Two Goats, One Car!

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.

coffee_cake

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

coffee_cake

ok i am going to get a headache soon :mad:

Wompa1

28064212 wrote: »

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


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

The way I would see it is, if he is giving you an option again that means it's a fresh selection. You pick between 2...so it's actually 50/50?

I get the theory behind it, that you look at it as being between the 3 and 1 is down therefor, if you change now it's a projected 66% ...but that's a theory. Like the double up method in roulette.

Earthhorse

bluewolf wrote: »

ok i am going to get a headache soon :mad:

I know, and if I reveal when, it'll increase your chances of getting a headache but only if you decide you won't get a headache soon.

pickarooney

Wompa1 wrote: »

The way I would see it is, if he is giving you an option again that means it's a fresh selection. You pick between 2...so it's actually 50/50?

If you ignore everything you know or toss a coin, yes, it's 50:50 as you're reselecting with new choices and statistics. In this case you are going to either pick the higher percentage door or pick the lower percentage door, but at random. If you consciously choose to pick the lower percentage door, you're shooting yourself in the foot, statistically.

28064212

Wompa1 wrote: »

The way I would see it is, if he is giving you an option again that means it's a fresh selection. You pick between 2...so it's actually 50/50?

It's only a fresh selection if the remaining goat and car are randomly assigned to new doors. They are not, so what happened beforehand affects the probabilities

Wompa1 wrote: »

I get the theory behind it, that you look at it as being between the 3 and 1 is down therefor, if you change now it's a projected 66% ...but that's a theory. Like the double up method in roulette.

It's not a projected theory, and nothing like the double up method. It's an absolutely definitive mathematical probability. Over an infinite (or very, very large) number of iterations of the game, switching will win 66% of the time

Wompa1

Earthhorse wrote: »

I know, and if I reveal when, it'll increase your chances of getting a headache but only if you decided you won't get a headache soon.

Yes but has there been a variable change? Has he/she started to feel a headache? Have they become more dehydrated?

I revert to the useless explanation in the movie Deja vu..it's possible when you view the universe as a piece of paper, we simply fold the paper....My God!

Tiddlypeeps

Wompa1 wrote: »

The way I would see it is, if he is giving you an option again that means it's a fresh selection. You pick between 2...so it's actually 50/50?

I get the theory behind it, that you look at it as being between the 3 and 1 is down therefor, if you change now it's a projected 66% ...but that's a theory. Like the double up method in roulette.

That's the point its not 50/50, its 33.33/66.66. The fact that you made the decision when the odds were 1/3 and that you carried the same decision over means its not a fresh selection.

And the double up method in roulette works as long as there is no maximum bet, its not just a theory its practically demonstrable. Same as the cups, if you play it out at home with cups and a ball you can see it work.

Sheeps

has anyone posted goatse yet?

Wompa1

pickarooney wrote: »

If you ignore everything you know or toss a coin, yes, it's 50:50 as you're reselecting with new choices and statistics. In this case you are going to either pick the higher percentage door or pick the lower percentage door, but at random. If you consciously choose to pick the lower percentage door, you're shooting yourself in the foot, statistically.

28064212 wrote: »

It's only a fresh selection if the remaining goat and car are randomly assigned to new doors. They are not, so what happened beforehand affects the probabilities


It's not a projected theory, and nothing like the double up method. It's an absolutely definitive mathematical probability. Over an infinite (or very, very large) number of iterations of the game, switching will win 66% of the time

Tiddlypeeps wrote: »

That's the point its not 50/50, its 33.33/66.66. The fact that you made the decision when the odds were 1/3 and that you carried the same decision over means its not a fresh selection.

And the double up method in roulette works as long as there is no maximum bet, its not just a theory its practically demonstrable. Same as the cups, if you play it out at home with cups and a ball you can see it work.

Yeah I gotcha. The Double up theory is valid if you have an infinite number of goes. Realistically you won't and you will either win or lose everything.

I'd still stay. I go with my gut.

Speaking of Game Shows anyone ever find it a cop out when on Who Wants to be a Millionare they ask the contestant which one's they are thinking about before they make it 50/50 and then the two they mention are left?

https://www.youtube.com/watch?v=GU0SpKIBDUw

pickarooney

Wompa1 wrote: »

Speaking of Game Shows anyone ever find it a cop out when on Who Wants to be a Millionare they ask the contestant which one's they are thinking about before they make it 50/50 and then the two they mention are left?

https://www.youtube.com/watch?v=GU0SpKIBDUw

I was thinking of exactly the same thing a minute ago