# Magellan's law (sp?)



## Kell (May 28, 2002)

Anyone heard of this? Not sure of the spelling and a quick google didn't pull anything up, but it's quite intersting (though you'll be the judge of that).

My FIL and the rest of us had a 'conversation' the other night after a game of cards and he put it to us that a mathmetician has come up with this theory which seems to be backed up by testing:

Its basically this...

There are, say, 10 doors and behind one of them is a million pounds. You get to pick one door as the oor you think it's behind. Then all the other doors are opened, but the one with the money behind it is opened before all the others are , you lose the cash.

So, eight doors are opened and it isn't behind any of those and you're then given an option. Stick with your origianl door or swap to the other door. Most people stick with their original choice, but the smart thing to do is swap...apparently.

While my FIL couldn't really articulate the rationale behind it, I THINK it's because of this.

When you pick the original door, you pick it at odds of 10/1. And despite the fact that other doors are opened and the odds for anyone picking once the other doors are opened get shorter, the fact that you picked when there were ten means the chances of it being behind your door are still 10/1.

Anyway, you get down to two doors and are given the choice. You know it's behind one of them but the chances of it being behind the one you didn't pick are now 2/1. However, because yours is still at 10/1, you should swap as the odds work much more in your favour.

We tried it with nine black cards and one red and you had to try and pick the red card then were given the chance to swap. Because we all thought it was bollox, we stuck with the origianl card, but out of, probably, 15 tries, we were wrong 14 times.

First of all, does anyone know the real theorist's name so I can read up on it a bit? and second is there any truth behind this or is it clever trickery with the cards when the 'proof' is shown?


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## Neil (May 7, 2002)

Bollox surely? :? Still 10/1 chance at the start of it being in either. Not a stats expert tho, so what do I know? :roll:

Been discussed on here b4 I'm sure. Can't remember what the conclusion was though.

If true though, someone should tell the Deal or no Deal contestants, for when they are offered a swap at the end


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## jampott (Sep 6, 2003)

Regardless of what the original odds were, surely it boils down to a simple matter of 50:50 chance.

Faced with Â£250k in one box and 1p in the other, I'd pretty much take whatever Noel and his imaginary "friend" offer me, as its likely to be in the region of Â£100k-Â£125k.

The card / money / whatever is either there or it isn't, and this isn't influenced by the order you do things or, indeed, getting down to a "face-off" 50/50 situation.


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## jampott (Sep 6, 2003)

Kell,

I think you are refering to the "Monty Hall Paradox"


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## jampott (Sep 6, 2003)

Warning.

The website link below will fuck with your head. Read it at your own risk.

http://en.wikipedia.org/wiki/Monty_Hall_problem


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## Rhod_TT (May 7, 2002)

This is mentioned a bit in the book "The Curious Incident of the Dog in the Night-time".


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## Kell (May 28, 2002)

So then my rationale was sort of correct - in that your odds don't change throughout the elimination of the other doors/cards?

I followed most of that until they started with the Venn Diagrams and tree charts... :roll:


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## Neil (May 7, 2002)

wiki wiki said:


> In the problem as stated above, it is because *the host must reveal a goat* and must make the offer to switch that the player has a 2/3 chance of winning by switching.


'ole Monty Hall is a different scenario then, surely? ie. he HAS to reveal a goat at the start, and not just reveal a random other box / door?

Therefore, Monty's answer (ie. yes, always switch) wouldn't be relevant to Kell's scenario? :?

ps. Bayes' theorem / Venn diagrams? WTF? [smiley=dizzy2.gif] [smiley=stop.gif]


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## Kell (May 28, 2002)

neil1003 said:


> wiki wiki said:
> 
> 
> > In the problem as stated above, it is because *the host must reveal a goat* and must make the offer to switch that the player has a 2/3 chance of winning by switching.
> ...


NOt really - the scenario is different only in that there is something behind the other doors (a goat), rather than nothing.

In the end though it boils down to the fact that you're less likely to pick the right door first time than you are to pick the wrong one. The more doors there are, the more likely it is you are to be wrong and therefore the more reason to switch.

Like it says though, you have to know that in the end the prize is either behind the one you chose origianlly or the other choice.


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## Neil (May 7, 2002)

Kell said:


> NOt really - the scenario is different only in that there is something behind the other doors (a goat), rather than nothing.


But in your scenario, the big prize could be revealed by the very 1st door / box opened after you made the choice, as it's just random. But with Monty, they deliberately avoid that box (he knows where the prize is).

Anyway, too complicated for me.... :roll:


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## jampott (Sep 6, 2003)

Kell said:


> neil1003 said:
> 
> 
> > wiki wiki said:
> ...


But - by that reasoning, the OTHER door (if left with 2) is just as likely to be wrong, as it, too, had the same chance as YOUR door of being "wrong" at the time you picked.

Yes, you ARE more likely to pick a "wrong" door the first time around, but if you go through and eliminate all other doors, leaving just the 2 (as per Deal or no Deal) the chances are equal.

Consider a situation where you have 10 boxes and a prize in just 1 of them. You choose a box (A) and stick with it. All others are opened, leaving just 2 boxes your (A) and the other - (B). At the start, the chance of picking the winning box was 1:10. However, the chance of picking box (B) was also 1:10. You can't suggest that switching to box (B) at that point gives you any advantage?


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## Neil (May 7, 2002)

jampott said:


> Consider a situation where you have 10 boxes and a prize in just 1 of them. You choose a box (A) and stick with it. All others are opened, leaving just 2 boxes your (A) and the other - (B). At the start, the chance of picking the winning box was 1:10. However, the chance of picking box (B) was also 1:10. You can't suggest that switching to box (B) at that point gives you any advantage?


I'd agree with that - in the "Deal" scenario, you would have no reason to switch.

The only way I can get my head round "Monty" (that you would switch) is due to the fact that Monty has deliberately eliminated a non-winning option.


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## PissTT (Apr 7, 2006)

The theory is absolutley correct .... it is to do with probabilty.. if you want I can explain it.

You should always swap on Deal or No Deal!

PissTT


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## Neil (May 7, 2002)

PissTT said:


> You should always swap on Deal or No Deal!


Why? :?


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## Kell (May 28, 2002)

> Consider a situation where you have 10 boxes and a prize in just 1 of them. You choose a box (A) and stick with it. All others are opened, leaving just 2 boxes your (A) and the other - (B). At the start, the chance of picking the winning box was 1:10. However, the chance of picking box (B) was also 1:10. You can't suggest that switching to box (B) at that point gives you any advantage?


I'm not sure if I explained quite what I meant in the first instance, but then it was explained to me in a drunken fashion too.

However you arrive at the final two doors/boxes, you know that the prize must be behind one of them - you know this because all others have gone and because the host also knows which box it's in. The likelihood of you being right the first time out is much lower than you being wrong.

Unfortunately, I don't have the mathematical background to explain myself properly - and at first I was thinking that it must be 50/50, but when you think about it it really isn't. It's not even 50/50 on box B in the final scenario - in the instance above there's a 90% chance that the prize will be in box B and only a 10% chance it will be in box A.


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## Neil (May 7, 2002)

Kell said:


> Unfortunately, I don't have the mathematical background to explain myself properly - and at first I was thinking that it must be 50/50, but when you think about it it really isn't. It's not even 50/50 on box B in the final scenario - in the instance above there's a 90% chance that the prize will be in box B and only a 10% chance it will be in box A.


Not getting the 90/10 thing - why?


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## jampott (Sep 6, 2003)

Kell said:


> > Consider a situation where you have 10 boxes and a prize in just 1 of them. You choose a box (A) and stick with it. All others are opened, leaving just 2 boxes your (A) and the other - (B). At the start, the chance of picking the winning box was 1:10. However, the chance of picking box (B) was also 1:10. You can't suggest that switching to box (B) at that point gives you any advantage?
> 
> 
> I'm not sure if I explained quite what I meant in the first instance, but then it was explained to me in a drunken fashion too.
> ...


Kell. Reverse the situation I put forward. You are now in posession of box (B) and you have the option to chose box (A) instead. Bearing in mind these are exactly the same boxes as you were offered in my original situation - are you seriously trying to tell me the odds change?

So if you pick box A and then have a choice of A or B, its most likely to be in B...

Yet if you pick box B and then have a choice of A or B, its most likely to be in A...?

That defies logic I'm afraid. Your choice of box has no bearing whatsoever on the overall probability, as you are picking "blind". You are left (again, "blind") with 2 "random" boxes. OK, so you've chosen one, and (in the case of DonD) you've "picked" how to end up with the remaining one - but they're still "random" inasmuch as nobody knows the contents.


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## PissTT (Apr 7, 2006)

The reason why is ...

Because despite human logic, mathematically the events are linked... each box opening is a connected event.

So by the time you are left with one box, your odds of getting the winning box are greater by swapping than by keeping your own.


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## jampott (Sep 6, 2003)

PissTT said:


> The reason why is ...
> 
> Because despite human logic, mathematically the events are linked... each box opening is a connected event.
> 
> So by the time you are left with one box, your odds of getting the winning box are greater by swapping than by keeping your own.


Yes, but they're also RANDOM. In the situation I suggested (along the lines of Deal or no Deal) you are essentially left with TWO boxes, BOTH of which you've "chosen".

It doesn't matter that things happen in sequence - the end result is the same, because there are no outside influences. Each box has the same chance of containing the biggest prize as any other.

The point is, you have no new information to make your decision to switch. You had the chance to take the 2nd box when presented with the initial 26, so why would it make a difference to switch to it now?


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## Kell (May 28, 2002)

Forget deal or no deal - I've not seen it, so don't know how it works.

Was thinking about htis on the way home and I *think* it must be important that the host gets rid of all the other boxes - in the same way that in the example given, he showed you the goat.

Maybe it's important, maybe it's not.

Neil - it becomes 90% because it has to be. In the link Tim posted it stated that once you get rid of the other option(s) all their probability stacks up behind the one that's left.

And of course, it has to add up to 100% - ergo if box A is 10% chance then box B HAS to be 90%.

And yes Tim, I do think if you reverse the situation, you'd still be better off (statistically) swapping boxes.


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## ag (Sep 12, 2002)

The Monty Hall Paradox is utter utter rubbish.

The probability of choosing the door behind which resides the car in the first instance is 1/3 assuming that it is totally blind, random and honest, i.e. no mind games and gentle persuasion by the host. Worked out by 1 desired outcome divided by 3 possible outcomes.

The host opens a door and finds a goat. The host had prior knowledge of the location of the car and was therefore able to locate either of the two goats. No probability involved.

There is now still 1 desired outcome and 2 possible outcomes. Therefore the probability of having the desired result is 1/2. This is better than the 1/3 at the time of the first guess. The probability that the car is behind door "A" is 1/2 and behind door "B" is also 1/2.

There is no improvement in probability in changing your mind. There is no paradox. The probability of it being behind the door you did not choose is not 2/3 because the number of possible outcomes has changed.

In this case the first guess as to the location of the car is totally irrelevant to the final outcome. One goat is always dismissed leaving a 50:50 chance every time.


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## Kell (May 28, 2002)

Well then why not prove it? Try it with 9 black cards and one red one. By your reasoning you'll be right 50% of the time.

I think the problem with the three cards one is that it all seems to work out nicely with everything being thirds. It's more obvious (to me anyway) when the numbers are higher, so...

10 boxes. One prize. no goats.

The chances of you picking the right box are 1/10. These odds don't change when you take away eight boxes.

The only way they'd change would be if you arrived at the two boxes after all the others had gone and then you'd have a 50:50 chance of picking the right one.


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## ag (Sep 12, 2002)

Kell, the odds do change.

10 boxes, 1 has a prize. You have a probability of 1/10 that the box you chose has the prize.

One box is opened, it is not the box you chose, it is empty. The box you chose is still closed along with the 8 other boxes. The probability that yours has the prize is now one box in nine, i.e. 1/9. Therefore the probability has changed.

If you get down to two boxes left and your chosen box is still unopened then the probability is 1/2, i.e. one chance in two.

As the number of alternative outcomes reduces the probability that your desired outcome is achieved increases.


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## Kell (May 28, 2002)

ag said:


> Kell, the odds do change.
> 
> 10 boxes, 1 has a prize. You have a probability of 1/10 that the box you chose has the prize.
> 
> ...


Well this is whare I was stumbling. When you get down to two boxes, you're thinking it must be in one or t'other so the odds are 50:50. But that disregards all the other boxes - which you can't do in this problem.

Yes when you get rid of one box, the reality is that it must be in one of the other nine, so your chances of being right go up, but the probability that you were right doesn't because you weren't asked to choose from nice boxes, you were asked to choose from 10. And that fact doens't change regardless.

I'm still not 100% sure I'm right, but it does make sense to me now.


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## ag (Sep 12, 2002)

Kell said:


> Yes when you get rid of one box, the reality is that it must be in one of the other nine, so your chances of being right go up, but the probability that you were right doesn't because you weren't asked to choose from nice boxes, you were asked to choose from 10. And that fact doens't change regardless.


It makes no difference how many boxes you were asked to choose from. If five boxes have been removed they can have no effect on the other boxes. You know that there is nothing in them so there is no probability that there is anything in them. The remaining five boxes each have an equal chance of containing the prize, therefore the probability that any one of them contains the prize is 1 in 5 or 1/5. So the probability that your guess was correct at the beginning was 1/10, it is now 1/5 if your box hasn't been opened. The fact that you started with 10 boxes is irrelevant when there are less than 10 boxes left.

The paradox is rubbish, sorry.


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## Kell (May 28, 2002)

I don't know how you can say that.

Put it another way - if you did this 10 times the likelihood is that nine times out of ten, you'd choose the wrong box first off. So only one time in ten would you be worse off by switching.

Perhaps it's the way it was phrased. When it was explained to me, all the other boxes/doors/cards are removed for you, not by you. And I'm not sure if makes any differnece whether they go one at a time or all at once.

You choose one from X amount. X-2 are removed. leaving your box and another. you know it's in one of them. And it's unlikely to be yours.


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## ag (Sep 12, 2002)

Kell said:


> And it's unlikely to be yours.


The liklihood is 50%!

Or, or, or. What you are tryig to say is that the probability that the prize was in your designated box was 1/10 at the beginning and still is. Taking that same logic a step further, the probability that it was in the other box was also 1/10. Therefore the probabilities are the same, and, as the sum of all the probabilities must equal 1, then as they are equal the probability of each event is 1/2.

Bingo.

Or put another way, the box that you chose has no effect on where the prize actually is, so how can choosing one box decrease the probability that it contains the prize.

I'm guessing that the paradox was demonstrated to you using empty beer glasses :wink:


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## Kell (May 28, 2002)

ag said:


> Kell said:
> 
> 
> > And it's unlikely to be yours.
> ...


I've just remembered about this...I was looking for the link Tim posted so I can send it to my FIL.

I'm still not 100% sure the figures add up, but all I can say is to try it out on someone. We tried it loads of times and only got it right once in maybe 15 tries - so it was even less than the odds.

If there are 10 cards and you're asked to pick a specific (ie the only red) one, then 9 times out of ten you'd be wrong.

Conversely, if you're then given the option of sticking with your original choice or another card (by someone who knows where the red card is don't forget) then you must be nines times better off by switiching.


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## Kell (May 28, 2002)

Having read through the link again.

It is vitally important to understand that the 'host's' actions have a large bearing on the Monty Hall scenario.

He has to make the offer to switch every time

He has to show you there's nothing behind whatever doors/cards he takes away



> Monty Hall example
> Remember the Monty Hall "Let's Make a Deal" TV show? A contestant is told that a substantial prize (a new car?) is behind one of three doors (named A, B, and C), and that small insignificant prizes (or nothing) are behind each of the other two doors. The contestant would choose a door, and then, after elaborate fanfare, Monty would open one of the other two doors to reveal an insignificant prize (a bar of soap) or nothing. Monty then would give the contestant a chance to change his/her mind about which door to choose: Stay with the original choice or switch? The studio audience would be screaming "Stay!" or "Switch" and the contestant would ultimately ...
> 
> What are the chances that the big prize is behind the other door? Sure thing? 50%? Some other number?
> ...


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## QuackingPlums (Mar 10, 2004)

Opening the door introduces a dependency (or conditional probability), which is why the final probability is not 50%.

Bayes' Theorem is used to calculate the probability of dependent events in terms of probabilities you DO have. 
eg: you want to know the probability that the car is behind door 1, given that the host has opened door 2. The rules of the game mean you know *he won't open the door with the car behind it*, so u can work out the probabilities of each INDIVIDUAL event using the probabilities you do know:
probability of him opening door 2 = 0.5
probability of the car being behind door 1 = 0.3333
probability of host opening door 2 GIVEN car behind door 1 = 1

Repeat with each combination of car/door and voila!

However, the assumption that switching ALWAYS works out better is also flawed because of the langauage used to express it mathematically.

If you look at the decision tree about half-way down that page, you can see all the possible outcomes charted with their respective probabilities.
The numbers are right, and it certainly does work out mathematically that switching gives you a higher chance of getting the right door. However, probability is merely that - it only gives you a higher chance of picking the right door, it doesn't guarantee it. 
"A higher chance" could be expressed as "if you repeated this 100 times, you would PROBABLY pick the right door more than 50 times". If you factor in standard deviation then it still wouldn't necessarily be abnormal to pick the wrong one 51 times.

Either way, in Deal Or No Deal, or any such game, it only works if you were allowed to repeat it several times!

It's a human condition (and a reason why we're so bad at analysing probability) to consider switching away from a correctly picked door to be worse than sticking with an incorrectly picked door from the off. So even if the odds *were* better, we wouldn't see it that way.

Don't get me started on waiting for buses...


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## t7 (Nov 2, 2002)

Kell it is 50:50 when you get down to the last 2 boxes.

When you make your choice the odds are 10% that you have it right. Every time a box is selected after that _if it doesnt contain the prize _the likelihood you have selected the right box increases. It reaches 50% when there are two boxes left. It cant stay at 10% bacuse you now know the other boxes are empty - their probability of having the prize dropped from 10% to zero at that point.

L


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## QuackingPlums (Mar 10, 2004)

t7 said:


> Kell it is 50:50 when you get down to the last 2 boxes.
> 
> When you make your choice the odds are 10% that you have it right. Every time a box is selected after that _if it doesnt contain the prize _the likelihood you have selected the right box increases. It reaches 50% when there are two boxes left. It cant stay at 10% bacuse you now know the other boxes are empty - their probability of having the prize dropped from 10% to zero at that point.
> 
> L


That's the common misunderstanding of this kind of probability puzzle!

The chances of you picking the right door do not increase with each door that the host opens - the host knows where the prize is, so he is deliberately confusing you by opening the ones without prizes.

The 100 door example is quite clear: if you selected door #1, u have a 1% chance of getting it right. The host now starts opening doors from 100 down, _BUT SKIPS DOOR #50_. When there are two doors left unopened (#1 & #50), your instinct would tell you that he's skipped that door on purpose - you'd be quite sure the prize was behind that door?

Alternatively, let's make it a two-player game. 3 doors, you get to choose one, but after the host opens the door containing a non-prize, player 2 gets the remaining one. 
The chances of you selecting the correct door is one in three (assuming we're repeating the experiment).
Therefore, because the host always opens a non-prize door, player 2 will win two in three times.


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## t7 (Nov 2, 2002)

QuackingPlums said:


> t7 said:
> 
> 
> > Kell it is 50:50 when you get down to the last 2 boxes.
> ...


Quacking - I was not talking about a game show scenario where the person selecting the boxes knows what is behind them. I meant in a truly blind and random selection process (where 4 out of 5 times the prize is selected before you get down to the final two) the probability does change to 50% IF you havent found the prize by the time you get to the last two.

If the person selecting the boxes knows where the prize is then of course when you get to two boxes it is more likely to be in the box you didnt select first.

I didnt understand the rationale behind goats though... 

L


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## Kell (May 28, 2002)

t7 said:


> If the person selecting the boxes knows where the prize is then of course when you get to two boxes it is more likely to be in the box you didnt select first.
> 
> I didnt understand the rationale behind goats though...
> 
> L


Yes - but that's the point, in the example I was trying to recall, the host always takes away the other choices - leaving you two.


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