Can you answer this veridical paradox?

[Doug Deep]

The crucial aspect of this is that the host (who, crucially, knows where the ferrari and goats are) deliberately opens a "goat door" - there always will be (at least) one which he can. You should switch doors.

If you still think you should stick, how about if there were 10 doors, 9 goats and a Ferrari. You pick one, then one by one the host opens 8 consecutive doors which he knows will each reveal a goat. That leaves your original selection and one other door. Do you still want to stick?

 

[rongreenblood]

It was easier when we tried to predict the result of the next Argyle game! :crazy:

 

[The Doctor]

The crucial aspect of this is that the host (who, crucially, knows where the ferrari and goats are) deliberately opens a "goat door" - there always will be (at least) one which he can. You should switch doors.

If you still think you should stick, how about if there were 10 doors, 9 goats and a Ferrari. You pick one, then one by one the host opens 8 consecutive doors which he knows will each reveal a goat. That leaves your original selection and one other door. Do you still want to stick?

That last paragraph is a cracking way to see that the correct answer is to switch.

 

[Argylegames]

That last paragraph is a cracking way to see that the correct answer is to switch.

Nope - I don't see it. whichever way your odds are 50/50

 

[AdelaideGreen]

I am struggling with the intuition on this one, but this is how I understand why it is better to switch.

You get to choose 1 door out of three. What is your probability of getting it right? 33.3%

Therefore, what is the probability that the car is behind one of the other 2 doors? 66.7%

What would you do if the host allowed you to swap your door for BOTH the other doors? If it is behind either one of them you win.

Then the host makes it even easier for you by showing you what is behind one of those doors - and it is a goat. By showing you that one of those doors is a goat does not change the original odds. It isn't random, there must be a goat behind at least one of those 2 doors and the host knows which one it is.

So the 66.7% chance is concentrated down to the one remaining door you did not pick.

In essence this is what the host does in the scenario, but he does the reveal and the option to change in a different order. This also does not change the original probabilities.

It is mind bending, but it makes some sort of sense ... I think.

 

[Bryan Tregunna]

I really don't get it. To me it's 50/50 all day!

However, the fact that the host chose No.3 and not No.2 does say something, doesn't it?

Or does it say sod all?

 

[Bryan Tregunna]

OK. I’ve just written a programme to do just this. (I really don’t have anything better to do!) And it is true!

I’ve run it 6 times and the percentages for correctly switching were: 64,70,64,70,62,71 (avge 66.83)

I think this now makes sense and this is my theory:

1. There is a 33% chance of choosing the car first time.

2. There is a 66% chance that the car will be in one of the other two options.

3. One of those other options is 100% going to be a goat.

4. Removing that goat doesn’t change the 66% chance that you selected wrongly first time around.[/list]

You gotta switch! (But don't bet your goat on it!)

 

[The Doctor]

OK. I’ve just written a programme to do just this. (I really don’t have anything better to do!) And it is true!

I’ve run it 6 times and the percentages for correctly switching were: 64,70,64,70,62,71 (avge 66.83)

I think this now makes sense and this is my theory:

1. There is a 33% chance of choosing the car first time.

2. There is a 66% chance that the car will be in one of the other two options.

3. One of those other options is 100% going to be a goat.

4. Removing that goat doesn’t change the 66% chance that you selected wrongly first time around.[/list]

You gotta switch! (But don't bet your goat on it!)

I love it. Now all you have to do is add in some code to model the chance of catching CV19 via contact with one or more of the goats/donkeys, the person or the Ferrari and you'll have the Government bashing down your door wanting you to join SAGE...

 

[The Doctor]

[This feels like one of those 'Only on PASOTI' moments!]

 

[GreenThing]

OK. I’ve just written a programme to do just this. (I really don’t have anything better to do!) And it is true!

I’ve run it 6 times and the percentages for correctly switching were: 64,70,64,70,62,71 (avge 66.83)

I think this now makes sense and this is my theory:

1. There is a 33% chance of choosing the car first time.

2. There is a 66% chance that the car will be in one of the other two options.

3. One of those other options is 100% going to be a goat.

4. Removing that goat doesn’t change the 66% chance that you selected wrongly first time around.[/list]

You gotta switch! (But don't bet your goat on it!)

That doesn’t fit the theory. The theory is that by switching you change your chances fro 33% to 50%.

Also, what results did you get fo sticking with your original choice?

 

[Lousy Pint]

Let's say there were seven green doors.

Behind three of the doors is a donkey.

Behind another three of the doors is a goat.

Behind the seventh door there is nothing.

My question is: was it a donkey or a goat that drove off in the Ferrari?

 

[Knibbsworth]

I have read this before in a book by John Douglas of the FBI. As another poster has identified, the OP is referring to the Monty Hall problem. Once you have established behind one door is a goat, you take it back to the start. You had 3 possibilities:
If you had Goat 1 originally, if you stick you keep Goat 1. You know the other door is Goat 2 now. If you twist, you get a Ferrari.
If you had the other goat it is the same. Stick, you keep Goat 2. You know behind the other door is Goat 1. Twist you get a Ferrari.
If you have a Ferrari, then stick you keep it. Twist you will get either Goat 1 or 2.

There were 3 scenarios, and in two of them twisting gets the star prize. If there was an even 33.3% chance before you were given a door, then there is a 66.6% chance that switching will benefit you.

 

[GreenThing]

I have read this before in a book by John Douglas of the FBI. As another poster has identified, the OP is referring to the Monty Hall problem. Once you have established behind one door is a goat, you take it back to the start. You had 3 possibilities:
If you had Goat 1 originally, if you stick you keep Goat 1. You know the other door is Goat 2 now. If you twist, you get a Ferrari.
If you had the other goat it is the same. Stick, you keep Goat 2. You know behind the other door is Goat 1. Twist you get a Ferrari.
If you have a Ferrari, then stick you keep it. Twist you will get either Goat 1 or 2.

There were 3 scenarios, and in two of them twisting gets the star prize. If there was an even 33.3% chance before you were given a door, then there is a 66.6% chance that switching will benefit you.

That’s all well and good in theory, but one goat and one door is irrelevant once opened. When the circumstances change, the odds change.

 

[Bryan Tregunna]

OK. I’ve just written a programme to do just this. (I really don’t have anything better to do!) And it is true!

I’ve run it 6 times and the percentages for correctly switching were: 64,70,64,70,62,71 (avge 66.83)

I think this now makes sense and this is my theory:

1. There is a 33% chance of choosing the car first time.

2. There is a 66% chance that the car will be in one of the other two options.

3. One of those other options is 100% going to be a goat.

4. Removing that goat doesn’t change the 66% chance that you selected wrongly first time around.[/list]

You gotta switch! (But don't bet your goat on it!)

That doesn’t fit the theory. The theory is that by switching you change your chances fro 33% to 50%.

Also, what results did you get fo sticking with your original choice?

The results are binary: you either switch or stick. The results I showed were for switching and as percentages for sticking are the remainder I didn't bother reporting these. But here they are. That is: 36,30.36,30,38,29 (33.16 avge).

If the host had shown the goat/donkey before you make the choice, then it would be 50/50. This would also be the case if the host revealed that your selection was a goat/donkey and invited you to select again.
The fact that you make the choice when it is one in three and the host removes one of the others is highly relevant and doesn't make it a new game.

 

[GreenThing]

OK. I’ve just written a programme to do just this. (I really don’t have anything better to do!) And it is true!

I’ve run it 6 times and the percentages for correctly switching were: 64,70,64,70,62,71 (avge 66.83)

I think this now makes sense and this is my theory:

1. There is a 33% chance of choosing the car first time.

2. There is a 66% chance that the car will be in one of the other two options.

3. One of those other options is 100% going to be a goat.

4. Removing that goat doesn’t change the 66% chance that you selected wrongly first time around.[/list]

You gotta switch! (But don't bet your goat on it!)

That doesn’t fit the theory. The theory is that by switching you change your chances fro 33% to 50%.

Also, what results did you get fo sticking with your original choice?

The results are binary: you either switch or stick. The results I showed were for switching and as percentages for sticking are the remainder I didn't bother reporting these. But here they are. That is: 36,30.36,30,38,29 (33.16 avge).

If the host had shown the goat/donkey before you make the choice, then it would be 50/50. This would also be the case if the host revealed that your selection was a goat/donkey and invited you to select again.
The fact that you make the choice when it is one in three and the host removes one of the others is highly relevant and doesn't make it a new game.

Surely asking someone if they want to change gives them 2 options, the one you chose or the one you didn’t. Choosing to stick is a new choice so it’s a new game with 2 options.