Two Goats, One Car!

OutlawPete

Nope, not that - although I do have that clip with some fine goats via PM if you want it

This is from the movie 21, where Kevin Spacey poses the following scenario:

https://www.youtube.com/watch?v=sXMVXxyIWzU

Many of you may know what the answer to this is and know what the name of this problem is, but would ask that you hold off posting that Wiki page for awhile, so that we can see what people would choose when presented with the problem for the first time. Even certain mathematicians have argued over this, so no shame in not selecting the right answer here and the Poll is private so as to prevent any condescending finger pointing.


So, to repeat: you are on a Game Show and given the choice of one of three doors, behind two of them stand: Goats .. and the other: a Car!

The Game Show host knows where the Car is and after you choose, he opens one of the other doors to reveal a Goat!

He then offers you the chance to change your mind and go for the other remaining door ..

So, if you were on that Game Show: would you take up the offer and switch doors or would stick with your original choice?

Is it in your advantage to switch in other words?

I myself like Goats, and so would be just as happy to lose.

extra-ordinary_

Atari jaguar?

Spunge

You should always switch, or so i've heard.

FatherLen

i dunno prob stay with the same door.



am i right?

Pj!

Monty Hall Problem


Switch

Screaminmidget

Sounds like a ****e gameshow tbh.
I think most people would switch, for the simle fat that you odds of winning the car have gone from 3/1 to 2/1??? or am i missing something??

hypervalve

Pj! wrote: »

Monty Hall Problem


Switch

You just couldn't resist could you?

Pj!

No

FatherLen

Pj! wrote: »

Monty Hall Problem


Switch

you ruined my day.









and this thread

This is psychology plus probability isn't it? I never became convinced of it on pure probability
So for the game show I'd switch but in pure theory I'd say you're ok to stick

Mr. Presentable

Pj! wrote: »

Monty Hall Problem


Switch

Only took four replies

KnifeWRENCH

I haven't learned to drive yet. I have no use for a car.

I'll have the goat, please.

Fremen

Fúck your honda civic. I want a goat.

Meeeh-eeeeh-eeeeh-eeeeeeh!

pickarooney

I reject all mathematical 'evidence' that it makes a difference either way.

TeddyTedson

Well there's 2 doors left so it's a 50/50 chance. Simples!:p

https://www.youtube.com/watch?v=c6IDO6M4Occ

flyton5

https://www.youtube.com/watch?v=mgRP7JlBfWQ

Is this the answer?

Earthhorse

Every time I see this again I think "This can't be right". But then I read the explanation and it appears it its. It feels wrong though.

pickarooney

I get the explanation but I think it's based on the flawed premise of 'changing' the original 1/3 choice. Essentially, once the goat is revealed you're faced with an entirely new choice between two doors and you pick A or B, 50:50 - what you originally chose is no longer directly relevant.

In fact, the whole first part - the original choice and the reveal, are completely irrelevant. You can't 'lose' as your choice is not revealed. Whatever you do, the host will simply reveal a goat and leave you with two doors, behind one of which is a car. The idea of a 2/3 probability for two options is false logic.

OutlawPete

Well, seen as it's now been revealed s 'Monty Hall' now

Here's the full clip:

https://www.youtube.com/watch?v=Zr_xWfThjJ0

Good film by the way and here's an interesting article on the Monty Hall problem from American Scientist:

Monty Hall Redux
How do you persuade yourself that a statement is true or an answer is correct? How do you persuade someone else? Various subcultures have evolved strategies and devices for dealing with these issues—scientific experiment, mathematical proof, trial by jury, consulting holy writ or the Guinness Book of Records. The trouble is, these methods are not perfectly reliable. Sometimes they convince people of a falsehood or fail to convince them of a truth.

In the July-August issue of American Scientist I reviewed Paul J. Nahin's Digital Dice: Computational Solutions to Practical Probability Problems, which advocates computer simulation as an additional way of establishing truth in at least one domain, that of probability calculations. To introduce the theme, I revisited the famous Monty Hall affair of 1990, in which a number of mathematicians and other smart people took opposite sides in a dispute over probabilities in the television game show Let's Make a Deal. (The game-show situation is explained at the end of this essay.)

When I chose this example, I thought the controversy had faded away years ago, and that I could focus on methodology rather than outcome. Adopting Nahin's approach, I wrote a simple computer simulation and got the results I expected, supporting the view that switching doors in the game yields a two-thirds chance of winning. But the controversy is not over. To my surprise, several readers took issue with my conclusion. (You can read many of their comments in their entirety here.) For example, Bruce Sampsell of Chapel Hill, N.C., wrote:

The initial statement of the problem as three doors with a prize randomly placed behind them does have a 1/3 probability of being the right choice prior to the opening of a door that doesn't contain a prize. Once a door is opened without the prize, the problem changes to two doors, each of which is equally likely to have the prize behind it. There is no advantage to switching doors.

It seems we are at an impasse. Neither the mathematical arguments given in my review nor the simulation results I reported there were enough to persuade Sampsell. On the other hand, Sampsell's assertion does not persuade me. How can we settle our differences? I could try further arguments and analysis; so could Sampsell. I could publish my simulation program so that others could run it for themselves; but my program could have an error or might be deliberately rigged to give the result I favor. There's an online version of the game created to accompany a John Tierney article in the New York Times; but in some quarters even the Times is not above suspicion.

Another correspondent, Ingrid Eisenstadter of New York City, put it this way:

So, it doesn't really matter that Mr. Hayes has written a program showing that you should switch doors after one has been eliminated, because it is based on the notion that one cannot reselect the original door, leaving arbitrary assumptions, word games and personal opinions. I would be happy to provide a similar computer program showing that the world is flat.

The mention of assumptions and word games introduces an important caveat. We can't expect to reach the same conclusions unless we all play by the same rules and understand the problem in the same way. A number of commentators—going back to the first wave of controversy in the early 1990s—have pointed out that certain assumptions are crucial to the analysis of the Monty Hall puzzle. In particular, it's important that Monty Hall must always open one door and offer the option of switching, and the door opened can never be the one initially chosen by the contestant, nor can it be the winning door.

Differences in the interpretation of the problem statement doubtless account for many of the continuing disputes over the Monty Hall puzzle. But there is a residuum of disagreement that is not so easily explained away. In another letter, David Lippmann of Austin, Texas, offers this analysis:

If Monty Hall always opens a door that does not conceal the prize and that the contestant has not chosen, there are only eight possible outcomes. Suppose that the doors are A, B and C and that the prize is behind C. The possible pairs of guesses are A followed by A, A followed by C, B followed by B, B followed by C, C followed by B, C followed by C, C followed by A, and C followed by C. Four of the second guesses are C, independent of what the first guess was. The probability that the contestant will win the prize is 1/2.

Lippmann and I apparently share the same assumptions about the game-show rules, and I even agree with his conclusion, in a limited sense: If the player chooses randomly whether to stay or switch, then the probability of winning is indeed 1/2. Nevertheless, I believe that Lippmann's analysis is incorrect, and a player who always switches doors has a two-thirds chance of winning. (The source of the error is that the eight cases do not all have the same probability. Monty Hall sometimes has two choices about which door to open, sometimes only one.)

The issue that concerns me here is not who is right and who is wrong about the odds of winning on Let's Make a Deal. The issue is how I can persuade anyone that my answer—or any particular answer—is correct. I have been stewing about this for several weeks, frustrated that I am powerless to communicate what I take to be a simple truth. But I've finally decided that what the episode demonstrates is the vigorous good health of the scientific enterprise.

Making progress in the sciences requires that we reach agreement about answers to questions, and then move on. Endless debate (think of global warming) is fruitless debate. In the Monty Hall case, this social process has actually worked quite well. A consensus has indeed been reached; the mathematical community at large has made up its mind and considers the matter settled. But consensus is not the same as unanimity, and dissenters should not be stifled. The fact is, when it comes to matters like Monty Hall, I'm not sufficiently skeptical. I know what answer I'm supposed to get, and I allow that to bias my thinking. It should be welcome news that a few others are willing to think for themselves and challenge the received doctrine. Even though they're wrong.

All the same, in the future I think I'll find some other example when I need to illustrate probability calculations.

The Monty Hall Probability Puzzle
The puzzle that has entered the lore of mathematics is not exactly the same as the game played on television. The situation in the mathematical puzzle is this: You are shown three doors and told that exactly one of them has a prize behind it. After you choose a door, Monty Hall (who knows where the prize is) opens one of the two unchosen doors, showing that the prize is not there. You are then offered a choice: Stick with your original door or switch to the remaining unopened door.

Lumen

Earthhorse wrote: »

Every time I see this again I think "This can't be right". But then I read the explanation and it appears it its.

Me too. Then someone posts this...

TeddyTedson wrote: »

Well there's 2 doors left so it's a 50/50 chance. Simples!:p

...and I'm back to square one.

OutlawPete

Lumen wrote: »

...and I'm back to square one.

Aye and as all independent events with 1/2 probability (such as flipping a coin) are always 50:50 chances, it appears that way.

But because of what has occurred previously (the host was going to show you a Goat no matter what Door you chose) the chances are increased by switching, as if he didn't open the other Door - the chances could be that the reason for that was because it contained the car.

dougal-maguire

Lumen wrote: »

Me too. Then someone posts this...



...and I'm back to square one.

maybe this is totaly wrong,but the way im lookin at it is chances of picking a goat.you have a 66% chance of picking a goat,if you pick the goat he has to eliminate the other goat,so you switch.if you stick to this you have a 66% chance of picking the car when you switch.

you have a 33% chance of picking the car from the start,he eliminates a goat,you now have a 50% chance that you picked the car.if you stick you'll win the car 50% of the time if you switch you'll win the car 66% of the time. :cool:


so would this logic be of any benefit to deal or no deal contestants?

AckwelFoley

The odds of winning are good.

2 goats one car.


You cant fcuk a car.


Jackpot

Earthhorse

OutlawPete wrote: »

But because of what has occurred previously (the host was going to show you a Goat no matter what Door you chose) the chances are increased by switching, as if he didn't open the other Door - the chances could be that the reason for that was because it contained the car.

Yes, and this is where I believe pickarooney is wrong. What goes on before the second choice is actually relevant in determining the probability. I don't know what practical application the problem has though.

I think if it was a, b, c and then one door was *randomly* picked & eliminated, you would then have 50/50 assuming the door *randomly* picked was not the car.

But because monty knows which door is the car, I think that takes it out of random probability realms, so the explanation probably does make sense for that situation.
"maybe this is totaly wrong,but the way im lookin at it is chances of picking a goat.you have a 66% chance of picking a goat,if you pick the goat he has to eliminate the other goat,so you switch" yeah, that

fabbydabby

A lot of people know about this problem because of that bestselling book the Curious Incident of the Dog in the night

pickarooney

dougal-maguire wrote: »

maybe this is totaly wrong,but the way im lookin at it is chances of picking a goat.you have a 66% chance of picking a goat,if you pick the goat he has to eliminate the other goat,so you switch.if you stick to this you have a 66% chance of picking the car when you switch.

you have a 33% chance of picking the car from the start,he eliminates a goat,you now have a 50% chance that you picked the car.if you stick you'll win the car 50% of the time if you switch you'll win the car 66% of the time. :cool:


so would this logic be of any benefit to deal or no deal contestants?

But imagine a slightly different game where there are three doors. One is open with a goat behind it and two are closed. One has a car. You choose. No game show host, no music, just the doors.

Obviously, nobody who wants the car is going to pick the open door. That leaves two doors with one goat and one car. 50:50.

Here's the crux of the difference of opinion and until soemone adequately explains the causality of the initial rigmarole with the guess and the non-random opened door there is no chance that the 50/50 camp will ever be swayed.

Earthhorse

pickarooney wrote: »

But imagine a slightly different game where there are three doors. One is open with a goat behind it and two are closed. One has a car. You choose. No game show host, no music, just the doors.

Obviously, nobody who wants the car is going to pick the open door. That leaves two doors with one goat and one car. 50:50.

Here's the crux of the difference of opinion and until soemone adequately explains the causality of the initial rigmarole with the guess and the non-random opened door there is no chance that the 50/50 camp will ever be swayed.

Not sure anyone can restate the problem, or solution, in such a way that it will sway you. Best way to convince yourself is to set up an experiment if you have enough people to do this with. Have one group follow a strategy of switching and the other of non-switching and see which group wins more.

OutlawPete

Earthhorse wrote: »

Best way to convince yourself is to set up an experiment if you have enough people to do this with. Have one group follow a strategy of switching and the other of non-switching and see which group wins more.

That's done here, very scripted though:

https://www.youtube.com/watch?v=o_djTy3G0pg

pickarooney

I first approached the problem with a completely open mind but I'm afraid I've become attached to my reasoning now. The last time I tried to prove something mathematically sound to myself through experimentation (the old shared birthday question) I ended up 'proving' it wrong over and over.

I'm sure someone has a way of explaining the link between part one and part two. It looks like bluewolf has been successfully won over, bless her little actuarial socks :P

pickarooney

pickarooney wrote: »

I first approached the problem with a completely open mind but I'm afraid I've become attached to my reasoning now. The last time I tried to prove something mathematically sound to myself through experimentation (the old shared birthday question) I ended up 'proving' it wrong over and over.

I'm sure someone has a way of explaining the link between part one and part two. It looks like bluewolf has been successfully won over, bless her little actuarial socks :P

Only because they're not unbiased random independent events :pac:
if it was random I'd still stick with 50 50


edit: interestingly, a friend once decided to experiment to test the hypothesis that it's 66%/switch - she did end up "proving" it in the majority of the runs

strobe

pickarooney wrote: »

until soemone adequately explains the causality of the initial rigmarole with the guess and the non-random opened door there is no chance that the 50/50 camp will ever be swayed.

It's more than that even Pick. It has been adequately explained, in several different ways but for some reason a large chunk of 'the 50/50 camp' will never be swayed.

I actually find this fact far more fascinating than the problem itself. I've often wondered if there has ever been a psychological study done to see if there are any kind of common traits amongst the hardcore 50/50ists.

I spent almost 4 drunken hours on and off one night before trying come up with a way to explain this to an ex girlfriend (probably part of the reason the ex part is in there? ) but no matter what she was unshakable in her conviction.

Lumen

Earthhorse wrote: »

Not sure anyone can restate the problem, or solution, in such a way that it will sway you. Best way to convince yourself is to set up an experiment if you have enough people to do this with. Have one group follow a strategy of switching and the other of non-switching and see which group wins more.

OK, here goes.

The idea that the odds are 50:50 is based on the idea that the probability of the car being in either is the same. This is false. Just because you have complete ignorance of two things, it doesn't make them equally probable (see Climate Change poll).

The car is more likely to be in the one that you didn't choose, because of the knowing intervention of the host, who has reduced the probability of the car being in the open door to zero. Before he opens that door, the odds are even.

Earthhorse

Lumen wrote: »

OK, here goes.

The idea that the odds are 50:50 is based on the idea that the probability of the car being in either is the same. This is false. Just because you have complete ignorance of two things, it doesn't make them equally probable (see Climate Change poll).

The car is more likely to be in the one that you didn't choose, because of the knowing intervention of the host, who has reduced the probability of the car being in the open door to zero. Before he opens that door, the odds are even.

Look, you don't need to tell me all that, it's pickarooney who won't believe!

Plus that explanation only confused me more.

pickarooney

OK, the combination of the video and the realisation that the 1/3 refers not to the chances of winning in a specific instance of selecting a cup but to to the chances of winning per three complete iterations of the game has me seeing it in a new light.

Now, I want to see a video from someone with no vested interest in proving the theory right doing the following:

The player picks a door/cup. The host reveals an animal. Repeat 20 times.
Now a second player, who hasn't seen the original choice, picks one cup from each of the twenty sets.

What do people guess the result would be in this case?

Kimono-Girl

i make bad choices, so by switching im swapping the bad choice for the good choice, under that logic how could i fail????





i'd get the goat...

dougal-maguire

pickarooney wrote: »

OK, the combination of the video and the realisation that the 1/3 refers not to the chances of winning in a specific instance of selecting a cup but to to the chances of winning per three complete iterations of the game has me seeing it in a new light.

Now, I want to see a video from someone with no vested interest in proving the theory right doing the following:

The player picks a door/cup. The host reveals an animal. Repeat 20 times.
Now a second player, who hasn't seen the original choice, picks one cup from each of the twenty sets.

What do people guess the result would be in this case?

both the same if i pick you up right.what your saying is,2 players,2 sets of 3 cups.can player 1 switch after the contents of 1 of the remaining 2 cups is revealed?im guessing player 2 picks a cup and is shown whats in it?

mcmoustache

o = car x = goat.


Each column is equivalent here so let's say we choose the middle one (b) always.
If we choose either column and stick, the chances are 1 in 3.


a b c

---

x o x -> I pick b: The host can choose a or c. Either way, if I switch, I lose.
x x o -> I pick b: The host must choose a. If I switch I win.
o x x -> I pick b: The host must choose c. If I switch I win.



So switching gives me a probability of winning of 2 in 3.

The argument is identical if I were to choose a or c instead of b at first so I'm not including it here.


[EDIT] - sorry for the formatting, I wanted this in courrier new.

Anyway, I just want to add, when you switch, you get the opposite of what you chose first time round.
Since the first choice has a 2 in 3 chance of being a goat, switching gives a 2 in 3 chance of being the car.

Rockn

This is what I think is an easy way to understand it:

If you choose a goat in the first round then the host reveals the other goat and you should switch to get the car. If you choose the car in the first round then obviously you should stick. But because there are two goats and one car you're twice as likely to choose a goat first time so if you switch you will win 2/3 of the time. If you stick you only win 1/3 of the time.

Oranage2

It's a 1 in 3 chance when you pick the first box of winning
Keeping your answer the same after the revel keeps it a 33%

so changing the box equals 66% chance of winning

still though I'd probably stick because I'd be sick if i changed and then lost!

pickarooney

dougal-maguire wrote: »

both the same if i pick you up right.what your saying is,2 players,2 sets of 3 cups.can player 1 switch after the contents of 1 of the remaining 2 cups is revealed?im guessing player 2 picks a cup and is shown whats in it?

In my theoretical example, player 2 is not in the room for the first set of picks.
He comes in when there are 20 sets of two cups and a revealed goat and picks a cup from each of the twenty sets. What percentage success rate is he likely to have?

This is to dispense with the notion of psychological influence and reduce it to pure maths.

28064212

pickarooney wrote: »

In my theoretical example, player 2 is not in the room for the first set of picks.
He comes in when there are 20 sets of two cups and a revealed goat and picks a cup from each of the twenty sets. What percentage success rate is he likely to have?

50:50

pickarooney wrote: »

This is to dispense with the notion of psychological influence and reduce it to pure maths.

It doesn't dispense with it at all. Or rather, it changes the game completely. Imagine Monty and the the 2 players are robots who are capable of making truly random choices. In your situation, it's a 50:50 shot.

However, if the 2nd robot is told what door the first robot chose and switches every time, he will have a 66% success rate (for an infinite number of games)

Micky Dolenz

Can we swap the goats for sheep?

dougal-maguire

player 2 will have a 50% success rate if he doesnt know what cup player 1 picked.
player 1 will have a 33% rate if he sticks all the time.
player 1 will have a 66% rate if he switches all the time.

OutlawPete

As said on another thread, part of the problem online is also because the way the question is sometimes phrased differently in different scenarios, and implied that it is not necessary at all to say that the Host knows where the car is, and that that is irrelevant to the question being put.

As is the case from this mathematician online, where he presents the solution as being best to switch, but doesn't emphasise that the host knows where the Car is or that he must always reveal a Goat before offering the choice to switch:

http://www.askamathematician.com/?p=787

Tiddlypeeps

The way it was explained to me that made me see sense was this scenario:

You have 100 doors, behind 1 is a car and behind 99 are goats. You pick a door randomly. Your odds of getting the car are 1/100. 98 doors that have goats behind them are removed. You have the choice to change door.

Its the same set up with more doors so the probabilities involved are much more drastic so its easier to understand, I think. In the above scenario its quite obvious that if you stick your chances are 1/100 and if you switch your chances increase to 1/2.

Tiddlypeeps

Tiddlypeeps wrote: »

Its the same set up with more doors so the probabilities involved are much more drastic so its easier to understand, I think. In the above scenario its quite obvious that if you stick your chances are 1/100 and if you switch your chances increase to 1/2.

Eh not really

pickarooney

28064212 wrote: »

50:50


It doesn't dispense with it at all. Or rather, it changes the game completely. Imagine Monty and the the 2 players are robots who are capable of making truly random choices. In your situation, it's a 50:50 shot.

However, if the 2nd robot is told what door the first robot chose and switches every time, he will have a 66% success rate (for an infinite number of games)

So the clincher is the informed choice - the unpicked cup has a statistically higher chance of having the car (I still have to grasp why) and the player in the orginal example is not simply choosing randomly from two cups over and over but choosing logically from two statistically weighted cups over and over.

Tiddlypeeps example makes a lot of sense to me actually. It better illustrates how the host is actually telling you that the leftover door is probably the one with the car.

Wompa1

Oranage2 wrote: »

It's a 1 in 3 chance when you pick the first box of winning
Keeping your answer the same after the revel keeps it a 33%

so changing the box equals 66% chance of winning

still though I'd probably stick because I'd be sick if i changed and then lost!

exact same as that I would stick with it. Maybe I'm wrong but you have a 66% chance that the one you already picked is right and personally I'd stick because if I changed and lost I'd kick myself.

Tiddlypeeps

bluewolf wrote: »

Eh not really

Sorry was totally wrong with the numbers there its much higher. 1/100 if you stick 99/100 if you switch. Obviously the chances have to add up to a whole.

28064212

pickarooney wrote: »

So the clincher is the informed choice - the unpicked cup has a statistically higher chance of having the car (I still have to grasp why) and the player in the orginal example is not simply choosing randomly from two cups over and over but choosing logically from two statistically weighted cups over and over.

Exactly. Obviously it's not easy to explain why, but bear with me

Wompa1 wrote: »

exact same as that I would stick with it. Maybe I'm wrong but you have a 66% chance that the one you already picked is right and personally I'd stick because if I changed and lost I'd kick myself.

You have a 33% chance that the one you picked originally is right. You'd be a fool not to switch