Logic Problem for GoT

[jwvdcw]

[ QUOTE ]
Before you there are two boxes, one white and one black. Each box contains money; one has twice as much as the other. You may choose either box and keep whatever money is inside.

You choose the white box and find $100 in it.

Okay, the first rule to solving logic puzzles is to read the intro thoroughly because there's often clues and tricks in it. I've read this through a couple times now carefully and this statement still seems to be impossible to me:

Proposition 2. The amounts [that you're able to gain or lose by swtiching] are the same.

I'll keep thinking...

GoT

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GoT:

A question for you then:

Suppose I'm all set to pick box A. After I pick box A, should I switch?

Well what if at the last second I change my pick to box B. Should I then switch back to A after I pick?

[aloiz]

This is faulty logic. You're essentially taking the average of two seperate variables, because your starting amounts are different. First case -x/2 second case +2*x. x is different in each case, so they can't just be averaged.

aloiz

[blackaces13]

If AFTER you picked the first box a guy flipped a coin and put either 2X or 1/2 the amount of the box you picked in another box, then you should ALWAYS switch because of the reasons that me and GoT thought it was a +$25 EV to switch.

If however the boxes are just laying there the whole time then you're just as likely to pick the box with the higher $ amount in it so switching or not is meaningless and the EV of switching is 0.

Cool question.

[maurile]

[ QUOTE ]
If AFTER you picked the first box a guy flipped a coin and put either 2X or 1/2 the amount of the box you picked in another box, then you should ALWAYS switch because of the reasons that me and GoT thought it was a +$25 EV to switch.

If however the boxes are just laying there the whole time then you're just as likely to pick the box with the higher $ amount in it so switching or not is meaningless and the EV of switching is 0.

[/ QUOTE ]
This is intuitively correct. Proving it is another matter.

Here's the technical explanation: link

Mike Caro, of all people, has provided the best simple explanation I've seen [I'm editing it a bit]:

The solution. Switching makes no sense. It's a waste of your time and effort. I'm going to give you two explanations. One is slightly sophisticated, the other is simple and has powerful poker applications.

Sophisticated comes first. Nobody's bankroll is unlimited, and mine is no exception. I decide that a range of $50 to $800 is about right for this experiment. No need to give away big money just to prove a point.

So, I prepare the following sets of boxes: $50/$100, $100/$200, $200/$400, and $400/$800. (Obviously, I could write other pairs of checks, such as $75/$150, but that would just complicate things without changing the explanation.)

When I'm finished, I've created four pairs of boxes. Now I randomly select one pair. I'm not going to make this difficult by breaking it down into what percent of the time you'll get which box and what you gain or lose by switching, but if you're so inclined, you should be able to map this out for yourself in a few minutes.

What I'm going to tell you is this: If you choose a $100, $200, or $400 box, you always gain by switching, because half the time you'll lose half the amount, and half the time you'll double. In the case of a $200 box, this means half the time you'll lose $100 (ending up with $100) and half the time you'll win $200 (ending up with $400). Since each is equally likely, you average a $100 gain for every two times you switch, so trading boxes is worth $50.

This is the crux of the paradox. It previously seemed as if you should always trade. But now that we know the secret size limits of the checks, we can look at it differently. Now we see that there are two exceptions to your lose-half-or-double expectation. If you open a $50 box, there's only one thing that can happen by switching: You gain $50. And if you open an $800 box, there's only one thing that can happen by switching: You lose $400.

So, by switching in both those "extreme" cases, you lose $350. And, wouldn't you know it, that exactly balances out all the gains from all the other choices. So if your strategy is to always switch, you gain nothing. If your strategy is to never switch, you lose nothing. Since you don't know what range your game show decided on for the checks (but that obviously he had to use some range), it does you no good (or no harm) to switch. If you had information letting you know when you were at the high or low end, then you could beat the system by always switching when low, never switching when high. But you don't have this information.

By the way, this explanation holds true no matter how long the sequence of choices you devise, how small the minimum, or how large the maximum.

A simpler explanation. When you think about switching boxes, don't try to figure out all the mathematical implications. Just ask yourself if your opponent wants you to switch. Imagine you offered the boxes. You shuffled them until you couldn't remember which was which. At that point, would you care which box was opened? Of course not! That's the point; look no further.

[GuyOnTilt]

Hey maurile,

Your question is COMPLETELY different from this one that Mike Caro asked. In your question, YOU SHOULD ALWAYS SWITCH. In Mike Caro's, you should only switch if you pick certain amounts. Knowing the value of the 4 pairs of boxes is what makes the difference. In your problem, there was only one pair, and you should always switch.

GoT

[blackaces13]

[ QUOTE ]
In your question, YOU SHOULD ALWAYS SWITCH. In Mike Caro's, you should only switch if you pick certain amounts. Knowing the value of the 4 pairs of boxes is what makes the difference. In your problem, there was only one pair, and you should always switch.




[/ QUOTE ]


[ QUOTE ]
Imagine you offered the boxes. You shuffled them until you couldn't remember which was which. At that point, would you care which box was opened? Of course not! That's the point; look no further.


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[GuyOnTilt]

I'm sorry, I worded my statment poorly. If you have a SERIES of boxes and do not know the range of values that was being offered, you shouldn't ever switch. If there are only 2 boxes, as in the original problem, YOU SHOULD ALWAYS SWITCH.

Caro's problem and solution isn't wrong, but it requires much different logic from the original one posted. I remain very convinced that in the original logic problem, you should always switch.

GoT

[blackaces13]

GoT,

When ALWAYS switching you have exactly a 50% chance of choosing either box. So how can this possibly be better than simply keeping the original box you choose which also has you choosing either of the 2 boxes exactly 50% of the time? All you are doing by switching is adding an extra step into a completely random process.

I'd never win a poker argument with you but I'm fairly confident about this one. [img]/images/graemlins/smile.gif[/img]

[jwvdcw]

Ok, well lets look at it this way then:

After you pick and box, and then you switch.

Then, the game show host(or whomever is giving you this chance) allows you the option of staying put or switching yet again! What do you do? You see, its the same problem. If you really believe that you're getting +EV to switch, then you could just go on switching forever and winning more and more money.

[maurile]

[ QUOTE ]
Your question is COMPLETELY different from this one that Mike Caro asked.

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No, they're exactly the same. (Edit: Caro made some simplifying assumptions [e.g., only four sets of possible box-combinations] to demonstrate the point more easily. For the full version without those simplifying assumptions see the link to the PDF file earlier in that post.)

Here's a different way of thinking about it. You and another fellow are each given a box, and told that one box has twice as much money in it as the other one.

Should you both want to trade?

Should you both be willing to pay me (say, a penny each) to allow you two to trade? This would obviously be +EV for me (to the tune of two cents) -- which means it has to be -EV for each of you.