Impeccable logic - part 2

[Softus]

...the actual question is quite basic so there is no need to get technical...

I disagree.

The question, and the Monty Hall paradox that you've cited as proof of your claim that you're right, is far from basic. It has baffled many people over many years, and is an excellent example of a problem where the answer is not the one that most people intuitively believe to be correct.

So Softus first you say the Monty scenario is far from basic but now your saying its quite simple.
Hmmm your contradicting yourself !!!

I can tell that you're intent on seeing it like that, so I won't try and stop you.

For everyone else's benefit, jackpot has taken those two quotes out of context.

My first statement was intended to reflect the fact that many people find the question a complex one. This view is supported by the debate currently underway, that has spanned two topics.

The second statement was intended to reflect the fact that I find the answer very simple to understand. It wasn't always so - when I first encountered the question I got it utterly wrong. Just as so many people are continuing to do. I'd like to think that those people will understand, one day, but they're being rather obstinate.
____________

Softus....... the whole essence of the Monty Hall game, is that the host always reveals the goat.

I view that as the essence of the show, but not the essence of the statistical problem.

Thus 2 doors, each with a 1 in 3 chance becomes 1 door with a 2 in 3 chance.

I agree with that.

Monty,s knowledge is irrelevant during the first pick, but his knowledge is the only factor which turns the game in your favour, when given the option to change.

If this were true, then the probability of winning the car after swapping would not be 2/3. However, it is 2/3, therefore the revealing of the goat has no bearing - it's just showmanship, and is a very effective way of confusing most observers.
____________

Please read all the following assertions very carefully. I've taken care in the wording, not to be devious but to be as plain and simple as possible, and without introducing any other games, or people, or goats, or doors, or cars.

Observing the first choice in the MH game, I believe, as do many people, that the probability of winning the car, at the moment of that choice, is 1/3.

Swapping, at this point, to the other two doors, will therefore yield a probability of 2/3 of winning the car.

[to be continued...]

If you disagree with the above, then please don't read on, but instead please explain which part you disagree with, and why.
__________

[continuing...]

If Monty doesn't know where the car is, the probability is still 2/3.

If Monty knows where the car is, the probability is still 2/3.

If Monty doesn't reveal anything, the probability is still 2/3.

If Monty reveals a goat after the swap, the probability is still 2/3.

If Monty reveals a goat before the swap, but I close my eyes and ears and go ahead with the swap anyway, then the probability is still 2/3.

So what, in the name of all that's Holy, is the effect of a goat being revealed?

This is a rhetorical question - the answer is: "nothing".

 

[joe-90]

I must be losing the plot here but I can't see where.

If door 1 hid a goat, door 2 hid a car and door 3 hid a sheep, then if the contestant picked the car (door 2) Monty would offer a goat or a sheep.
Whichever he offered would leave a 'car or sheep' or 'car or goat'. 50-50?

If the contestant picked the goat, then Monty would reveal the sheep which would leave car or goat 50-50?

If the contestant picked the sheep, then Monty would reveal the goat. Which would leave the sheep and the car 50-50?

Is that wrong? If so, I can't see where. Can anyone point out where I've got this wrong?

 

[Softus]

I must be losing the plot here but I can't see where.

You're not losing the plot - I think you might just be starting to fully understand.

Before I answer your specific points, please try and imagine how to work out the odds if the act of revealing the goat/sheep never happened.
_____

If door 1 hid a goat, door 2 hid a car and door 3 hid a sheep, then if the contestant picked the car (door 2) Monty would offer a goat or a sheep.
Whichever he offered would leave a 'car or sheep' or 'car or goat'. 50-50?

True, but it isn't 50:50. The reason you think it is is because you've given undue prominence to the fact that two doors remain unopened.

If the contestant picked the goat, then Monty would reveal the sheep which would leave car or goat 50-50?

If the contestant picked the sheep, then Monty would reveal the goat. Which would leave the sheep and the car 50-50?

As above for both of those scenarios.

Is that wrong? If so, I can't see where. Can anyone point out where I've got this wrong?

The descriptions are perfect, but where it's wrong is that the probability isn't calculated according to whether or not doors are opened, but where the car might be.

I assure you that the opening of the door isn't relevant. It's a distraction.

Imagine if no doors were opened until after you'd stuck/swapped. You can still work out the car-winning odds for sticking and for swapping.

 

[Trazor]

If this were true, then the probability of winning the car after swapping would not be 2/3. However, it is 2/3, therefore the revealing of the goat has no bearing - it's just showmanship, and is a very effective way of confusing most observers.

I totally disagree with this one point, for the following reason....

If the goat is never revealed the following odds now apply...

1) Your first choice is a 1 in 3 chance, you will win the car on average 33 times out of 99 if you stick.

2) You are now offered the swap with no reveal, OK so you swap, all you are doing is moving your 1 in 3 chance to another door, ie. you are still selecting 1 door out of 3.
Think about it, you could be offered the swap 50 times, but you still have ONE chance in 3 of settling on the correct door.

If the show was run on these lines, 33 people out of 99, on average, would win the car, 66 out of 99 people, on average, would lose.

 

[Softus]

If the goat is never revealed the following odds now apply.

I've not said that the goat would never be revealed. What I suggested to joe-90 was that he approach working out the odds as if the goat were never revealed.

If you think that my explanation of the solution involves not revealing the goat, then rest assured that's not what I intended, so I apologise if I made it confusing.

I note that you "disagree with this one point" - does that mean you agree with all of the other points?

 

[megawatt]

So, anyway, back to the OP's original point ... What are the odds of a swap in DoND when down to the last two boxes.

I'll start ... 50:50

WHO THE F*** EVER MENTIONED MONTY F***ING HALL

MW

 

[jackpot]

Agreed with Megawatt yet again.

Softus, you are 100% correct on your experiment/scenario, however lets get back and argue the original post. No goat, no car, when there is 3 boxes left the contestant can only pick to discard 1 of the 2 outside boxes unlike Montys where he has a choice of 3 to pick from.

Hardly the same experiment

 

[joe-90]

If the goat is never revealed the following odds now apply.

I've not said that the goat would never be revealed. What I suggested to joe-90 was that he approach working out the odds as if the goat were never revealed.

If you think that my explanation of the solution involves not revealing the goat, then rest assured that's not what I intended, so I apologise if I made it confusing.

I note that you "disagree with this one point" - does that mean you agree with all of the other points?

I'm willing to admit I'm being thick, but I still can't see it.
If Monty knows where the car is then he can either offer the car or hide the car. Let's assume he doesn't reveal a goat.

Contestant picks one door out of three, Odds are 1-3

If contestant picks door with car - then Monty (who knows that) can offer a goat. Odds 50-50
If contestant picks a door with a goat (which Monty knows) then he can offer a swap with another goat. (no chance to win whatsoever)
How on earth can that make it sensible to switch?
Where does the 2 in 3 chance come in?

Thicko-90

 

[blondini]

So, anyway, back to the OP's original point ... What are the odds of a swap in DoND when down to the last two boxes.

I'll start ... 50:50

WHO THE F*** EVER MENTIONED MONTY F***ING HALL

MW

First mention of MH was in the OP wasn't it?

If you believe that swapping on the MH game scenario makes the odds better than 50:50, but you also believe that swapping makes no difference on the OP DOND scenario, then there is a zero probability of you fully understanding why the MH swap changes the odds.

In both MH and OP DOND there is an ultimate choice between two boxes or doors and there are two possible outcomes, so it looks llike 50:50. But in both, the odds are already weighted in favour of the hosts final box or door concealing the big prize.

Understand MH and be enlightened.

 

[blondini]

If the goat is never revealed the following odds now apply.

I've not said that the goat would never be revealed. What I suggested to joe-90 was that he approach working out the odds as if the goat were never revealed.

If you think that my explanation of the solution involves not revealing the goat, then rest assured that's not what I intended, so I apologise if I made it confusing.

I note that you "disagree with this one point" - does that mean you agree with all of the other points?

I'm willing to admit I'm being thick, but I still can't see it.
If Monty knows where the car is then he can either offer the car or hide the car. Let's assume he doesn't reveal a goat.

Contestant picks one door out of three, Odds are 1-3

If contestant picks door with car - then Monty (who knows that) can offer a goat. Odds 50-50
If contestant picks a door with a goat (which Monty knows) then he can offer a swap with another goat. (no chance to win whatsoever)
How on earth can that make it sensible to switch?
Where does the 2 in 3 chance come in?

Thicko-90

When I first saw this I thought it must be 50:50, but there was a nagging doubt that something wasn't right.

Do what I did. Get pen and paper.

Following the rules of the gameplay, list the possible choices or permutations that you could play.
There are six. Three of them are stick permutations and three are swap.

Now pick any door for the car to be behind and note which of your six possible choices would have won it. There will be three winning permutations but two out of those three will be swaps. 2/3.

Doing this made me realise that the swap is not 50:50. Realising this is the first step to understanding. Understanding why took me a little longer and truly understanding longer again.

 

[megawatt]

Blondini: The original post asked the following question ...

OK, in a particularly moronic moment you apply to go on Deal or no Deal, and surprisingly you are accepted. When the moment comes, you chose a box (out of 20?) and by a combination of blind chance and good fortune, whittle the lot down to two boxes, yours and one out there. The boxes have in them £250,000 and £1. The banker, being possessed of malicious humour, offers you the swap. You think for a few moments, this time using impeccable logic. What should you do to maximise the chance of choosing the £250k box?

Now ... Go get your pen and paper (as it seems to work for you) and list the possible permutations of this scenario ... No car, no doors, no f***ing goat and a banker who has no idea what is behind where.

Then come back and try and convince us all that, when faced with the final two boxes, the odds are something other than 50:50.

And, when you can't, take a look at my earlier post which clearly explains the probability for this scenario.

MW

 

[oompah]

But if you have two boxes left containing £250.000 and £1 the banker then offers you a certain figure to quit does this alter the odds in any way?

 

[Softus]

Agreed with Megawatt yet again.

I don't know what megawatt wrote, but I can assure you that it's wrong.

Softus, you are 100% correct on your experiment/scenario, however lets get back and argue the original post. No goat, no car, when there is 3 boxes left the contestant can only pick to discard 1 of the 2 outside boxes unlike Montys where he has a choice of 3 to pick from.

Hardly the same experiment

OK. So if you understand the MH paradox and agree that the odds favour swapping, then the only point of disagreement is whether Kes' OP on the first topic is like, or unlike, a MH decision.

I shall tell you why I think it's the same. Perhaps you would extend me the courtesy of opening your mind and letting go of your insistence that it's different.

Firstly, Kes' scenario is not representative of every DoND game. The number of sequences in which n-2 boxes are opened is a large number - it isn't important to calculate it here. All we need to know is that there are only two permutations, of a great many, in which the only two remaining unopened boxes contain £1 and £250K.

Let the box chosen by the contestant be Box A.
Let the sole remaining unopened box be Box B.

There are two now points that form the basis of the next conclusion:

1. The probability of the £250K being in Box A, before any box was opened, is 1/n (i.e. if 20 boxes in total, then 1/20).

2. The probability of the £250K being in Box A never changes, just like in the Monty Hall scenario.

Now I foresee that you'll agree with point #1, and disagree with point #2, in which case we will need to resolve that before moving on.

If you agree with both points, then we're just about there....

Since the probability of Box A containing £250K is 1/n, the probability of it being somewhere else is n-1/n.

Thus, if you're offered the opportunity to abandon Box A, and you don't, then you'll win the £250K only once out of n such scenarios.

If you do swap, then then n-1 times out of n, you will win the £250K.

 

[jackpot]

So, anyway, back to the OP's original point ... What are the odds of a swap in DoND when down to the last two boxes.

I'll start ... 50:50

WHO THE F*** EVER MENTIONED MONTY F***ING HALL

MW

First mention of MH was in the OP wasn't it?

If you believe that swapping on the MH game scenario makes the odds better than 50:50, but you also believe that swapping makes no difference on the OP DOND scenario, then there is a zero probability of you fully understanding why the MH swap changes the odds.

In both MH and OP DOND there is an ultimate choice between two boxes or doors and there are two possible outcomes, so it looks llike 50:50. But in both, the odds are already weighted in favour of the hosts final box or door concealing the big prize.

Understand MH and be enlightened.

I fully understand. Its a different experiment for a number of reasons.

1. With 3 boxes left the contestant can only choice to get rid of 1 of 2 boxes. Where as Monty he is picking from 3 boxes.

2. Let play 22 boxes on Montys game. The contestant has a 21 in 22 chance of not have the box. 20 boxes are discarded knowing that the top prize is not in any of the boxes. So 100% of the time this will happen. As thats the game. Its impossible for the top prize to be in 1 of these 20 discarded boxes.

However in the deal or no deal. Its not impossible as proved in the show on many occasions. The top prize can be eliminated at any time but on this occassion the top prize is in one of the remaining 2 boxes.

3. Prity much the same as above which proves its a different experiment

Lets play with 3 boxes again and the top prize can be anywhere. If the contestant picks a box (1 of 2) as he cant touch is own. He hits the 250k and that box is eliminated. Game over he is left with another prize.

Now in Montys he picks 1 of 3 boxes as above but he can pick his own box now. And picks the same box. A goat is revealed in one of the other boxes but the contestant still does not know where the top prize is. Then he gets a choice to swap.

The person hardly has the same choice of options does he? The experiment is hardly under the same circumstances. Therefore you can not compare the 2 experiments together as they are different experiments for different scenarios.

 

[jackpot]

Agreed with Megawatt yet again.

I don't know what megawatt wrote, but I can assure you that it's wrong.

Softus, you are 100% correct on your experiment/scenario, however lets get back and argue the original post. No goat, no car, when there is 3 boxes left the contestant can only pick to discard 1 of the 2 outside boxes unlike Montys where he has a choice of 3 to pick from.

Hardly the same experiment

OK. So if you understand the MH paradox and agree that the odds favour swapping, then the only point of disagreement is whether Kes' OP on the first topic is like, or unlike, a MH decision.

I shall tell you why I think it's the same. Perhaps you would extend me the courtesy of opening your mind and letting go of your insistence that it's different.

Firstly, Kes' scenario is not representative of every DoND game. The number of sequences in which n-2 boxes are opened is a large number - it isn't important to calculate it here. All we need to know is that there are only two permutations, of a great many, in which the only two remaining unopened boxes contain £1 and £250K.

Let the box chosen by the contestant be Box A.
Let the sole remaining unopened box be Box B.

There are two now points that form the basis of the next conclusion:

1. The probability of the £250K being in Box A, before any box was opened, is 1/n (i.e. if 20 boxes in total, then 1/20).

2. The probability of the £250K being in Box A never changes, just like in the Monty Hall scenario.

£250K.

But there is also a probablility of 1:20 that the 250k can be in box B before any box has been opened, and the same for your second point. The probability of the 250k in box B will never change.