to switch or have it not matter?

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phillip1882
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to switch or have it not matter?

Postby phillip1882 » Tue Sep 09, 2014 11:13 pm UTC

here's a puzzle that's one of my personal favorites. you have two boxes. it is known that one of the boxes has 1/3 the dollar value of the other box. you open a box, see 9 dollars, and are allowed to switch. should you do so? or is it no better?
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Re: to switch or have it not matter?

Postby ahammel » Wed Sep 10, 2014 12:56 am UTC

Naive answer, which I expect somebody will enjoy telling me why it's wrong:

Spoiler:
Expected value of switching = 0.5($3) + 0.5($27) = $10, which is more than $9, so yes you should switch.

Similar logic would seem to suggest that you should switch no matter how much money is in the box, which doesn't sound right.
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Re: to switch or have it not matter?

Postby phlip » Wed Sep 10, 2014 12:58 am UTC

We've had this puzzle before... the short version is:
Spoiler:
You need to know, a priori, which is more likely (and by how much): that the boxes contain ($3, $9), or that the boxes contain ($9, $27). There's no such thing as a uniform distribution over all numbers (and it wouldn't make any practical sense for the amount of money in the boxes to be potentially-arbitrarily-large, there's going to be some practical limit there), so there is some a-priori probability curve over different possibilities of what's in the boxes, you can't just say it's 50/50 without more info. Typically the puzzle is presented in the context of a game show, in which case you can estimate the probability curve by watching the show for a long time and recording what sort of values have been in the boxes in the past, and running stats over that.

If the ($3, $9) configuration is more than 3 times as likely as the ($9, $27) configuration, then by pure EV it makes more sense to keep the $9. If it's less, then it makes more sense to switch. If it's exactly 3 times, then your EV is the same either way. Utility curves and risk-averseness and whatnot will move the exact cutoff around, but it'll be around there.

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Re: to switch or have it not matter?

Postby Diemo » Wed Sep 10, 2014 1:03 am UTC

Spoiler:
Assuming you open randomly, there is 50% chance of getting 3$ and 50% cnace of 27$. If you do this a hundred times, switching each time you do it expected value is 27*50 +3*50>900; Therefore you whould switch.
Ninja, my solution assumes a equal probability of getting either a bigger or smaller box (50-50)
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Re: to switch or have it not matter?

Postby Derek » Wed Sep 10, 2014 8:26 pm UTC

phlip wrote:We've had this puzzle before... the short version is:
Spoiler:
You need to know, a priori, which is more likely (and by how much): that the boxes contain ($3, $9), or that the boxes contain ($9, $27). There's no such thing as a uniform distribution over all numbers (and it wouldn't make any practical sense for the amount of money in the boxes to be potentially-arbitrarily-large, there's going to be some practical limit there), so there is some a-priori probability curve over different possibilities of what's in the boxes, you can't just say it's 50/50 without more info. Typically the puzzle is presented in the context of a game show, in which case you can estimate the probability curve by watching the show for a long time and recording what sort of values have been in the boxes in the past, and running stats over that.

If the ($3, $9) configuration is more than 3 times as likely as the ($9, $27) configuration, then by pure EV it makes more sense to keep the $9. If it's less, then it makes more sense to switch. If it's exactly 3 times, then your EV is the same either way. Utility curves and risk-averseness and whatnot will move the exact cutoff around, but it'll be around there.

It's possible to construct a distribution such that P((X/3, X)) < 3*P((X, 3X)) for all X > 0, but I guess any such distribution has an undefined/infinite mean.

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Re: to switch or have it not matter?

Postby phlip » Thu Sep 11, 2014 12:52 am UTC

Derek wrote:It's possible to construct a distribution such that P((X/3, X)) < 3*P((X, 3X)) for all X > 0, but I guess any such distribution has an undefined/infinite mean.

Well, sure, the St. Petersburg Paradox is always an option if you want to go into that realm...

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Re: to switch or have it not matter?

Postby gmgm » Thu Sep 11, 2014 8:42 am UTC

phillip1882 wrote:you have two boxes. it is known that one of the boxes has 1/3 the dollar value of the other box. you open a box, see 9 dollars

This is either inconsistent or it is underdefined. I see that from the fact that it is given that the amounts contained in both boxes are set beforehand, yet you have also given us here the value (presumably representing a constant) in a particular box that "I open", implying choice and/or volition on my part. However, you fail to specifically define how was the box chosen. If it was an act entirely dependent on our own choice (knowing that different people may make their choices differently), then you would not be able to give us the result of that choice specifically, in this case as a dollar amount. Therefore the choice of the box must have been done in some other way, or if there was an element of our own decision making in this step then it would be a part of our answer.
If it was chosen from the flip of a fair coin then it would have the same inconsistency - you wouldn't be able to give us a specific constant dollar amount, only an example amount or a parameter, e.g. n together with the condition that we cannot infer the possibility or impossibility of the other value being 1/3 the revealed value.
If, however, the initial choice was made in a way that I cannot be certain to have been at least fair and impartial,
Spoiler:
or better yet, with a known skew towards the greater or the lower value, or revealing information about the second value, then it would be better for me to make a choice about switching based on the flip of a fair coin. Expected value or other considerations based on probabilities of two different values for the other box would not apply in this case, since it would not be a random variable.

You've also not explicitly stated that we don't know anything about the value of the second box. It is only implied and I assumed this in the above.
Please clarify or rectify these aspects and then you can have a definitive solution.

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Re: to switch or have it not matter?

Postby thefargo » Tue Sep 23, 2014 9:41 pm UTC

I argue you do have a bit of information - specifically, whether or not the money can be divided into thirds. If I open the first box, and the amount in there is $9.01, I can infer that this is the smaller of the two (since there is no way the other box contains 1/3 of this amount), so I always switch.

In the case that the money IS divisible by 3, you have a 1/3 chance that it is the smaller amount and coincidentally divisible by three (since 1/3 of all numbers are divisible by three), and therefore 2/3 chance that it is the larger of the two. In this case, the expectation value of switching is (2/3)*(1/3)+(1/3)*3=$1.22 (relative to a dollar)

So... I guess you still switch? I thought this was going to somehow go somewhere more interesting...

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Re: to switch or have it not matter?

Postby thefargo » Tue Sep 23, 2014 9:42 pm UTC

So... I guess you still switch? I thought this was going to somehow go somewhere more interesting...


By this I mean my argument, not the general discussion

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Re: to switch or have it not matter?

Postby Trebla » Thu Sep 25, 2014 6:52 pm UTC

thefargo wrote:
So... I guess you still switch? I thought this was going to somehow go somewhere more interesting...


By this I mean my argument, not the general discussion


It becomes far more interesting (and more similar to the linked thread) if you don't know the value of either envelope prior to opening either. Knowing that one contains $9 and the other is equally likely to be $3 or $27 greatly simplifies the decision tree (though I'm struggling to figure out exactly why).

Borrowing from St. Pete's Paradox... let a fair coin be flipped until tails comes up. Call "k" the number of times heads showed up first. In the envelopes put 3^k and 3^(k+1). When you pick an envelope, you don't know if you have k or k+1, both are equally likely to have occurred. What is your expected prize if you switch? What is the expected prize if you don't switch?

Of course, this is pretty much just re-stating the problem in the other thread with powers of 3 instead of 2.


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