There are two boxes with Money

[Tomcat]

There are two sealed boxes on the table and i tell you that one box contains twice as much as the other.

I give you the choice and you choose Box A.

You look inside and there is a cheque for 10,000 USD.

I offer you the chance to switch boxes but you say no point. But is this the best strategy..

switching boxes improves your chances by 25 percent it seems . In other words in a long run of choices you would average 12,500 USD each time by switching boxes

Actually it appears that logic dicates that you should switch without even looking in the box first. anyone come accross this one before

The proof seems to be logical enough ( ill post it later if anyone wants to know)

[pacman]

OK, I'll bite. Just how can a 50/50 proposition favour one outcome over another?

I seem to remember some similar type of conundrum with the same bias. It was shown to be based on a false premise.

What am I missing here TC? Let's see what game you are playing at... ;)

[KenW]

I haven't come across it before Tomcat, but the logic you refer to seems to rely on your initial statement that it is a long run process.

That is, any switch after looking will yield, as you say, the punter 1.25 of the hidden purse over the long run. Proved by your initial statements, if A = a, then the switch is either 2a or a/2. Long run average = 1.25a. (Given your initial amount of a = 10).

But you also seem to imply the punter is only offered one punt.

Whereby surely the one-off punter will not win in this way, but only 50% of the time.

For them it does seem indeed more sensible to stop after 1 punt when the other amount is unknown (20 or 5), and cannot be hunted long run.

However, as you also seem to be talking long run, if the punter switches without looking, there is no initial amount (10,000), and the punter, once switched, gets from B an amount, let's say b which is over the long run n X b, which if in any state where b > a, must over the long haul be larger than n X a. But, if b < a, ...

That's a bit muddled, I know.

Please post the proof (eventually after enough replies).

+1 for a top idea for a thread.

[johnnydiver]

The math here is way beyond me as I learned that "math sucs" from singer Jimmy Buffett a long time ago! Me I would open the first box to determine the amount inside and if the second box was leee then I would sue his ass in court for breach of Contract!

[Tomcat]

Thanks Ken

you have it spot on. it seems like s switch is the best bet as if you find 10,000 one would think well i can get 20K or 5K then its food for thought as 1/2 X and 2X averages out to 1.25X which means you should switch.

as you state Ken

It is is the way the conundrum is posed that leads to false logic. On the net it is widely known as the " Two envelope problem/paradox" and probably best for those interested to look it up. Its still a good paradox puzzle though and requires rigid thinking to get to grips. Actually the Maths boffins still argue about this today .

[pdogg]

Good one Tomcat!

I believe one has a 50/50 shot of initially choosing the best box and that doesn't change after looking at the first box. I'm finding it difficult to come up with the right words to explain my logic but I'll give it a try.

This is not the same as a game where you get 5000 bucks if you a roll a die and it comes up 1,2, or 3 but 20,000 bucks if it comes up a 4,5, or 6. Given that scenario, it would take 12,500 to walk away from the game without rolling. I wouldn't accept 10,000.

Now if there were 3 boxes containing 5, 10, and 20 and you then chose the 10K box, if you were then offered a chance to switch to one of the other two boxes which contain either 5K or 20K then of course you would.

But if you were told in advance that there were two boxes and one contained. 5K and the other contained 10K and you picked the 10K you sure wouldn't change.

There's not a 50/50 shot that the second box contains half or double after you look at the first box. Rather the box contains what it contains.

The posterior probabilty of the second box having the other amount of whatever the two original amounts were is 100% after one looks at the first box.

[Tomcat]

The other famous conundrum is this one.. the Game Show

There are three doors, A B And C and behind two of them are Goats and behind the third is a brand new Ferrari car . So you have a one in three chance of winning the Car given the choice of one door

But heres the thing , you choose say door A .. I then open Door B and show you a goat and offer you the chance to switch to Door C... should you stick with Door A ,or switch to C

Do the odds change by switching to C ( you bet they do , you have a two in three chance by switching to C)

[Manarak3]

Thanks Ken

you have it spot on. it seems like s switch is the best bet as if you find 10,000 one would think well i can get 20K or 5K then its food for thought as 1/2 X and 2X averages out to 1.25X which means you should switch.

as you state Ken

It is is the way the conundrum is posed that leads to false logic. On the net it is widely known as the " Two envelope problem/paradox" and probably best for those interested to look it up. Its still a good paradox puzzle though and requires rigid thinking to get to grips. Actually the Maths boffins still argue about this today .

In the first case, PigDogg got it right, a switch brings nothing.

But proving it is a bit more complex than I thought at the beginning...

The problem seems to stem from the self-referencing, i.e. Box B contains either half or double of box A.

I'll research a bit and come up with something.

About the game show, the answer is yes, people should switch.

The reason will become apparent to anyone if you change the number of doors to 1000.

You pick one and the game master opens 998 doors with the goats.

So do you think you picked the correct gate from 1000, or do you think there is a higher chance that the Ferrari was behind one of the 999 other gates?

[Manarak3]

addendum:

try applying logic on the "unexpected tiger" problem

[KenW]

The other famous conundrum is this one..

There are three doors, A B And C and behind two of them are Goats and behind the third is a brand new Ferrari car . So you have a one in three chance of winning the Car given the choice of one door

But heres the thing , you choose say door A .. I then open Door B and show you a goat and offer you the chance to switch to Door C... should you stick with Door A ,or switch to C

Do the odds change by switching to C ( you bet they do , you have a two in three chance by switching to C)

Ah, the Monty Hall problem. This one I can't reply to Tomcat for I know it and its solution well.

However, I have to say that I remain among the sceptics (and there are many of us) concerning the standard solution.

I have read elegant proofs and understand them.

But have yet to see someone explain why the posterior probability relating to door C can change (after the goat has been revealed by Monty behind door but the same cannot be said for option A.

Why is posterior probability necessarily a non-democratic property?

If it were democratic then as prob C changes then so is prob A allowed to change.

That is, as the sceptics avow, as prob C changes from 1/3 then so may prob A change from 1/3. Each to a half in this case.

Which leaves us with posterior probabilities, given a goat behind B, in each case (A & C) of 1/2.

O well, the argument goes on and on...

[Tomcat]

ref Monty Hall Problem

I dont think there is any argument here or dispute

lets say u always choose box A , it always has a 1/3 chance

But B and C are really the same box as human intervention removes part of the bad luck and they morph into box X which always has 2/3 chance. Remember the goat is always removed not by chance but by design

So in the case of box C as u stated it is really box X ( 2/3 chance) by design. The goat may have been shown in box C and then box B would have the 2/3 chance as we have manipulated the situation)

[Manarak3]

I think I got it for the 2 boxes problem.

The thing is, to calculate probabilities, you need to know in which game you are playing in.

But in this case, you don't know.

You cannot consider the probabilities to have A/2 or 2A together, because these values are not part of the same universe, these cases are mutually exclusive.

we need to pose the game as follows:

2 boxes, containing X and 2X.

Then we open a box and assign the value A to the box.

p(A=X) = 1/2

p(A=2X) = 1/2

Esperance of A is (X+2X)/2

Let's calculate the Esperance of the other box, let's call it B

p(B=X) = 1/2

p(B=2X) = 1/2

Esperance of B is (X+2X)/2

Since we don't know the universe we are playing in, I don't see any reason or even possibility to use the value of A to determine B's esperance, it just doesn't compute.

QED.

[Lefty]

I would just ask PDogg what to do, and since he gave his answer, that's it for me too. The boy is a friggin' math and chess genius. :bow:

[pdogg]

Esperance

New word for me. Easier to say than "expected value" (EV).

[KenW]

New word for me. Easier to say than "expected value" (EV).

Me too. I thought it was a town in Western Australia.