Can you answer this veridical paradox?

[Meerkat Man]

This reminds me of a similar well known conundrum.

There are only 2 doors this time. Behind one is a donkey and behind the other is a Ferrari. In front of each door is a guard. One of the guards always tells the truth and the other always lies but you don’t know which is which. You are only allowed to ask one question, which may be asked of either of them (but not both).

What question should you ask and which door should you subsequently open?

 

[warney]

It's true that at the start there is a 33% chance of picking the Ferrari door. However, when one "donkey" door is opened you now have a 50% chance of the Ferrari whether or not you decide to change. so it will make no difference. It's the opening of the first door that has changed the odds - nothing you do subsequently will.

 

[The Doctor]

This is a classic problem, and a mind-bending one at that. There's a good wikipedia page about it. It's called the Monty Hall problem: https://en.wikipedia.org/wiki/Monty_Hall_problem

 

[mervyn]

I’d keep to my original selection as you wouldn’t have offered me a swap if my first choice was wrong.

But if you stick to your first choice you halve your chances of winning.

 

[mervyn]

This is a classic problem, and a mind-bending one at that. There's a good wikipedia page about it. It's called the Monty Hall problem: https://en.wikipedia.org/wiki/Monty_Hall_problem

If you enjoy these things doc, I’ve just received an interesting book by Jim Al-Khalili (one of my heroes)
entitled Paradox.

 

[warney]

I’d keep to my original selection as you wouldn’t have offered me a swap if my first choice was wrong.

But if you stick to your first choice you halve your chances of winning.

No, you don't. It's 50/50 whether or not you change your mind. Statistically what has occurred before does not influence what happens in the future, assuming an unbiased model.
If you throw a dice you have a one in six chance of throwing a six. So if you throw it sixty times you would expect to throw a six around 10 times (on average - if you repeated the exercise 100 times, say). However, if you've thrown the dice 59 times and not got a single six, that does not mean you're more likely to throw a six with the final throw - the odds are still one in six. The odds for the whole exercise do not affect the odds for a single part of that exercise.
(Of course, in reality it's so unlikely to get to 59 throws without a six that you'd suspect a weighted dice but's that another matter!)

 

[GreenThing]

This reminds me of a similar well known conundrum.

There are only 2 doors this time. Behind one is a donkey and behind the other is a Ferrari. In front of each door is a guard. One of the guards always tells the truth and the other always lies but you don’t know which is which. You are only allowed to ask one question, which may be asked of either of them (but not both).

What question should you ask and which door should you subsequently open?

Ask the first guard “What would the 2nd guard tell me if I asked him if there is a Ferrari behind his door”, then open the other door as you will always get a lie as a reply.

 

[The Doctor]

I’d keep to my original selection as you wouldn’t have offered me a swap if my first choice was wrong.

But if you stick to your first choice you halve your chances of winning.

No, you don't. It's 50/50 whether or not you change your mind. Statistically what has occurred before does not influence what happens in the future, assuming an unbiased model.
If you throw a dice you have a one in six chance of throwing a six. So if you throw it sixty times you would expect to throw a six around 10 times (on average - if you repeated the exercise 100 times, say). However, if you've thrown the dice 59 times and not got a single six, that does not mean you're more likely to throw a six with the final throw - the odds are still one in six. The odds for the whole exercise do not affect the odds for a single part of that exercise.
(Of course, in reality it's so unlikely to get to 59 throws without a six that you'd suspect a weighted dice but's that another matter!)

It's certainly counter-intuitive that you should switch but it IS the right thing to do and it DOES increase your odds of winning the car. Have a look at the first few lines of this part of the wikipedia entry: https://en.wikipedia.org/wiki/Monty_Hal ... _solutions
When I read that it made sense to me.

 

[GreenThing]

I’d keep to my original selection as you wouldn’t have offered me a swap if my first choice was wrong.

But if you stick to your first choice you halve your chances of winning.

No, you don't. It's 50/50 whether or not you change your mind. Statistically what has occurred before does not influence what happens in the future, assuming an unbiased model.
If you throw a dice you have a one in six chance of throwing a six. So if you throw it sixty times you would expect to throw a six around 10 times (on average - if you repeated the exercise 100 times, say). However, if you've thrown the dice 59 times and not got a single six, that does not mean you're more likely to throw a six with the final throw - the odds are still one in six. The odds for the whole exercise do not affect the odds for a single part of that exercise.
(Of course, in reality it's so unlikely to get to 59 throws without a six that you'd suspect a weighted dice but's that another matter!)

It's certainly counter-intuitive that you should switch but it IS the right thing to do and it DOES increase your odds of winning the car. Have a look at the first few lines of this part of the wikipedia entry: https://en.wikipedia.org/wiki/Monty_Hal ... _solutions
When I read that it made sense to me.

It would make sense to change but the game is reset once the choices have been reduced to two. What happens if the host opens the middle door to reveal the Ferrari? Should you still have 1/3 chance of you keep the same choice and a 1/2 chance if you change your mind? No, you’d go home with a goat because the game has been reset.

 

[mervyn]

I’d keep to my original selection as you wouldn’t have offered me a swap if my first choice was wrong.

But if you stick to your first choice you halve your chances of winning.

No, you don't. It's 50/50 whether or not you change your mind. Statistically what has occurred before does not influence what happens in the future, assuming an unbiased model.
If you throw a dice you have a one in six chance of throwing a six. So if you throw it sixty times you would expect to throw a six around 10 times (on average - if you repeated the exercise 100 times, say). However, if you've thrown the dice 59 times and not got a single six, that does not mean you're more likely to throw a six with the final throw - the odds are still one in six. The odds for the whole exercise do not affect the odds for a single part of that exercise.
(Of course, in reality it's so unlikely to get to 59 throws without a six that you'd suspect a weighted dice but's that another matter!)

It's certainly counter-intuitive that you should switch but it IS the right thing to do and it DOES increase your odds of winning the car. Have a look at the first few lines of this part of the wikipedia entry: https://en.wikipedia.org/wiki/Monty_Hal ... _solutions
When I read that it made sense to me.

It would make sense to change but the game is reset once the choices have been reduced to two. What happens if the host opens the middle door to reveal the Ferrari? Should you still have 1/3 chance of you keep the same choice and a 1/2 chance if you change your mind? No, you’d go home with a goat because the game has been reset.

What the doc says is right, you must read Monty Hall, which explains it.

 

[GreenThing]

I’d keep to my original selection as you wouldn’t have offered me a swap if my first choice was wrong.

But if you stick to your first choice you halve your chances of winning.

No, you don't. It's 50/50 whether or not you change your mind. Statistically what has occurred before does not influence what happens in the future, assuming an unbiased model.
If you throw a dice you have a one in six chance of throwing a six. So if you throw it sixty times you would expect to throw a six around 10 times (on average - if you repeated the exercise 100 times, say). However, if you've thrown the dice 59 times and not got a single six, that does not mean you're more likely to throw a six with the final throw - the odds are still one in six. The odds for the whole exercise do not affect the odds for a single part of that exercise.
(Of course, in reality it's so unlikely to get to 59 throws without a six that you'd suspect a weighted dice but's that another matter!)

It's certainly counter-intuitive that you should switch but it IS the right thing to do and it DOES increase your odds of winning the car. Have a look at the first few lines of this part of the wikipedia entry: https://en.wikipedia.org/wiki/Monty_Hal ... _solutions
When I read that it made sense to me.

It would make sense to change but the game is reset once the choices have been reduced to two. What happens if the host opens the middle door to reveal the Ferrari? Should you still have 1/3 chance of you keep the same choice and a 1/2 chance if you change your mind? No, you’d go home with a goat because the game has been reset.

What the doc says is right, you must read Monty Hall, which explains it.

“ The fact that the host subsequently reveals a goat in one of the unchosen doors changes nothing about the initial probability.” The word ‘initial’ is paramount here. It initial chances are unaffected, but the initial chances are null and void as removing one option resets the game. You will be starting again and not changing your choice will actually be making a new choice, the same one as last time.

 

[Lousy Pint]

Never mind the Ferrari. I want to know how the donkeys were changed into goats???

 

[GreenThing]

I’d like a goat as much as I’d like a Ferrari, just making sure I’m a winner in all cases.

 

[Bryan Tregunna]

I made my choice. Anyone want to buy a donkey?

 

[Fergy]

https://youtu.be/1hFvxzTMw6k