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talking about the Monty Hall Problem the other day brought us on to the two box problem

Consider this

I have two boxes, one has a cheque in it that is twice the size the size of the other. You make a choice of one box and find £10,000 in it. You are given the chance to change afterwards... should you change.. yes you should according to some

say box one has a cheque for £10,000.. Box two therefore must have a cheque either £5000 or £20,000 pounds. If you average it out you would get 25% more by switching every time. using maths you would state that you have a box with X and then another box with either 2X or 1/2X


In fact you should switch without even looking in the box ( according to some)

Scrodhinger would disagree*, but we're getting a bit quantum here.
I can see the maths in this falls apart just as in the Monty Hall problem and your final sentence actually sums the final situation up quite well. Conditional probability throws a lot of people!

I still go back to Deal Or No Deal and state the average pay out would be a lot higher if the game was simply "i'll have that box"


I have one for you:
If every man in one town that doesn't shave his own beard gets his beard shaved by the town's one barber, who shaves the barber?


*or agrees (did you see that I did there?)

Lawrence wrote:I have one for you:
If every man in one town that doesn't shave his own beard gets his beard shaved by the town's one barber, who shaves the barber?

Either...
1. No-one, as the Barber is growing out a beard.
2. No-one, as the Barber is female and isn't a Bearded Lady.
3. Himself (Presuming that he is male and not part of the group that does not shave themselves)

[I can see the maths in this falls apart just as in the Monty Hall problem ]

The maths in the Monty Hall problem stacks up deosnt it. If you change doors you have a 2/3 chance of winning. If you dont then you have 1/3
I know many dont get it but the math is secure

Ya the Barber.. ha ha..i never figured that out and i guess its the way its worded

Antera wrote:In fact you should switch without even looking in the box ( according to some)

If you switch, you're back to where you were and by applying the same logic, you shoud swap again .... ad infinitum .....

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

The difference between the Monty Hall problem, and the two envelopes problem, is in the Monty Hall problem the person offering you the chance to switch is doing so knowing what you currently have, even though you don't.
With the two envelopes problem, it is inferred that nobody knows what you have, which is why as soon as you switch, it is now also just as advantageous to switch right back again, as the same rules apply to each scenario.

Antera wrote:talking about the Monty Hall Problem the other day brought us on to the two box problem

Consider this

I have two boxes, one has a cheque in it that is twice the size the size of the other. You make a choice of one box and find £10,000 in it. You are given the chance to change afterwards... should you change.. yes you should according to some

say box one has a cheque for £10,000.. Box two therefore must have a cheque either £5000 or £20,000 pounds. If you average it out you would get 25% more by switching every time. using maths you would state that you have a box with X and then another box with either 2X or 1/2X


In fact you should switch without even looking in the box ( according to some)

It doesn't matter whether you switch or not. You're getting at least £5,000!

That's not the right answer, is it?

DenmarkKilo wrote:

Lawrence wrote:I have one for you:
If every man in one town that doesn't shave his own beard gets his beard shaved by the town's one barber, who shaves the barber?

Either...
1. No-one, as the Barber is growing out a beard.
2. No-one, as the Barber is female and isn't a Bearded Lady.
3. Himself (Presuming that he is male and not part of the group that does not shave themselves)

Ah, i see a point here; assume the barber to be male (obviously this was originally wirtten in a time of less equality).
Other points:
1. Everyone is shaven
3. Only those that don't do it themself get it done at the barber's. IE, if he does it himself then therefore the barber doesn't do it.


It's set theory; and is therefore largely a load of balls.

Lawrence wrote:It's set theory; and is therefore largely a load of balls.

That was exactly my finding after a 3 hour lecture at Reading Uni during my OU maths foundation summer school. Maybe it was the lecturer's fault, but it looked a bit overcomplicated. Paradigm is an ugly word...

It's an evens gamble but it pays 2:1
Is the influence of the size of the cheque factored into the model? I'm guessing the larger the amounts get the less likely most people would be to take the gamble. Wait wait we're back at Noel Edmonds again aren't we...

Lawrence wrote:

DenmarkKilo wrote:

Lawrence wrote:I have one for you:
If every man in one town that doesn't shave his own beard gets his beard shaved by the town's one barber, who shaves the barber?

Either...
1. No-one, as the Barber is growing out a beard.
2. No-one, as the Barber is female and isn't a Bearded Lady.
3. Himself (Presuming that he is male and not part of the group that does not shave themselves)

Ah, i see a point here; assume the barber to be male (obviously this was originally wirtten in a time of less equality).
Other points:
1. Everyone is shaven
3. Only those that don't do it themself get it done at the barber's. IE, if he does it himself then therefore the barber doesn't do it.


It's set theory; and is therefore largely a load of balls.

Does the barber shave himself? I can't see a contradiction in that. He's doing it himself so he doesn't have to go to the barber. Is that right?

Ted wrote:Does the barber shave himself? I can't see a contradiction in that. He's doing it himself so he doesn't have to go to the barber. Is that right?

The problem being that if he is doing it himself then the barber isn't doing it, but he IS the barber...
And if the barber is doing it then he isn't doing it himself

It's a set of all sets that are not members of themselves: i.e. if the barber is in the set then he is not in the set, also if he is not in the set then he is in the set.
As I say, set theory is a load of balls.
I prefer mathematical problems that can be visualised with goats behind doors!

Lawrence wrote:The problem being that if he is doing it himself then the barber isn't doing it, but he IS the barber...
And if the barber is doing it then he isn't doing it himself

Paradox, innit.

Or massively recursive, or something...

Tomo wrote: or something...

/MATHS

Lawrence wrote:

Ted wrote:Does the barber shave himself? I can't see a contradiction in that. He's doing it himself so he doesn't have to go to the barber. Is that right?

The problem being that if he is doing it himself then the barber isn't doing it, but he IS the barber...
And if the barber is doing it then he isn't doing it himself

It's a set of all sets that are not members of themselves: i.e. if the barber is in the set then he is not in the set, also if he is not in the set then he is in the set.
As I say, set theory is a load of balls.
I prefer mathematical problems that can be visualised with goats behind doors!

OK, I get what the puzzle is trying to do but I would have thought, in terms of logic, that...

...if he is shaving himself then the barber *is* shaving him, because he is the barber.

There were stipulations that if the men don't shave themselves then the barber does it, but that does not mean that they can't shave themselves and have the barber do it too.