3 doors - One contains a brand new car, the other two contain a goat. Choose.

PARlance · 24-09-2017 10:22PM

As my grandfather once said. Stick with your goat.

Wonderful man, never had any kids.

Ficheall · 24-09-2017 10:26PM

shaunr68 wrote: »

I think I understand

So let's say, again because it makes no difference, that the host shows you door 3 and that contains a goat.

We now know that case 3 cannot be true, and therefore at this point there are two possible outcomes with 2 possible rewards

Of course this works the same way if he shows you door 2 which discounts case 2

You're not looking at it in a convenient way.

Pretend the doors on the show aren't numbered yet.
You pick a door - call THAT door 1.
Label the other doors 2 and 3.
Then look at the table again.
The three cases listed show the three possible scenarios in which you now find yourself. In each scenario, pretend Monty opens one of doors 2 or 3 to show you a goat. Then consider what you ought to do, bearing the table in mind.

Hannibal_Smith · 24-09-2017 10:36PM

I just got totally distracted by the very cute visual example

Burial. · 24-09-2017 10:37PM

Switch every time. The only way you lose is if you picked the car initially.

I love that movie and seeing that maths problem rattled my brain as a young fella and ultimately made me take the math of poker seriously and made me lots and lots of money down the road.

Your Face · 24-09-2017 10:42PM

'Let's take a look at what you could have won'

Avatar MIA · 24-09-2017 10:46PM

Burial. wrote: »

Switch every time. The only way you lose is if you picked the car initially.

Which you may have, which was a 331/3% chance at the time. The other two choices being 66 2/3 %.

So, as it stands the car is mostly likely in the subset you've not chosen.

The logic I've seen so far even though one of those choices is proven not to contain the car (when the presenter opens one of the boxes) is that this subset is still 66 2/3% likely to contain the car.

THIS is the bit I don't get .

Burial. · 24-09-2017 10:47PM

shaunr68 wrote: »

I think I understand

So let's say, again because it makes no difference, that the host shows you door 3 and that contains a goat.

We now know that case 3 cannot be true, and therefore at this point there are two possible outcomes with 2 possible rewards

Of course this works the same way if he shows you door 2 which discounts case 2

You pick door 1 with a goat, he shows you the other goat behind door 3, switch to 2 you win. You pick door 2 with the car, he shows you goat behind 3, you switch to 1 you lose. You pick door 3 with the goat, he shows you door 1 with the other goat, you switch to door 2 you win. You win 2 out of 3 times when you always switch, thus 66.6%.

Initially you have a 33.3% chance, pretty straight forward. Host shows you one of the goats. To consider it now 50/50 to get the car is discounting the information he's given you. If you pick the car initially the host has a free choice of choosing which goat to show you. However if you pick a goat (66.6% chance initially) he's no choice but to show you the other goat. He has no choice 66% of the time, he can pick a door at random 33%. You're effectively trying to find out if you gave him a choice to choose or not. 66% of the time he has no choice.

shaunr68 · 24-09-2017 10:49PM

Ficheall wrote: »

You're not looking at it in a convenient way.

Pretend the doors on the show aren't numbered yet.
You pick a door - call THAT door 1.
Label the other doors 2 and 3.
Then look at the table again.
The three cases listed show the three possible scenarios in which you now find yourself. In each scenario, pretend Monty opens one of doors 2 or 3 to show you a goat. Then consider what you ought to do, bearing the table in mind.

Thanks, I still can't see it but it's been a long day, I can't get past my conclusion that this teaser doesn't take into account that probability will change based on new information. Fascinating thread, though where's the "Makes no difference" option?

24-09-2017 10:57PM

shaunr68 wrote: »

However at the point the contestant is asked to choose, one of those outcomes has now been discounted and this of course changes the odds.

It doesnt though, that is the whole trick - and the reason it messes with peoples heads so bad.

shaunr68 wrote: »

Thanks, I still can't see it but it's been a long day, I can't get past my conclusion that this teaser doesn't take into account that probability will change based on new information.

Ok here is another way to phrase the explanation that usually gets people to see the "trick" for me in the end.

Forget for a moment the door numbers. Just imagine you pick a door. Call your door "over here" and all the other doors "over there".

Now the chances you picked the right door are 1/3. That means that the chances the right door are in the _other_ doors "over there" is 2/3.

Now it does not matter if the host takes away a wrong door - it changes nothing. The chance still is 2/3 that your original choice was wrong and that therefore the right door is more likely not the one you picked and is in the group "over there".

Imagine the same thing with 100 doors. You pick one. There is a 1/100 chance your door is right. There is a 99/100 chance the right door is "over there".

Now he can take away 98 wrong doors from "over there" just leaving one. But the chances the right door is "over there" is still 99/100. Solely because the chances your original choice was right is 1/100.

Avatar MIA · 24-09-2017 11:03PM

Deleted User wrote: »

Now he can take away 98 wrong doors from "over there" just leaving one. But the chances the right door is "over there" is still 99/100. Solely because the chances your original choice was right is 1/100.

Yes, I get it now. And that's why it's key that the presenter knows which door holds the car, and which one doesn't.

Pheeeeeeeeeeew :pac:

Pherekydes · 24-09-2017 11:05PM

What type of car?

Are the goats brand new also?

Switch is the answer. You can write a simple simulator in your preferred language. If you switch you win 2/3 of the time. If you don't switch you win 1/3 of the time.

Spudmonkey · 24-09-2017 11:14PM

The Monty hall problem is easier to understand if you increase the doors.

If you had a hundred doors and you made your choice. Then 98 were opened to reveal the goats, you are almost always better off switching.

If you are now down to two doors, one with a goat and one with a car. Of course the likelihood of you picking the correct door in the first place was 1/100. Now that 98% of the wrong doors have been revealed to you, the likelihood that it's still behind your door is still 1/100. In that case you're almost always better to switch.

Whether it's 3 or 100 doors the principle remains the same.

Avatar MIA · 24-09-2017 11:19PM

Spudmonkey wrote: »

The Monty hall problem is easier to understand if you increase the doors.

If you had a hundred doors and you made your choice. Then 98 were opened to reveal the goats, you are almost always better off switching.

If you are now down to two doors, one with a goat and one with a car. Of course the likelihood of you picking the correct door in the first place was 1/100. Now that 98% of the wrong doors have been revealed to you, the likelihood that it's still behind your door is still 1/100. In that case you're almost always better to switch.

Whether it's 3 or 100 doors the principle remains the same.

Sometimes when you know something too well it can be hard to tell it correctly.

To improve your explanation above, I'd humbly add, that when they are showing you the 98 empty/goat hiding doors they know that the car/prize isn't behind them. So, when they open the 98 doors there's zero chance they'll reveal the car/prize.

shaunr68 · 24-09-2017 11:29PM

Spudmonkey wrote: »

Now that 98% of the wrong doors have been revealed to you, the likelihood that it's still behind your door is still 1/100. In that case you're almost always better to switch.

Whether it's 3 or 100 doors the principle remains the same.

I thought you were being asked to stick or twist, essentially to choose between the two remaining doors once all the others (whether one or 998) had all been proven to not contain the car? At that point then surely you have a one in two chance of making the right choice.

Anyway, I will ponder this tomorrow and maybe put it to my two expert mathematician pygmy goats, Ted and Dougal! Await their expert analysis! Anybody want to swap them for a car?

Avatar MIA · 24-09-2017 11:33PM

shaunr68 wrote: »

I thought you were being asked to stick or twist, essentially to choose between the two remaining doors once all the others (whether one or 998) had all been proven to not contain the car? At that point then surely you have a one in two chance of making the right choice.

Anyway, I will ponder this tomorrow and maybe put it to my two expert mathematician pygmy goats, Ted and Dougal! Await their expert analysis! Anybody want to swap them for a car?

When you figure it out, you'll kick yourself. I KID you not

Outlaw Pete · 25-09-2017 01:06AM

shaunr68 wrote: »

I thought you were being asked to stick or twist, essentially to choose between the two remaining doors once one of the others has been proven to not contain the car? At that point then surely you have a one in two chance of making the right choice.

Well, the choice would absolutely be a 'one in two' chance if the host was not aware of what was behind each of the doors and where it only accidentally becomes apparent that there is a Goat behind Door No.3. Under those circumstances, there would be no advantage of switching at all.

You see, when the host purposefully reveals that one of the doors has a goat behind it, they really aren't showing you anything you don't already know yourself and so that action therefore doesn't alter the mathematical odds in any way shape or form. They might as well have showed you nothing and just offered you the chance to swap Door No.1 for Door No.2 and Door No.3 combined.

Which is why I have always maintained that the Monty Hall dilemma is really more of a trick question than an actual mathematical problem.

25-09-2017 02:03AM

Me mate Anto got caught out with something similar before and they made him bring it home.

somefeen · 25-09-2017 02:18AM

This doesnt make any sense.

Nothing changes if you switch. You played a game with three doors and lost and now you are playing a game with two doors. The probability of the car being behind the door you choose is the same as the other remaining door.

I'm no stats genius, but all this seems to do is show the limitations of probability calculations.

Digital Solitude · 25-09-2017 02:46AM

somefeen wrote: »

This doesnt make any sense.

Nothing changes if you switch. You played a game with three doors and lost and now you are playing a game with two doors. The probability of the car being behind the door you choose is the same as the other remaining door.

I'm no stats genius, but all this seems to do is show the limitations of probability calculations.

Nothing changes if you switch, but its more likely that the door you originally chose was a goat than car.

I think this explanation makes sense.

There's a 1/3 chance you choose the car first time, and a 2/3 chance its in one of the other 2 doors. After one of the other doors is opened, there's a 2/3 chance the other door has the car. As there's still a 1/3 chance you chose correctly.

Its not a maths problem at all, just a brain teaser imo.

greencap · 25-09-2017 03:22AM

Digital Solitude wrote: »

Nothing changes if you switch, but its more likely that the door you originally chose was a goat than car.

I think this explanation makes sense.

There's a 1/3 chance you choose the car first time, and a 2/3 chance its in one of the other 2 doors. After one of the other doors is opened, there's a 2/3 chance the other door has the car. As there's still a 1/3 chance you chose correctly.

Its not a maths problem at all, just a brain teaser imo.

its weird though, because how does reality decide youre still playing the original game.

for example what if you were on the stage, having chosen the first door and gotten a goat, and then you said to Monty 'hang on I'll be back in 5 minutes' and then you go have a drink and take a sht.

Then you come back and theres a stage, a host, 3 doors (one which happens to have a goat standing in it).

Now your job is to choose a door. Door 1 or door 2.
Aren't you a different person now, different from the you of 5 mins ago.
Or do your prior decisions follow you forever?

Ok .... now if Monty says and does nothing, and you just walk up and choose a door, then thats a 50/50 choice surely, right?

But if while you're walking up to the doors Monty says 'would you like to switch' then the maths goes all funny.

How does reality keep track of your odds.

Like if I'm playing and get the first goat and monty says 'stick or switch' I'm now facing mathematical laws cause ->I<- chose a certain door.

But say my mate bob is in the audience and I'm standing there faced with a goat and 2 doors and I wave at bob and say 'hey bob pick a door' then that HAS to be a 50/50 for bob, thats the very definition of a 50/50.

dat iz mental innit.

Digital Solitude · 25-09-2017 04:31AM

greencap wrote: »

its weird though, because how does reality decide youre still playing the original game.

for example what if you were on the stage, having chosen the first door and gotten a goat, and then you said to Monty 'hang on I'll be back in 5 minutes' and then you go have a drink and take a sht.

Then you come back and theres a stage, a host, 3 doors (one which happens to have a goat standing in it).

Now your job is to choose a door. Door 1 or door 2.
Aren't you a different person now, different from the you of 5 mins ago.
Or do your prior decisions follow you forever?

Ok .... now if Monty says and does nothing, and you just walk up and choose a door, then thats a 50/50 choice surely, right?

But if while you're walking up to the doors Monty says 'would you like to switch' then the maths goes all funny.

How does reality keep track of your odds.

Like if I'm playing and get the first goat and monty says 'stick or switch' I'm now facing mathematical laws cause ->I<- chose a certain door.

But say my mate bob is in the audience and I'm standing there faced with a goat and 2 doors and I wave at bob and say 'hey bob pick a door' then that HAS to be a 50/50 for bob, thats the very definition of a 50/50.

dat iz mental innit.

You wouldn't pick another door when you return from your ****, you've already chosen. You pick a door, are offered a switch and then go for a ****e. When you return, your options are still stick or switch.

Its 50/50 for Bob because he chooses from two doors, not three.

In both cases, you've a higher likelihood of getting the car if the other door is chosen.

dclifford · 25-09-2017 06:58AM

I can see why the OP used goats instead of booby prizes. Imagine where the thread would go, on after hours if he had said booby prize.

Bishopsback · 25-09-2017 08:02AM

Fabriel wrote: »

It's not weird at all, you just don't understand it. Imagine if it were one thousand doors.

You pick one door then the host open 998 doors with goats leaving your chosen door and one otjer door unopened. Do you really think you have a 50/50 chance of having the door with the car? In reality it is still one in a thousand that your door has the car. You should switxh doors.

Nope, every door he opens with a goat reduces the odds, if it then comes down to two doors in the end its 50/50, if you switch and get the car, hurrah, if you get the goat, you look like a goat, or vice versa.
Either way good luck!

thisistough · 25-09-2017 08:24AM

I'd imagine the rate of depreciation on a goat is far less than on a new car, they'd also provide a better return - I'd take the goat

corner of hells · 25-09-2017 08:36AM

Is the goat diesel or petrol ?

Outlaw Pete · 25-09-2017 09:00AM

Bishopsback wrote: »

Nope, every door he opens with a goat reduces the odds..........

Nope, the odds would only change to 50/50 if it had been randomly or accidentally revealed where one of the Goats were........ that's the crucial part of the paradox / trick question.

Imagine that the host offers you the option to switch what's behind Door 1 for what's behind Door 2 & Door 3 combined and WITHOUT revealing to you where one of the Goats was........... that would be a no-brainer for people and pretty much everyone would switch as they'd (correctly) see it as getting to swap what's behind one door for what's behind two doors.

But switching Door 1 for Door 2 (after Door 3 is revealed to have a Goat behind it) is the exact SAME as switching Door 1 for Door 2 & Door 3 combined.... simply because it being revealed that there was a Goat either behind Door 2 or 3 did nothing....... as we already knew there was at least one Goat behind those doors. Nothing changed and so the probability didn't change.

Therefore, switching gives you a 66.6% chance of winning but sticking only gives you a 33.3% chance.

25-09-2017 09:06AM

I never liked this one. I always just chalk it up to not being pure chance, it's a probability given prior information which is another kettle of fish
Bayes law n ****
I do wish people had showed me that proof in the first place instead of tormenting me with stuff about 98 doors

razorblunt · 25-09-2017 09:08AM

somefeen wrote: »

This doesnt make any sense.

Nothing changes if you switch. You played a game with three doors and lost and now you are playing a game with two doors. The probability of the car being behind the door you choose is the same as the other remaining door.

I'm no stats genius, but all this seems to do is show the limitations of probability calculations.

No, but the odds have moved in your favour. You now stand a better chance of winning by switching.

25-09-2017 09:10AM

Somefeen: http://angrystatistician.blogspot.ie/2012/06/bayes-solution-to-monty-hall.html?m=1

Chancer3001 · 25-09-2017 09:10AM

Imagine a million doors.

Host knows where the car is.

You pick one at random

Host says , ok cool, the car is either behind that door you selected or this one here... (one of them does indeed have the car)

The chances of it being behind his door is way higher. He knows where the car is

3 doors - One contains a brand new car, the other two contain a goat. Choose.

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