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Topic: Boxing Clever (Read 2514 times) |
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ThudnBlunder
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Boxing Clever
« on: Jun 19th, 2004, 7:34am » |
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Assume I flip a fair coin n times until it comes up Heads. Now I take two large boxes, in one of which I place $3n and in the other I place $3n+1. I then give you the two closed seemingly identical boxes and allow you to open one box and count the money. At this point you can either take the money in the box you chose or take the unknown amount of money in the other box. Is there a strategy which you may employ in order to maximize your expected gain?
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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| « Last Edit: Jun 21st, 2004, 4:03am by ThudnBlunder » |
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rmsgrey
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Re: Boxing Clever
« Reply #1 on: Jun 19th, 2004, 7:45am » |
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It looks like the best strategy is ::always switch:: though I'd want to go away and do some serious modelling before locking that in as my final answer...
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Grimbal
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Re: Boxing Clever
« Reply #2 on: Jun 19th, 2004, 10:01am » |
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I'd really like to play that game, since the expected prize is infinite!
Funnily, it seems it is always good to switch. And this, whatever you see in the box. So in fact, you can decide to switch before even looking inside. And since it is always good to switch, it is still better to switch twice, which is the same as not switching. Hm.... 
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towr
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Re: Boxing Clever
« Reply #3 on: Jun 20th, 2004, 7:11am » |
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Switching only really seems to make sense if you count $3 in the firts box, because that gives you information about what's in the other box. Otherwise either one could be the 'high' box
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ThudnBlunder
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Re: Boxing Clever
« Reply #4 on: Jun 20th, 2004, 7:31am » |
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Deja vu, Grimbal?
Hey, maybe this puzzle should be in Easy.
Quote:| Funnily, it seems it is always good to switch. |
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But trouble is, if you switch you will always be disappointed.
(Welcome back, towr. Gee, that must be a really tough problem that Icarus is stuck on!)
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| « Last Edit: Jun 21st, 2004, 4:07am by ThudnBlunder » |
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Leon
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Re: Boxing Clever
« Reply #5 on: Jun 20th, 2004, 8:01am » |
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on Jun 20th, 2004, 7:11am, towr wrote:| Switching only really seems to make sense if you count $3 in the firts box, because that gives you information about what's in the other box. Otherwise either one could be the 'high' box |
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If the box you pick is anything but $3, it's a 2/3 chance that you did pick the high box.
The reason I think so is:
if you pick a box with $27 in it, then the other box has either $9 (coin was flipped twice ( Tails Heads) making $27 the +1 box) or $81 (coin was flipped 3 times ( T T H) making $81 the +1 box).
Since the chance of the flips having been T H is 1/4, and the chance of the flips having been T T H is 1/8, then there is a 2:1 chance that the $27 you picked is the +1 box ( 1/4 : 1/8 ).
The part I can't figure out is the "maximize your expected gain" and the comment "you will always be disappointed" since even with the 2/3 chance that you have the highest amount already, the extension of 1/3 x possible gain ($27 to $81) less the 2/3 chance of lossing ($27 to $9) the expected outcome is still positive.
Personally, if it was higher than $243 (or $729 maybe), I'd stay since there is only a 6.25% chance that the $243 box is the $3n box (and therefore lower than the other box). But I suppose that technically speaks more to my utility of having $243 in hand vs. losing $162 or gaining $486 than the "maximization" of the gain.
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| « Last Edit: Jun 20th, 2004, 8:23am by Leon » |
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towr
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Re: Boxing Clever
« Reply #6 on: Jun 20th, 2004, 8:11am » |
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on Jun 20th, 2004, 7:31am, THUDandBLUNDER wrote:| (Welcome back, towr. Gee, that must be a really tough problem that Icarus is stuck on!) |
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welcome back? I don't think I've missed a day here for months  Well, maybe at weekends, but not recently.
Who I'm really wondering about is James Fingas, he's been missing since new year..
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towr
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Re: Boxing Clever
« Reply #7 on: Jun 20th, 2004, 8:19am » |
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on Jun 20th, 2004, 8:01am, Leon wrote:| If the box you pick is anything but $3, it's a 2/3 chance that you did pick the high box. |
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There are only two boxes. The chance you picked the high one when it isn't a 3 is infinitesimally (basicly not) higher than picking the low one.
Does knowing what is in one of the boxes give you any information about what's in the other, given you have a pair of preselected boxes?
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Leon
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Re: Boxing Clever
« Reply #8 on: Jun 20th, 2004, 8:30am » |
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on Jun 20th, 2004, 8:19am, towr wrote:
The chance you picked the high one when it isn't a 3 is infinitesimally (basicly not) higher than picking the low one. |
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Simplifying a bit, I read: The chance you picked the +1 box is no higher than picking the low one.
By this you mean that it's 50 - 50? I agree with that statement before you have opened it.
on Jun 20th, 2004, 8:19am, towr wrote:| Does knowing what is in one of the boxes give you any information about what's in the other, given you have a pair of preselected boxes? |
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Given that the $ put into the boxes was methodically determined, I think so. Why isn't it more likely the one you picked is the +1 box? If it's not, why did T&B saying you'll always be disappointed by switching?
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towr
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Re: Boxing Clever
« Reply #9 on: Jun 20th, 2004, 8:52am » |
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There is a problem with both 'allways switch' and 'never switch', because if either is true, then it doesn't matter what's actually in the box, it doesn't help to see it.
'never switch' has the added problem that you obviously should switch if you get $3 in the first box.
I'm still not quite sure what the difference is with the other thread..
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Grimbal
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Re: Boxing Clever
« Reply #10 on: Jun 20th, 2004, 3:33pm » |
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The difference is that in the other thread the possible amounts and the probability of these amounts being in the boxes is not specified.
The probability that you have the high box is not 50% any more after you looked inside, because the amount inside gives you some information. If you see $3, you know it is not the high box. If you see any other number 3n, that number could be the higher of the pair n or the lower of the pair n-1, but since the pair n-1 is twice as likely, it is more likely that it is the high value of pair n-1.
But I don't really see how it can be that it is always good to switch, regardless of what is inside.
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pedronunezmd
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Re: Boxing Clever
« Reply #11 on: Jun 20th, 2004, 8:57pm » |
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Quote:| I'm still not quite sure what the difference is with the other thread... |
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I assume the other thread referred to is ENVELOPE GAMBLE I?
I think the only real difference here is that you know, in this game, what the absolute lowest amount of money in the box can be, namely "3 dollars", so if you see that amount, always switch...
Otherwise, every other amount you see, you don't know if you are looking at the "x" in the one box or the "3x" in the other box. The exponents are just a fancy way of saying the second box has 3 times the amount in the first box. The puzzle is otherwise the same: expected gain from switching is 0.5 (3x-x) + 0.5 (x-3x) = 0.
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Leon
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Re: Boxing Clever
« Reply #12 on: Jun 20th, 2004, 9:28pm » |
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on Jun 20th, 2004, 8:57pm, pedronunezmd wrote:| The puzzle is otherwise the same: expected gain from switching is 0.5 (3x-x) + 0.5 (x-3x) = 0. |
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Don't you mean 0.5 (3x-x) + 0.5(x/3 - x) = 2/3x? Using $27 as the box you opened you will either lose $18 (to $9) or gain $54 (to $81).
0.5 ($81 - $27) + 0.5 ($9 - $27) = $18 ($27 x 2/3).
Why do you feel that the way T&B populated the boxes does not come into play after you know how much is in a given box?
Even then though, the expected outcome is still positive: 1/3 (3x-x) + 2/3(x/3 - x) = 2x/9
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pedronunezmd
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Re: Boxing Clever
« Reply #13 on: Jun 20th, 2004, 9:35pm » |
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on Jun 20th, 2004, 9:28pm, Leon wrote:| Don't you mean 0.5 (3x-x) + 0.5(x/3 - x) = 2/3x? |
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No, I definitely don't mean that. See other thread for long discussion regarding labelling the boxes this way.
on Jun 20th, 2004, 9:28pm, Leon wrote:| Why do you feel that the way T&B populated the boxes does not come into play after you know how much is in a given box? |
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Once the boxes have been populated, if you open the box and see $3, you know you have the low box, so you switch. If you open the box and find anything other than $3, then you don't know if you have the $x from the ($x, $3x) possibility, or the $3x from the ($x, $3x) possibility.
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Leon
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Re: Boxing Clever
« Reply #14 on: Jun 20th, 2004, 9:43pm » |
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on Jun 20th, 2004, 9:35pm, pedronunezmd wrote:
No, I definitely don't mean that. See other thread for long discussion regarding labelling the boxes this way. |
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Fair enough, sorry for putting words into your mouth.
But your equation doesn't work since the expected outcome isn't always $0. This is shown with an example.
If I open the $27 box and assume their is a 50% chance the other box is $9 and a 50% chance the other box is $81, then the expected outcome is 0.5 x $18 loss + 0.5 x $54 gain = $18.
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pedronunezmd
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Re: Boxing Clever
« Reply #15 on: Jun 20th, 2004, 9:48pm » |
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What you are trying to do is label the box that you have now counted as containing x dollars, and then saying that there is a 50% chance that the other box contains x/3 dollars and a 50% change that the other box contains 3x dollars. This is not a correct way to analyze the problem.
The correct way is to realize that you are either in possession of the x or the 3x box from a set of boxes which are labeled x and 3x. Do the calculations from there.
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BNC
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Re: Boxing Clever
« Reply #16 on: Jun 20th, 2004, 10:20pm » |
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on Jun 20th, 2004, 9:48pm, pedronunezmd wrote:
What you are trying to do is label the box that you have now counted as containing x dollars, and then saying that there is a 50% chance that the other box contains x/3 dollars and a 50% change that the other box contains 3x dollars. This is not a correct way to analyze the problem.
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And why is that "not a correct way"? Do you have a reason?
I can present you with another game (you tell me if it's similar). I'll flip a fair coin. If it's head, you give me $6. If tails, I'll give you $18. Want to play?
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towr
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Re: Boxing Clever
« Reply #17 on: Jun 21st, 2004, 1:08am » |
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Given a pair of boxes ($3n, $3n+1)
you pick one of them, and based on it's value decide wether you switch.
If you picked $3, then you know which of the two you have, otherwise you don't. You might have picked $3n, in which case you want to switch, or you might have picked $3n+1 in which case you wouldn't want to switch. But it's not like there's a chance $3n-1 or $3n+2 are involved, because there are only two given boxes.
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BNC
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Re: Boxing Clever
« Reply #18 on: Jun 21st, 2004, 2:33am » |
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It is beoming a duplicate thread, is it not? 
You opened a box, finding $3n. Therefore, you know there were at least n tails. Now the question is: was the next one heads or tails (50/50). If heads, you "shouldn't switch" but the penalty for switching is $2*3n-1. If tails, you should switch, gaining 2*3n. Your gain (50%) is larger than your possible loss (50%), hence you should switch.
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towr
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Re: Boxing Clever
« Reply #19 on: Jun 21st, 2004, 3:11am » |
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But then you should switch regardless of what is in the box, so you needn't look in the box before deciding to switch, which doesn't make any sense as both boxes are then equivalent.
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BNC
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Re: Boxing Clever
« Reply #20 on: Jun 21st, 2004, 3:19am » |
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on Jun 21st, 2004, 3:11am, towr wrote:| But then you should switch regardless of what is in the box, so you needn't look in the box before deciding to switch, which doesn't make any sense as both boxes are then equivalent. |
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Yes, indeed.
I'll repeat something I wrote in the other thread. I'm not sure if it's true, but some years ago, when I took a basic statistics couse at the uni, the teaching assistant told us that this is a recognized paradox.
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pedronunezmd
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Re: Boxing Clever
« Reply #21 on: Jun 21st, 2004, 6:33am » |
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on Jun 20th, 2004, 10:20pm, BNC wrote:I can present you with another game (you tell me if it's similar). I'll flip a fair coin. If it's head, you give me $6. If tails, I'll give you $18. Want to play?
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Yes, I will play this game, and in fact, I will play this game with a $12 bet. That is the expected payoff of this game, so I would play for $12 and expect to break even in the long run.
on Jun 21st, 2004, 3:19am, BNC wrote:| I'll repeat something I wrote in the other thread. I'm not sure if it's true, but some years ago, when I took a basic statistics couse at the uni, the teaching assistant told us that this is a recognized paradox. |
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I agree that this is becoming a duplicate thread. However, I disagree that this is a paradox. There is a flaw in labeling one box as containing x, and the other box has a 50% chance of (3x or x/3). It is not a paradox if the reasoning is flawed.
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towr
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Re: Boxing Clever
« Reply #22 on: Jun 21st, 2004, 8:14am » |
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It's a paradox in that it's seemingly contradictory. not actually contradictory.
You can f.i. get an answer from simulation. Or by using an approriate markov model.
In more classical analysis the problem lies in asking the right question, because otherwise you won't get the right answer.
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pedronunezmd
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Re: Boxing Clever
« Reply #23 on: Jun 21st, 2004, 11:05am » |
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on Jun 21st, 2004, 8:14am, towr wrote:| It's a paradox in that it's seemingly contradictory. not actually contradictory. |
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I have never heard of that definition of a paradox. "Seemingly contradictory" when you use incorrect reasoning is better called "incorrect" rather than "paradox". A paradox is better defined as seemingly contradictory statement derived from otherwise acceptable premises.
For example, the riddle where it is "proved" that "1 = 0" is not really a paradox, it is just an invalid proof. derived from unacceptable premises.
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Leon
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Re: Boxing Clever
« Reply #24 on: Jun 21st, 2004, 11:14am » |
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on Jun 20th, 2004, 9:43pm, Leon wrote:
But your equation doesn't work since the expected outcome isn't always $0. This is shown with an example.
If I open the $27 box and assume their is a 50% chance the other box is $9 and a 50% chance the other box is $81, then the expected outcome is 0.5 x $18 loss + 0.5 x $54 gain = $18. |
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What about the above is incorrect?
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