind the other two doors are goats.

You are allowed to choose one of the doors but it isn’t opened yet.
Instead, the host of the show opens one of the other doors, to reveal a goat.
Two doors remain closed.
Host then gives you a choice:
A ) stick with your original door
or
B ) switch to the other unopened door instead.
The first is that the car is always placed randomly amongst the three doors.
Second, the host always knows where the car is.
The third is that the host will always open a door to reveal a goat.
The last important detail is that the contestant (that’s you) knows all of this information ahead of time."

My rules: once you start you are committed to a minimum of 99 games. We woupd of course simulate the goats and cars with other objexts

If you come down bring at least $2000 cos statistically I stand to win $1650. It could be more if I get lucky. If you play the minimum 99 games I will refund your airfare, so long as its a normal domestic air fare, no sneaky travelling via Europe.

Of course if you are right then we should break even on the game but I will still pay your air fare.

Switt775 COME ON DOWN!

Cheers

Doug

[robbygard]

actually, not on his own ... i'm with switt ... i concede my time with maths is probably now antique, but i DO favour maths over logic ever since primary school in the 1960s when a teacher tried to explain that one couldn't shoot a running rabbit with an arrow fired from a bow because by the time the arrow reached the position of the rabbit, it would have moved on a bit etc

of course if the rabbit was smart enough to run sideways, it might get away with it. :) but that's not how logic works

regards david

[Wongo]

Chance of getting a good prize is 2/3. It must be a NZ game show.

:D :D

[doug3030]

Chance of getting a good prize is 2/3. It must be a NZ game show.

:D :D

More than a game show, it is proof that people think emotionally not logically. Look at the 100 doors example referred to in TT's explanation. The contestant picks one box hoping it is the winner. The host then starts to open the other 99 boxes to reveal only losing boxes but he knows which are winners and which are losers. Every time the host opens a new box it is increasing the chance that the prize is still in one box that he has not opened but the contestant who has an emotional need to feel lucky and smart, sees it as reinforcement that he has chosen the right box right from the start. He does not consider that the chance of the host revealing the prize is ZERO because he knows where it is. He will only have the right box one time in 100 and the rest of the time the right box will be the last one left unopened by the host. Despite the odds against it it is human nature for people to think that with two boxes left to go its still 50/50 so I will hold on to what I have. "Wouldn't it be terrible if I picked the right box from the start and gave it away at the end?"

[Wongo]

It makes no difference.

You are left with 2 choices and the odds are 50/50

Your 2 choices are,
1 - pick the one you did not pick
2 - pick the one you did pick

1 & 2 have equal chance so how do you get a better change by swtiching?

[doug3030]

:banghead: :banghead: :club: :club:

I am not gong to explain it any more. If you wish to disprove it let me know and take up the offer I made to Switt above and bring your money.

[Wongo]

I understand how it works now. I missed the fact the host will always show the goat.

[AlexS]

...but one goat...

And was it a good looking goat?

[CAG]

In the 100 door scenario, the two remaining doors both had a 1 in 100 chance of initially concealing the car. As the other 98 doors are opened, BOTH doors have their odds of concealing the car increase to.... well, what do you know?.....50% each.

The host cannot open your door if you have the car (he cannot open your door full stop) and he cannot open a door you did not choose if it conceals the car (prematurely stops the game.) Therefore, every show without fail, the host works his way through to two doors remaining. Half of the time, one door will have the car and you will have won; half of the time, still one door will have the car and you will have won...a goat. WIN WIN!!!

[labr@]

Actually there are only 10 possibilities no matter how many doors: either you win or you lose, so really the odds are 1 in 10.

(Of the 10 types of people in the world 1 will understand this deviation from topic and the other won't.:D)

[doug3030]

Actually there are only 10 possibilities no matter how many doors: either you win or you lose, so really the odds are 1 in 10.

(Of the 10 types of people in the world 1 will understand this deviation from topic and the other won't.:D)

Labr@, that is very naughty of you, bringing up that old thread (ok I admit I thought of doing that but some of them are having enough trouble with this one).

Congratulations on how smoothly you introduced it.

Cheers

Doug

EDIT: At least this way I wont get the blame this time :D

[labr@]

EDIT: At least this way I wont get the blame this time :D

Dang! I forgot about that. :doh:

[CAG]

Isn't anyone going to pull me up on my logic? I thought this would spur you guys on for sure! :U

[doug3030]

Isn't anyone going to pull me up on my logic? I thought this would spur you guys on for sure! :U

I'm not going to correct your logic, just come on down and bring your money :D

Cheers

Doug

[_fly_]

OK
..................................a=b
multiply by a...........a^2 = ab
subtract b^2. a^2 - b^2 = ab - b^2
factorize........(a+b)(a-b)=b(a-b)
cancel a-b................a+b=b

since a = b............b + b = b
................................2b = b
cancel b......................2 = 1.

therefore all you 50% people can give me $2 and I'll give you $1 back, cos its the same thing, I just proved it to you.

[doug3030]

OK
..................................a=b
multiply by a...........a^2 = ab
subtract b^2. a^2 - b^2 = ab - b^2
factorize........(a+b)(a-b)=b(a-b)
cancel a-b................a+b=b

since a = b............b + b = b
................................2b = b
cancel b......................2 = 1.

therefore all you 50% people can give me $2 and I'll give you $1 back, cos its the same thing, I just proved it to you.

I'm not sure what it proves really, but whatever he is drinking, can I have a double?

Doug

[Wongo]

That is good maths and I love good maths. But don't waste your time, just say 1=2? :D

[enelef]

So many zeros - my brain hurts.

By the way - why are you trying to factorise zero for anyway? (ab-b^2=0) because a=b and a^2 = ab then ab also = b^2

Or did i miss the joke?

[bosox]

In Philippines, they eat goats. Just saying.