Is My Masterplug RCD Safe? Please Help!

[JohnW2]

So it tells him the probability of it being better to swap, and the probability of it not being better, and one of those numbers is larger than the other, and that doesn't tell him whether it is better to swap or not swap? Not sure I see that....

Sorry, I wasn't at all clear. The 'statement of the obvious' to which I was refering was that it would be right to switch if he'd originally chosen the goat and wrong to switch if he had initially chosen the car!

The probabilities you mention are where all the hair starts being pulled out! They were, indeed, true at the time the initial choice was made, but that was at a time when there were three 'unknowns' to choose between. Once one 'goat door' has been taken out of the equation, it is very hard (even if one should!) to get one's head away from the thought that there are now just two doors, one with a car and the other with a goat, and that there 'must' (famous last words!) be an equal probability of the car being behind either.

You are, as you know, right in your conclusion/answer, but it isn't anything like as obvious as you are suggesting. Can you, for example, answer this .... once one door has been revealed to have a goat behind it, why does the remaining situation differ from one which would have existed if that door never existed, and one simply had chosen between two doors, one with a car and the other with a goat?

I had been hoping to avoid getting into 'the discussion', but (I suppose inevitably) it seems to have happened!

Kind Regards, John

 

[JohnW2]

Isn't it that if you are doing it many times it is better to swap every time - law of averages - and therefore it must follow that it is better to swap even if you are doing it only once - even though, obviously, you may still lose?

That's obviously not a general truth. If you were, say, just choosing between two doors (one good and one bad), it wouldn't make an iota of difference as to whether or not you changed your initial choice (before either had been opened!), whether you were doing it once or a million times.

Kind Regards, John

 

[plugwash]

You are, as you know, right in your conclusion/answer, but it isn't anything like as obvious as you are suggesting. Can you, for example, answer this .... once one door has been revealed to have a goat behind it, why does the remaining situation differ from one which would have existed if that door never existed, and one simply had chosen between two doors, one with a car and the other with a goat?

I think the key is in the non-randomness of which door is opened.

If the host ignored your initial choice and opened a goat door at random (which may have been the door you initially chose) then the probability of either of the remaining doors having the car would be 50/50. However the host can't do that. He is constrained to open a goat door and he is constrained not to open the door you chose initiallt.

If you initially (33% chance) picked the car then the host can show you either goat. On the other hand if you initially picked a goat (66%) then which door the host chooses to show you is forced. By doing this the host merges the winning chances of the two doors you did not initially pick onto the "twist" choice.

 

[ban-all-sheds]

The remaining situation is not the same as if there never was a 3rd door, because you are being asked whether to change the choice you made when there were 3 doors, and because the removed door is not random - the presenter knows which is which.

Before the reveal, you had a 1/3 chance of getting the right door, and a 2/3 chance of getting a wrong one. Nothing about your choice is changed by the reveal, you still have the same door, only now it might as well have a label on it saying "1 in 3 chance of a car, 2 in 3 chance of a goat" - clearly better to not keep it.

doh! Too slow

 

[JohnW2]

I think the key is in the non-randomness of which door is opened.

Indeed - it really has to be, otherwise it gets ridiculos!

If the host ignored your initial choice and opened a goat door at random (which may have been the door you initially chose) then the probability of either of the remaining doors having the car would be 50/50. However the host can't do that. He is constrained to open a goat door and he is constrained not to open the door you chose initiallt.

Indeed.

If you initially (33% chance) picked the car then the host can show you either goat. On the other hand if you initially picked a goat (66%) then which door the host chooses to show you is forced....

That's clearly all true.

...By doing this the host merges the winning chances of the two doors you did not initially pick onto the "twist" choice.

That's the bit which people will be less able to 'get their heads around'. Many will think that, no matter whether he chose one goat from two or only had one goat to chose from, he's chosen a goat - and hence just left one goat and one car!

Kind Regards, John

 

[ebee]

DELETED

 

[ebee]

Isn't it that if you are doing it many times it is better to swap every time - law of averages - and therefore it must follow that it is better to swap even if you are doing it only once - even though, obviously, you may still lose?

Yep.
I believe that`s what the producers said.
After a few weeks it was found pretty much in line with theory that people who switched won about twice as many times as people who did not switch.

 

[JohnW2]

Before the reveal, you had a 1/3 chance of getting the right door, and a 2/3 chance of getting a wrong one. Nothing about your choice is changed by the reveal, you still have the same door, only now it might as well have a label on it saying "1 in 3 chance of a car, 2 in 3 chance of a goat" - clearly better to not keep it.

[This is obviously all a bit difficult for me, since I 'know the answers' and am trying to represent (at Devil's Advocate, one might say) the difficulties and counter-arguments which many people come up with.]
I think that one of the problems which 'they' have with what you say above is they find it hard to accept that, even though you've put the "1 in 3 chance of a car, 2 in 3 chance of a goat" label on their first-choice door, that those probabilities remain unchanged once one of the three original options has been taken out of the equation.

In fact, this sort-of leads in to one of the approaches one can take to trying to get 'them' to understand. One changes the situation to one in which there are, say, 100 (or even 1,000 or more) doors, with one car and 99 goats. After the contestant has selected one door (1% change of it being a car), the presenter (who knows where the car is) then opens 98 'goat doors', leaving either the car or one goat behind the final unopened door. Many people can then accept that the probability of that one remaining door hiding the car is much greater than their 1% chance of having selected it in the first place.

Kind Regards, John

 

[JohnW2]

Isn't it that if you are doing it many times it is better to swap every time - law of averages - and therefore it must follow that it is better to swap even if you are doing it only once - even though, obviously, you may still lose?

Yep. I believe that`s what the producers said. After a few weeks it was found pretty much in line with theory that people who switched won about twice as many times as people who did not switch.

Sure, but that's because that is the 'answer' in this particular case (due, as being discussed, to the non-random choice of door by the presenter). The point I was making to EFLI is that this is true because of the particular circumstances of this game - it is not (as I think he was suggesting ) any general truth relating to 'the law of averages'. For any given 'game' of this sort, switching either will, or will not offer an advantage in terms of probabilities of winning (depending on the details of the game) - and if it does, then that probability will be just as applicable if you play the game once as if you take the average of playing it a million times. If the latter, then you obviously would actually 'win' roughly the number of times indicated by the probability (i.e. you would win prob x 1 million games) - which is precisely what is meant by the probability your of winning if you play just one game.

Kind Regards, John

 

[EFLImpudence]

I was only referring to the game.

 

[JohnW2]

I was only referring to the game.

OK, I misunderstood. So, as I implied, I suppose what you were effectively doing was defining the meaning of probability - i.e. if the probability of winning one game (say, with switching) is X%, that obviously means that if you play the game N times (with switching), you will win N*(X/100) of those games - but the probability of winning will be the same for every game.

Kind Regards, John

 

[EFLImpudence]

I was only referring to the game.

OK, I misunderstood. So, as I implied, I suppose what you were effectively doing was defining the meaning of probability - i.e. if the probability of winning one game (say, with switching) is X%, that obviously means that if you play the game N times (with switching), you will win N*(X/100) of those games - but the probability of winning will be the same for every game

More the other way round to persuade people that it is better to switch.

If you play many times you 'will'/should, by switching, win two thirds of the games - by luck.

Therefore the chance of winning only one game is still 66.67% even though you may lose - i.e. you have won 0% or 100% of your game(s).

 

[JohnW2]

More the other way round to persuade people that it is better to switch. If you play many times you 'will'/should, by switching, win two thirds of the games - by luck. Therefore the chance of winning only one game is still 66.67% even though you may lose - i.e. you have won 0% or 100% of your game(s).

Well, as I said, you are merely explaining to them what probability (of winning a single game) means - but, yes, probably the most obvious way to explain that is by saying that it means that if the play a lot of games, they will win "X%" of them (and loose (100-X)% !).

However, that is all obviously very much seconday to the primary question, which is that of establishing the numerical value of "X%", with and without switching!

Kind Regards, John

 

[EFLImpudence]

However, that is all obviously very much seconday to the primary question, which is that of establishing the numerical value of "X%", with and without switching!

Not really, surely.

If it is accepted that it is two thirds when switching then it must be and obviously is one third when not switching.

I didn't think there was any doubt about the 'not switching' chance.

 

[JohnW2]

However, that is all obviously very much seconday to the primary question, which is that of establishing the numerical value of "X%", with and without switching!

Not really, surely. If it is accepted that it is two thirds when switching then it must be and obviously is one third when not switching. I didn't think there was any doubt about the 'not switching' chance.

I fear you have missed the 'issue' The countless thousands (or millions) of (wo)man-hours that have gone into the discussions, debates and arguments over the years about the MH problem have all been about the probability of winning if one switches (or doesn't switch). In other words, "If it is accepted that it is two thirds when switching..." is very far from 'being accepted' without a big fight!!

This is where the problem with BAS's approach of putting a label on the first-choice door saying (probability of car = 0ne-third) runs into its dificulties. If people believed that the one-third probability (i.e. the probability of winning without switching) remained unchanged after one of the goats had been revealed, then the whole thing would become totally trivial (not requiring thousands/millions of hours of debate) - once the third door with a goat has been revealed, one of the other two must have the car, so if the probability of it being a car without switching is still one-third, then the probability with switching obviously has to be two-thirds. The difficulties/debates/whatever arise because the first impression of most people, including many statisticians, is not that the one-third probability remains unchanged but, rather, that the probabilities with and without switching both become 50% once one of the goats has been revealed.

Kind Regards, John