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For example, if you open a box and see $3 you should always switch because there will definitely be $9 in the other box. However, if you open a box and see $9, switching might get you $3 or might get you $27.

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- Thread starter Ursole
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For example, if you open a box and see $3 you should always switch because there will definitely be $9 in the other box. However, if you open a box and see $9, switching might get you $3 or might get you $27.

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AKG

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[tex](1/2)^{x-1} = 4 \times (1/2)^{x+1}[/tex]

But if it contains more money, it contains 9x more than a box with less money. Let [itex]a = 3^x[/itex]. Currently, you have $a. Switching boxes is expected to give:

(a/3)(4/5) + (3a)(1/5)

= 4a/15 + 3a/5

= 4a/15 + 9a/15

= 13a/15

< a

Of course, this doesn't work in the case where x=1. If you want to have n flips, with only the nth being a head, the probability of that happening is (1/2)^n, and that fact was used as the basis for claiming that the probability of the other box having less money was 4 times more than it having more. However, the probability that all flips are tails except the nth, for n=0 is undefined, so the equation doesn't make sense when x=1, because then we're calculating (1/2)^{1-1} = (1/2)^0 = 1, which doesn't really tell us the probability of all tails, and the zeroeth being heads, since that doesn't even make sense and should be undefined.

So, if the box has $3, change boxes, otherwise don't.

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Ursole said:......I then give you the two closedseeminglyidentical boxes and allow you to open one box and count the money......

They seem identical but once you open them there is a noticable difference.

The Bob (2004 ©)

- #4

Gokul43201

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If the underlying probability distribution (that $X is in a box) is flat, it would seem the the expectation value is always higher for the "other box". However, it is not possible to normalize a flat distribution from 0 to infinity. I think the paradox is resolved by understanding the underlying probabilities of putting money in the boxes.

EDIT : Didn't read the coin flip part. Ignore above discussion.

EDIT : Didn't read the coin flip part. Ignore above discussion.

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The Bob (2004 ©)

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AKG said:Let's say you open the box, and it has $[itex]3^x[/itex] in it. Now, the chances that the box contains less money is 4 times greater than the chances it contains more, simply because:

[tex](1/2)^{x-1} = 4 \times (1/2)^{x+1}[/tex]

But if it contains more money, it contains 9x more than a box with less money. Let [itex]a = 3^x[/itex]. Currently, you have $a. Switching boxes is expected to give:

(a/3)(4/5) + (3a)(1/5)

= 4a/15 + 3a/5

= 4a/15 + 9a/15

= 13a/15

< a

So, if the box has $3, change boxes, otherwise don't.

While it's true that the probability of picking the box with the larger amount is 1/2, the conditional probability that the other box contains more money given that the first box contained a dollars is 1 if a = 3 and 1/3 otherwise.

So if you find a (a > 3) in the box and stick, you get a. If you switch, you get a/3 2/3 of the time, and 3a 1/3 of the time, for an expected value of 11a/9. So should you always switch? We don't need to open the boxes at all then.

Well, that's one way of looking at it, but there's a bit more to the puzzle than that.

.

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- #7

AKG

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First of all, where did I go wrong? Second of all, why is the conditional proability 1/3?Ursole said:While it's true that the probability of picking the box with the larger amount is 1/2, the conditional probability that the other box contains more money given that the first box contained a dollars is 1 if a = 3 and 1/3 otherwise.

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AKG said:First of all, where did I go wrong? Second of all, why is the conditional proability 1/3?

Let n = number of times the coin is flipped.

I open a box and see [tex]3^a[/tex] dollars, where a > 1.

I know n = a or a-1, and Prob(n = a) is half of Prob(n = a-1);

Prob(n = a) = 1/3

Prob(n = a-1) = 2/3

By the way, an 'optimum' strategy of always switching violates symmetry.

.

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- #9

AKG

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That explains why the conditional probability is 1/3, but then where did I go wrong?Ursole said:Let n = number of times the coin is flipped.

I open a box and see [tex]3^a[/tex] dollars, where a > 1.

I know n = a or a-1, and Prob(n = a) is half of Prob(n = a-1);

Prob(n = a) = 1/3

Prob(n = a-1) = 2/3

By the way, an optimum strategy of always switching violates symmetry.

- #10

NateTG

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Ursole said:By the way, an 'optimum' strategy of always switching violates symmetry.

.

So let's say that I choose some value with an even distribution of 0 and 1, and put [tex]1000^x[/tex] in one box, and [tex]3^{x+1}[/tex] in another.

Now, there are four equally likely possibilities when the first box has been pulled:

Flipped 0, pulled 0, Flipped 0, pulled 1

Flipped 1, pulled 1, Flipped 1, pulled 2.

Obviously, in half the cases, the optimal behavior is clear. That leaves the situation where the box contains 1000 dollars.

Clearly this is one of the two cases, and those two cases must be equally likely, so, the expected payout from switching is $499,000.5, but according to you, the violation of symetry indicates that the player should gain no benefit from switching. However it's easy to see that the expected value is *much* higher for switching.

Now, let's take a look at the problem probabilities:

Flipped once, pulled low box (1/4)

Flipped once, pulled high box (1/4)

Flipped twice, pulled low box (1/8)

Flipped twice, pulled high box (1/8)

.

.

.

Flipped n times, pulled low box ([tex]\frac{1}{2*2^n}[/tex])

Flipped n times, pulled high box ([tex]\frac{1}{2*2^n}[/tex])

Flipped n+1 times, pulled low box ([tex]\frac{1}{2*2^{n+1}}[/tex])

Flipped n+1 times, pulled high box ([tex]\frac{1}{2*2^{n+1}}[/tex])

So, lets assume you pull a box with [tex]3^x[/tex] dollars where [tex]x>1[/tex]

So the possibilities are:

flipped [tex]x-1[/tex] times and pulled the high box [tex]\frac{1}{2*2^{x-1}}[/tex]

and

flipped [tex]x[/tex] times and pulled the low box [tex]\frac{1}{2*2^{x}}[/tex]

Renormalizing yields that the probability that the box is high is [tex]\frac{1}{3}[/tex].

So the expected gain from switching is [tex]\frac{2}{3} * 3^{x-1} + \frac{1}{3} * 3^{x+1} - 3^x[/tex]

or

[tex]2 \times 3^{x-2}[/tex]

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