Well, seeing as how I just joined, thought I might contribute a few logic problems I've got. Just one for now though. It's been bugging me for a long time.
Set up:
=========================================
Situation:
You are given a choice. In front of you are 2 boxes. Both contain money. The first contains X amount of money, and the other contains either twice that or half that.
Further:
You choose a box. Inside it you find a check for some given amount of money. You are then given a second choice. You can, without knowing anything more, discard the money you have gotten from the first box and take the money from the second.
Question:
Do you?
===========================================
Problem:
The obvious answer is it doesn't matter.
However, look at this line of reasoning. This is where the real logic problem comes into play.
POINT: X CAN VARY FROM CASE TO CASE
p1) Box 1 contains X money
p2) Box 2 contains either 2X or 1/2X money
(good so far)
p3) Assuming you never switch, you will always end up with X money in each case (x being whatever you get)
p4) Assuming you always switch, you will end up with 2X half the time, and 1/2X half the time.
(good so far)
p5) This can be rewritten as 50% chance to get half X and 50% chance to get twice X.
(still good)
p6) This can be rewritten as (.5 * .5X) + (.5 * 2X) = (.25X) + (1X) = (1.25X)
(still good)
C) Since staying with box 1 gives out an average of X, and always switching gives out an average of 1.25X, it is correct to ALWAYS switch.
Now this is the opposite conclusion of what is intuitive. And pretty obviously wrong.
My question is not which answer is right, because it's obvious that it doesn't matter if you switch or not. My question is WHY does the stated logic problem leads to the wrong conclusion? At what point, at what premise (or between what 2 premises) does my logic fail me? No matter how many times I look over it, I can't figure out where that reasoning fails.
Also, don't think I have the answer, I have no idea.
I would appreciate any help.
Problem seems to be at p6 where you are assuming you get the contents of the box with .5 AND the box with 2 X in.
I'm assuming you are looking at an average over time, so half the time you would get 50%X and half the time you would get 200%X
Since there are 2 possibilities and both are equally likely, multiply both by .5 and add them to get the average output, 1.25X
Since there are 2 possibilities and both are equally likely, multiply both by .5 and add them to get the average output, 1.25X
I think you were right the first time - if it was 0.5 or 1.5 then it wouldn't matter - always choose again.
The fallacy is this assumption...
"Whatever my starting amount is, I have a 50/50 chance of the other amount being larger"
This is not true. The larger the amount you start with, the smaller the probability of gaining by swapping.
The catch lies in the fact that the envelopes were filled using some method or other, and there is no way of filling the boxes/envelopes that break the rule above.
For example, the usual way of assuming how the boxes were filled goes like this: Put a random real number (up to a limit L) in one box, and put twice that in the second one.
In that case, the link between the value of my box (x) and the probability that it is the smaller one is as follows
All of the small envelopes, but only half of the large ones have a value from 0 to L, therefore, for an envelope containing x dollars...
- If x bigger than L, the probability of gaining by swapping is zero (I must have the larger envelope)
- If x is smaller (or equal to) L, the probability of gaining by swapping is 2/3 (for every large envelope in this range, there are two small ones).
I admit that this breaks down if L is infinity. That is, if you put a random number from 0 to infinity in the first envelope, and double that in the second, then for any envelope x, it DOES have a 50/50 chance of being the smaller one, regardless of the value of x, and the paradox holds true.
Unfortunately..... it is meaningless to talk of a random number from zero to infinity. The expression is as meaningless as "divide by zero".
I repeat - there is no way of making the probabiity of winning (by swapping) independant of the value you start with. Do a simulation on a spreadsheet (however you fill the envelopes), and you will see that in the long run, the value of the winning envelopes is less than the value of the losing envelopes (by just enough to make your total winnings come to zero in the long run). Not surprising really put this way is it, since the smaller envelope "wins" (by swapping) every time ?
"Whatever my starting amount is, I have a 50/50 chance of the other amount being larger"
This is not true. The larger the amount you start with, the smaller the probability of gaining by swapping.
The catch lies in the fact that the envelopes were filled using some method or other, and there is no way of filling the boxes/envelopes that break the rule above.
For example, the usual way of assuming how the boxes were filled goes like this: Put a random real number (up to a limit L) in one box, and put twice that in the second one.
In that case, the link between the value of my box (x) and the probability that it is the smaller one is as follows
All of the small envelopes, but only half of the large ones have a value from 0 to L, therefore, for an envelope containing x dollars...
- If x bigger than L, the probability of gaining by swapping is zero (I must have the larger envelope)
- If x is smaller (or equal to) L, the probability of gaining by swapping is 2/3 (for every large envelope in this range, there are two small ones).
I admit that this breaks down if L is infinity. That is, if you put a random number from 0 to infinity in the first envelope, and double that in the second, then for any envelope x, it DOES have a 50/50 chance of being the smaller one, regardless of the value of x, and the paradox holds true.
Unfortunately..... it is meaningless to talk of a random number from zero to infinity. The expression is as meaningless as "divide by zero".
I repeat - there is no way of making the probabiity of winning (by swapping) independant of the value you start with. Do a simulation on a spreadsheet (however you fill the envelopes), and you will see that in the long run, the value of the winning envelopes is less than the value of the losing envelopes (by just enough to make your total winnings come to zero in the long run). Not surprising really put this way is it, since the smaller envelope "wins" (by swapping) every time ?
Here's an equivalent problem: There are two boxes, one with an amount y, and one with 2y (however you look at your problem, one box always has twice the amount of the other). Now, there are two possibilities for your first choice, each with prob .5
1)you pick y.
Stay: y Switch: 2y
2)you pick 2y.
Stay: 2y Switch: y
Expectation value for staying: .5(y+2y)=(3/2)y
Expectation value for switching: .5(2y+y)=(3/2)y
So it doesn't matter what you do, in line with your "obvious answer"
1)you pick y.
Stay: y Switch: 2y
2)you pick 2y.
Stay: 2y Switch: y
Expectation value for staying: .5(y+2y)=(3/2)y
Expectation value for switching: .5(2y+y)=(3/2)y
So it doesn't matter what you do, in line with your "obvious answer"
physicsliz:
I have no problem understanding the logic behind the right answer, thanks though.
Chris in Sheffield:
Yes, that is why I didn't put the problem into the context of a game show, because in that case the limit would not be infinity. The trick to the normal problem with some finite limit is exactly as you pointed out.
However, while knowing that the problem lies within the lack of a limit to the amount of money, I still can't see how that would affect any logic about how to handle the choice. I mean, even with an infinite limit on what the boxes could contain, you still HAVE to find a finite amount in them. It's not like it's possible to actually recieve a check or something for an infinite amount.
(Is this where my logic fails? That you have to recieve a finite amount?)
And if you recieve a finite amount (again, which it is impossible not to), then there is a 50% chance the other box is half and a 50% chance it is twice, which means it is best to switch.
Perhaps someone could expain how dealing with infinity destroys the logic of problems like this? I understand how dividing by zero does, so I figure it's something similar, but knowing that might help.
I have no problem understanding the logic behind the right answer, thanks though.
Chris in Sheffield:
Yes, that is why I didn't put the problem into the context of a game show, because in that case the limit would not be infinity. The trick to the normal problem with some finite limit is exactly as you pointed out.
However, while knowing that the problem lies within the lack of a limit to the amount of money, I still can't see how that would affect any logic about how to handle the choice. I mean, even with an infinite limit on what the boxes could contain, you still HAVE to find a finite amount in them. It's not like it's possible to actually recieve a check or something for an infinite amount.
(Is this where my logic fails? That you have to recieve a finite amount?)
And if you recieve a finite amount (again, which it is impossible not to), then there is a 50% chance the other box is half and a 50% chance it is twice, which means it is best to switch.
Perhaps someone could expain how dealing with infinity destroys the logic of problems like this? I understand how dividing by zero does, so I figure it's something similar, but knowing that might help.
Not a problem.....If you just cash the cheque and spend it on 7.5% diamond/white/frosty Cider then you're not gonna give a toss! 
Hi Mr. Revenge,
Since the problem is only about money... cheat. You will win every time.
That is what is called "thinking outside the box".
Cheers
Since the problem is only about money... cheat. You will win every time.
That is what is called "thinking outside the box".
Cheers
I would expect nothing else from an elf. Too much time spent wrapping up presents leads to loss of plot. In the real world you get no presents if you don't behave yourself. For morality pills at prices you won't believe - contact <link removed by moderator>
Hi Confused2,
You can't change the odds by trying to out think "chance". Read the problem carefully and you will see that it is the only option to always win. If your life depended on this process you would cheat then.. wouldn't you? Lets put bigger stakes on it... the fate of Earth depends on it.... what then? I didn't say that you needed to hurt anyone - obviously the "benefactor" of this game is not that concerned how you play since there will be something in it for him as well... everyone wins (he is trying to cheat out of the bigger reward using your "enhanced" greed and you cheat him... fairly... by carefully reading the rules first).
For instance if he was filling the box with loose change then after a couple of rounds you could simply "heft" the boxes and always choose the heavier box. It depends on how much this process means to you as to how much technology you "invest" in the outcome.
Stage Magicians invest fortunes in some of their tricks so that their deceptions are concealed from the audience... I suggest you ask one of them and shove a large bundle of high denomination notes in his hand for the answer. For instance... you could bribe the "box packer" to follow a pre-arranged pattern for a piece of the action. ... he he he!
Cheers
PS: I have consistently told all of you what we elves are like... do not show any false surprise there
You can't change the odds by trying to out think "chance". Read the problem carefully and you will see that it is the only option to always win. If your life depended on this process you would cheat then.. wouldn't you? Lets put bigger stakes on it... the fate of Earth depends on it.... what then? I didn't say that you needed to hurt anyone - obviously the "benefactor" of this game is not that concerned how you play since there will be something in it for him as well... everyone wins (he is trying to cheat out of the bigger reward using your "enhanced" greed and you cheat him... fairly... by carefully reading the rules first).
For instance if he was filling the box with loose change then after a couple of rounds you could simply "heft" the boxes and always choose the heavier box. It depends on how much this process means to you as to how much technology you "invest" in the outcome.
Stage Magicians invest fortunes in some of their tricks so that their deceptions are concealed from the audience... I suggest you ask one of them and shove a large bundle of high denomination notes in his hand for the answer. For instance... you could bribe the "box packer" to follow a pre-arranged pattern for a piece of the action. ... he he he!
Cheers
PS: I have consistently told all of you what we elves are like... do not show any false surprise there
You have two different scenarios mixed together as if they are the same thing so this is a bit long winded, but here it goes...
If you pick up the first box and get X, and say that under the second box is either .5X or 2X, then you have 3 variables, I believe your looking at the problem as if there are only two, but solving your equations with 3. (Am I right in understanding that you don’t need to be shown how there is an equal chance either way of getting the 2X whether you stick with the first box or pick the second box as well?...You already know this, you just want to know where the misconception is in the seemingly logical solution that gives the illogical answer…right?)
If you do your problem with two variables....that is define it with two variables as you did, then you have to solve it with two variables....maybe like this:
(or you can do it with 3 as well...further below)
Situation A)-two variables
The two boxes having either X or 2X in them. If there are only two variables ie. X amount of money, or 2X amount of money, then there is no 0.5X amount of money. (Now if you want to use an average value for box one, the average between X and 2X would be 1.5X, this 1.5X however isn't used as an actual possible variable because you can't actually draw the 1.5X(this is where your 3rd variable is coming from)....also note that X : 1.5X : 2X is not the same relationship as your 0.5X : X : 2X. In yours the two extreme possibilities are different by a factor of 4, where as in the two variable problem (only 1 of 2 possible variables in each box) the difference is a factor of 2. Even if you make the average value “X” as you did in yours, that makes the low value 0.666X and the high value 1.333X that’s the only way you can get X as your average of two variables where one of your variables is double the other. This is the one of the main problem with your equations. I think you’ll find that if you put the 0.666X and the 1.333X into your equation for the second box, with a 50% chance at both, you’ll get the expected answer.but.....
Situation
-3 variables
The two boxes have either 0.5X, X or 2X. In your answer the first box always contains X, but if X is to be the average between 0.5X and 2X (which violates your proposition in the question, assuming box 1 and 2 both have an equal chance of having X or 2X in them…one of two possibilities), then X(first box) should actually be 1.25X(average of 0.5X and 2X). This would also be a way of balancing your equation.
Also, in a different scenario which is kind of (accidentally?) woven in to your problem, you really do always have X in the first box, and there is a 50/50 chance that the second will contain 0.5X or 2X
Scenario : first box always contains $100(X). There’s a 50% chance your playing the game with a 100(X) and a 200(2X), and a 50% chance your playing with a 100(X) and a 50(0.5X). This leaves a 50% chance that the 2nd box contains a 50, and a 50% chance it contains a 200.
Box 1=(1*X)=100
Box 2=(0.5*0.5X)+(0.5*2X)=125
In this case then Yes…you always choose the second box because the first is fixed at X while the second is an average of 0.5X and 2X and X is lower than the average of 0.5X and 2X.
So your equations don’t match because you have 2 different scenarios mixed together.
If you pick up the first box and get X, and say that under the second box is either .5X or 2X, then you have 3 variables, I believe your looking at the problem as if there are only two, but solving your equations with 3. (Am I right in understanding that you don’t need to be shown how there is an equal chance either way of getting the 2X whether you stick with the first box or pick the second box as well?...You already know this, you just want to know where the misconception is in the seemingly logical solution that gives the illogical answer…right?)
If you do your problem with two variables....that is define it with two variables as you did, then you have to solve it with two variables....maybe like this:
(or you can do it with 3 as well...further below)
Situation A)-two variables
The two boxes having either X or 2X in them. If there are only two variables ie. X amount of money, or 2X amount of money, then there is no 0.5X amount of money. (Now if you want to use an average value for box one, the average between X and 2X would be 1.5X, this 1.5X however isn't used as an actual possible variable because you can't actually draw the 1.5X(this is where your 3rd variable is coming from)....also note that X : 1.5X : 2X is not the same relationship as your 0.5X : X : 2X. In yours the two extreme possibilities are different by a factor of 4, where as in the two variable problem (only 1 of 2 possible variables in each box) the difference is a factor of 2. Even if you make the average value “X” as you did in yours, that makes the low value 0.666X and the high value 1.333X that’s the only way you can get X as your average of two variables where one of your variables is double the other. This is the one of the main problem with your equations. I think you’ll find that if you put the 0.666X and the 1.333X into your equation for the second box, with a 50% chance at both, you’ll get the expected answer.but.....
Situation
-3 variablesThe two boxes have either 0.5X, X or 2X. In your answer the first box always contains X, but if X is to be the average between 0.5X and 2X (which violates your proposition in the question, assuming box 1 and 2 both have an equal chance of having X or 2X in them…one of two possibilities), then X(first box) should actually be 1.25X(average of 0.5X and 2X). This would also be a way of balancing your equation.
Also, in a different scenario which is kind of (accidentally?) woven in to your problem, you really do always have X in the first box, and there is a 50/50 chance that the second will contain 0.5X or 2X
Scenario : first box always contains $100(X). There’s a 50% chance your playing the game with a 100(X) and a 200(2X), and a 50% chance your playing with a 100(X) and a 50(0.5X). This leaves a 50% chance that the 2nd box contains a 50, and a 50% chance it contains a 200.
Box 1=(1*X)=100
Box 2=(0.5*0.5X)+(0.5*2X)=125
In this case then Yes…you always choose the second box because the first is fixed at X while the second is an average of 0.5X and 2X and X is lower than the average of 0.5X and 2X.
So your equations don’t match because you have 2 different scenarios mixed together.
sorry...long winded guest was me
This answer is much simpler than some of you are making it. The reason why it is better in the long run to bet every time is because the difference between x and 2x is not the same as x and x/2. The common ratio does not matter in this circumstance since we are concerned here about the sum of money, not its rates. Try the problem with x and x/2, then x and 3x/2 and you will have a true zero sum game. I think staggerbot and confused2 was getting at the same idea, but they could've been a lot more concise.
Wouldn't the odds change dramatically if the box you selected had an odd number of bills enclosed, or an odd denomination (i.e., 'value')?
QUOTE (KAHruzer+Dec 3 2005, 02:19 AM)
Wouldn't the odds change dramatically if the box you selected had an odd number of bills enclosed, or an odd denomination (i.e., 'value')?
x/x/3=3x/x
In an arbitrarily long series of choices, nx = (x+y)n/2 + (x-y)n/2 That is, if n represents number of choices, preferably an infinite number
(so n results of x = [x+y]n/2 + [x-y]n/2 which are the probabilities of equally likely mystery choices). y represents common difference.
The most reduced version of the problem only asks whether a person will keep or risk his prize for a maybe better/worse deal. In this sort of a problem, the only things that modify the overall equation is the difference between the originial and the two options (the existence of a single value y versus two or more values). You could complicate this by requiring one to analyze probabilities of more than 2 options, or adding subvariations in the way bills or whatever are chosen, but please don't for the sake of my brain
x/x/3=3x/x
In an arbitrarily long series of choices, nx = (x+y)n/2 + (x-y)n/2 That is, if n represents number of choices, preferably an infinite number
(so n results of x = [x+y]n/2 + [x-y]n/2 which are the probabilities of equally likely mystery choices). y represents common difference.The most reduced version of the problem only asks whether a person will keep or risk his prize for a maybe better/worse deal. In this sort of a problem, the only things that modify the overall equation is the difference between the originial and the two options (the existence of a single value y versus two or more values). You could complicate this by requiring one to analyze probabilities of more than 2 options, or adding subvariations in the way bills or whatever are chosen, but please don't for the sake of my brain
QUOTE (JavaTool+Dec 3 2005, 03:40 AM)
QUOTE (KAHruzer+Dec 3 2005, 02:19 AM)
Wouldn't the odds change dramatically if the box you selected had an odd number of bills enclosed, or an odd denomination (i.e., 'value')?
x/x/3=3x/x
In an arbitrarily long series of choices, nx = (x+y)n/2 + (x-y)n/2 That is, if n represents number of choices, preferably an infinite number (so n results of x = [x+y]n/2 + [x-y]n/2 which are the probabilities of equally likely mystery choices). y represents common difference.
The most reduced version of the problem only asks whether a person will keep or risk his prize for a maybe better/worse deal. In this sort of a problem, the only things that modify the overall equation is the difference between the originial and the two options (the existence of a single value y versus two or more values). You could complicate this by requiring one to analyze probabilities of more than 2 options, or adding subvariations in the way bills or whatever are chosen, but please don't for the sake of my brain
ahh..ok. I'm assuming this is the case but just out of curiosity and for future reference, is x/x/3=3x/x a mathematical colloquialism for "no"?
Yes (sorry), the equation simplifies to 3 - the common ratio. Odd numbers wouldn't make a difference: 10 choices of amount x money would yield 10x money, while 5 choices of 3x money would yield 15x money and 5 choices of x/3 money would yield (5/3)x money. It's still a "contradiction" in that, in a long enough series of choices, the amount of money gained by switching is always higher than for keeping the money. Notice that the profit made from switching is even higher here than for the original problem, but this is merely because the difference between 3x and 2x is greater than the difference between x/3 and x/2 (only for x>1 of course). In fact, the profits will just get higher and higher as you go up the number line. Clearly, we can't rely on common ratios here.
I... don't think so.
x/x/3=3x/x
In an arbitrarily long series of choices, nx = (x+y)n/2 + (x-y)n/2 That is, if n represents number of choices, preferably an infinite number
(so n results of x = [x+y]n/2 + [x-y]n/2 which are the probabilities of equally likely mystery choices). y represents common difference.The most reduced version of the problem only asks whether a person will keep or risk his prize for a maybe better/worse deal. In this sort of a problem, the only things that modify the overall equation is the difference between the originial and the two options (the existence of a single value y versus two or more values). You could complicate this by requiring one to analyze probabilities of more than 2 options, or adding subvariations in the way bills or whatever are chosen, but please don't for the sake of my brain
ahh..ok. I'm assuming this is the case but just out of curiosity and for future reference, is x/x/3=3x/x a mathematical colloquialism for "no"?
QUOTE
ahh..ok. I'm assuming this is the case but just out of curiosity and for future reference, is x/x/3=3x/x a mathematical colloquialism for "no"?
Yes (sorry), the equation simplifies to 3 - the common ratio. Odd numbers wouldn't make a difference: 10 choices of amount x money would yield 10x money, while 5 choices of 3x money would yield 15x money and 5 choices of x/3 money would yield (5/3)x money. It's still a "contradiction" in that, in a long enough series of choices, the amount of money gained by switching is always higher than for keeping the money. Notice that the profit made from switching is even higher here than for the original problem, but this is merely because the difference between 3x and 2x is greater than the difference between x/3 and x/2 (only for x>1 of course). In fact, the profits will just get higher and higher as you go up the number line. Clearly, we can't rely on common ratios here.
In this case doubling your money with a 50 percent chance of loosing half you should always pick the second box (if there is truly a 50/50 chance). The odds are that you will increase at a rate of 1.25. Your numbers are correct. If you get the half, you have only lost half. But if you get double you have gained a whole.
QUOTE (Mr. Revenge+Dec 1 2005, 07:43 AM)
My question is not which answer is right, because it's obvious that it doesn't matter if you switch or not. My question is WHY does the stated logic problem leads to the wrong conclusion? At what point, at what premise (or between what 2 premises) does my logic fail me? No matter how many times I look over it, I can't figure out where that reasoning fails.
I'm confused.
What conclusion are you lead to believe?
I'm confused.
What conclusion are you lead to believe?
There are actually only 2 amounts of money - Not 3 - only 2.
If you pick the lower valued amount, the other must have double of it and vice-versa.
Then the probaility of getting 'M' amounts of money is 0.5 and the probability of its double is again 0.5. So the problem is symmetrical.
But the way the problem has been presented makes it look asymmetrical.
Do do not trade your initial choice. There is equal probalilty of making or loosing money.
If you pick the lower valued amount, the other must have double of it and vice-versa.
Then the probaility of getting 'M' amounts of money is 0.5 and the probability of its double is again 0.5. So the problem is symmetrical.
But the way the problem has been presented makes it look asymmetrical.
Do do not trade your initial choice. There is equal probalilty of making or loosing money.
The odds of getting more and loosing are equal (50/50). But the amount of gain and loss are not. Lets say the amounts are 100, 50, 200. If you get half you would have lost 50 and if you get double you have gained 100. This is why the odds are 1.25. If you kept doing this, your average gain would be 25 percent.
Is there some kind of larger point I should be thinking about here? The confusion here seems to be psychological, not logical. Is there any instance of this logic problem where the answer is either undecidable or favors keeping?
I think it depends how many times you get to choose a box. If it's only once, then you have a 50/50 chance of being better off, regardless of the amount. If it's many times, then... well.. that's different.
QUOTE
I think it depends how many times you get to choose a box. If it's only once, then you have a 50/50 chance of being better off, regardless of the amount. If it's many times, then... well.. that's different.
I... don't think so.
Here's how I figure it:
You've got two boxes, one of which has $x and the other of which has $2x. You pick one, examine the contents, and then have the option of switching.
There are four possible outcomes:
1. You pick x, and don't switch. You get x.
2. You pick x, and switch. You get 2x.
3. You pick 2x, and don't switch. You get 2x.
4. You pick 2x, and switch. You get x.
In the "switch" cases, you have one x and one 2x, for an expected result of 1.5x.
In the "don't switch cases, you have one x and one 2x, for an expected result of 1.5x.
It doesn't matter whether you switch or not.
You've got two boxes, one of which has $x and the other of which has $2x. You pick one, examine the contents, and then have the option of switching.
There are four possible outcomes:
1. You pick x, and don't switch. You get x.
2. You pick x, and switch. You get 2x.
3. You pick 2x, and don't switch. You get 2x.
4. You pick 2x, and switch. You get x.
In the "switch" cases, you have one x and one 2x, for an expected result of 1.5x.
In the "don't switch cases, you have one x and one 2x, for an expected result of 1.5x.
It doesn't matter whether you switch or not.
Here's a more interesting variant:
You are presented with three boxes, each of which contains ten bills. In two boxes, the bills are $1. In one box, the bills are $1000.
You select a box. Before you open the box to check its contents, the owner of the boxes opens one of the boxes you didn't pick, revealing a stack of $1 bills. You are then given the option of keeping your first pick or taking the remaining box.
Is it better to keep your first choice or to switch boxes?
You are presented with three boxes, each of which contains ten bills. In two boxes, the bills are $1. In one box, the bills are $1000.
You select a box. Before you open the box to check its contents, the owner of the boxes opens one of the boxes you didn't pick, revealing a stack of $1 bills. You are then given the option of keeping your first pick or taking the remaining box.
Is it better to keep your first choice or to switch boxes?
QUOTE (Zeropoint+Dec 12 2005, 07:52 PM)
Here's how I figure it:
You've got two boxes, one of which has $x and the other of which has $2x. You pick one, examine the contents, and then have the option of switching.
There are four possible outcomes:
1. You pick x, and don't switch. You get x.
2. You pick x, and switch. You get 2x.
3. You pick 2x, and don't switch. You get 2x.
4. You pick 2x, and switch. You get x.
In the "switch" cases, you have one x and one 2x, for an expected result of 1.5x.
In the "don't switch cases, you have one x and one 2x, for an expected result of 1.5x.
It doesn't matter whether you switch or not.
You are right. I misunderstood the problem as allowing box 2 to vary between 2x and x/2 money (absurd, I know )
You've got two boxes, one of which has $x and the other of which has $2x. You pick one, examine the contents, and then have the option of switching.
There are four possible outcomes:
1. You pick x, and don't switch. You get x.
2. You pick x, and switch. You get 2x.
3. You pick 2x, and don't switch. You get 2x.
4. You pick 2x, and switch. You get x.
In the "switch" cases, you have one x and one 2x, for an expected result of 1.5x.
In the "don't switch cases, you have one x and one 2x, for an expected result of 1.5x.
It doesn't matter whether you switch or not.
You are right. I misunderstood the problem as allowing box 2 to vary between 2x and x/2 money (absurd, I know
)
Sorry Zeropoint,
Your problem is the famous Monty Hall problem, and the slightly counter-intuitive but mathematically sound answer, is that you do switch. The envelopes problem is a completely different paradox and although its easy to see that its pointless to swap, its not easy to see the fault in the logic.
If i offer you £5 or the opportunity to flip a coin to either win half or double the £5, you should always gamble.
Similarly, if i give you an envelope that has £5 in it and offer you the chance to swap it for one that has either £2.50 or £10 in, with equal probability of either, you should swap.
Equally obviously, if i offer you one one of 2 envelopes, one of which contains twice as much as the other, after accepting one at random its clearly pointless to swap them.
The paradox is why is it relevant if somebody tells me how much money is in the envelope that is in my hand or not. Either way I have a definite amount of money in my hand.
Understanding precisely what the paradox is crucial to understanding the solution.
Lets suppose the big honcho puts £2.50 and £5 in one set of envelopes, and £5 and £10 in the other set. Then, suppose he flips a coin to decide which set to use, and, using that set, he gives one envelope to each player.
Now, taking it from Player A's perspective, there are 3 amounts of money he could have in his envelope, namely £2.50, £5 or £10, but, sneakily, 4 scenarios.
1. He has £2.50. By swapping, he wins £2.50, because the other guy must have £5 in his envelope.
2. He has £10. By swapping, he loses £5, because the other guy must have £5 in his envelope.
3. He has £5 and the other guy has £10. By swapping he gains £5.
4. He has £5 and the other guy has £2.50. By swapping, he loses £2.50.
Since all scenarios are equally likely, each scenario happens exactly 1/4 of the time. Thus, by swapping 1/4 of the time he gains £5 and 1/4 of the time he gains £2.50, but, 1/4 of the time he loses £2.50 and 1/4 of the time he loses £5. Thus by using the strategy of always swapping, he ends up with a net expected gain of exactly £0 as you would expect.
The same of course applies to Player B, so both players gain and lose nothing by swapping, just as you would expect.
The statement, "The player can double or halve his money with equal probability" is clearly true, since he can always consider himself to be in the middle, the equal probability referring to scenarios 3 and 4 above, implying that he stands to make a net gain of £1.25.
The fallacy is that we're ignoring scenarios 1 and 2, which imply a net loss of £1.25, i.e a £0 net gain over all 4 scenarios.
The paradox works by saying that there's only 2 possible scenarios, either you're in the higher valued game with the smaller amount of money or you're in the lower valued game with the higher amount of money, with a 50:50 chance of either, leading you a net gain by swapping.
However, in reality, there's a 50% chance that the previous sentence is true, but a 50% chance that you're in the higher valued game with the larger amount of money or in the lower valued game with the smaller amount of money, leading to a net loss by swapping; a net loss that exactly cancels out the net game from the other scenarios.
Its true that half the time you double you're money, but but half the time you double, you double the smaller amount, and half the time you double the larger amount. Similarly for losing, half the time you lose half the larger amount and half the time you lose half of the smaller amount.
Thus, if nobody knows how much is in either envelope, the potential gains are exactly matched by the potential losses.
However, if you're now told how much you have in your envelope, i.e. £5, you've eliminated scenarios 1 and 2, and you should swap, (although of course the other person should refuse to swap if he knows that you have £5.)
Your problem is the famous Monty Hall problem, and the slightly counter-intuitive but mathematically sound answer, is that you do switch. The envelopes problem is a completely different paradox and although its easy to see that its pointless to swap, its not easy to see the fault in the logic.
If i offer you £5 or the opportunity to flip a coin to either win half or double the £5, you should always gamble.
Similarly, if i give you an envelope that has £5 in it and offer you the chance to swap it for one that has either £2.50 or £10 in, with equal probability of either, you should swap.
Equally obviously, if i offer you one one of 2 envelopes, one of which contains twice as much as the other, after accepting one at random its clearly pointless to swap them.
The paradox is why is it relevant if somebody tells me how much money is in the envelope that is in my hand or not. Either way I have a definite amount of money in my hand.
Understanding precisely what the paradox is crucial to understanding the solution.
Lets suppose the big honcho puts £2.50 and £5 in one set of envelopes, and £5 and £10 in the other set. Then, suppose he flips a coin to decide which set to use, and, using that set, he gives one envelope to each player.
Now, taking it from Player A's perspective, there are 3 amounts of money he could have in his envelope, namely £2.50, £5 or £10, but, sneakily, 4 scenarios.
1. He has £2.50. By swapping, he wins £2.50, because the other guy must have £5 in his envelope.
2. He has £10. By swapping, he loses £5, because the other guy must have £5 in his envelope.
3. He has £5 and the other guy has £10. By swapping he gains £5.
4. He has £5 and the other guy has £2.50. By swapping, he loses £2.50.
Since all scenarios are equally likely, each scenario happens exactly 1/4 of the time. Thus, by swapping 1/4 of the time he gains £5 and 1/4 of the time he gains £2.50, but, 1/4 of the time he loses £2.50 and 1/4 of the time he loses £5. Thus by using the strategy of always swapping, he ends up with a net expected gain of exactly £0 as you would expect.
The same of course applies to Player B, so both players gain and lose nothing by swapping, just as you would expect.
The statement, "The player can double or halve his money with equal probability" is clearly true, since he can always consider himself to be in the middle, the equal probability referring to scenarios 3 and 4 above, implying that he stands to make a net gain of £1.25.
The fallacy is that we're ignoring scenarios 1 and 2, which imply a net loss of £1.25, i.e a £0 net gain over all 4 scenarios.
The paradox works by saying that there's only 2 possible scenarios, either you're in the higher valued game with the smaller amount of money or you're in the lower valued game with the higher amount of money, with a 50:50 chance of either, leading you a net gain by swapping.
However, in reality, there's a 50% chance that the previous sentence is true, but a 50% chance that you're in the higher valued game with the larger amount of money or in the lower valued game with the smaller amount of money, leading to a net loss by swapping; a net loss that exactly cancels out the net game from the other scenarios.
Its true that half the time you double you're money, but but half the time you double, you double the smaller amount, and half the time you double the larger amount. Similarly for losing, half the time you lose half the larger amount and half the time you lose half of the smaller amount.
Thus, if nobody knows how much is in either envelope, the potential gains are exactly matched by the potential losses.
However, if you're now told how much you have in your envelope, i.e. £5, you've eliminated scenarios 1 and 2, and you should swap, (although of course the other person should refuse to swap if he knows that you have £5.)
Lets make X = 10. There are two configurations
X and 2X (10 and 20)
X and X/2 (10 and 5)
If Pick1 = X(10) then you have the following posibilites for pick2:
Pick2 = 2X (20 - gain 10) or X/2 (5 - loss 5)
If Pick1 = 2X(20) then you must have the following posibility
Pick2 = X(10 - loss 10)
If Pick1 = X/2 (5) then you must have the following posibility
Pick2 = X(10 - gain 5)
So look at the possible gains and losses:
Gain10 , Loss 5, Loss10, Gain 5.
It's split right down the middle. You have a 1 in 2 chance of winning/losing if you pick the second box.
If however you only make one choice, you have the following chances:
Pick1 = X (10 - no gain, no loss)
Pick1 = 2X (20 - gain 10)
Pick1 = X/2 (5 - loss 5).
Results:
No gain No loss, 10 gain, 5 loss.
Only 1 out of 3 will be negative, so you have a higher chance winning if you do not pick the second box.
Therefore, pick one and stick with it
X and 2X (10 and 20)
X and X/2 (10 and 5)
If Pick1 = X(10) then you have the following posibilites for pick2:
Pick2 = 2X (20 - gain 10) or X/2 (5 - loss 5)
If Pick1 = 2X(20) then you must have the following posibility
Pick2 = X(10 - loss 10)
If Pick1 = X/2 (5) then you must have the following posibility
Pick2 = X(10 - gain 5)
So look at the possible gains and losses:
Gain10 , Loss 5, Loss10, Gain 5.
It's split right down the middle. You have a 1 in 2 chance of winning/losing if you pick the second box.
If however you only make one choice, you have the following chances:
Pick1 = X (10 - no gain, no loss)
Pick1 = 2X (20 - gain 10)
Pick1 = X/2 (5 - loss 5).
Results:
No gain No loss, 10 gain, 5 loss.
Only 1 out of 3 will be negative, so you have a higher chance winning if you do not pick the second box.
Therefore, pick one and stick with it
(2+.5)> 1
2
Although the odds are even that you will ge either 2x or .5x, over the long term the amount rtecovered will be greater by always switching.
ex: $10x20= $200 vs. ($5x10 +$20x10)=$250
2
Although the odds are even that you will ge either 2x or .5x, over the long term the amount rtecovered will be greater by always switching.
ex: $10x20= $200 vs. ($5x10 +$20x10)=$250
What on earth are you talking about?
Which riddle are you trying to explain?
Which riddle are you trying to explain?
There are actually three pages to this riddle conversation. Start from the beginning page. 
Trust me, three pages is REALLY short for these riddles...
Trust me, three pages is REALLY short for these riddles...
Looks like Mr. Revenge long ago lost interest here. But since some activity continues, I'll answer the question that Mr. Revenge asked. He wanted to know where the logic mistake is...
There are two problems, in p3 and p4, both dealing with the definition of the variable X. The variable X is poorly defined all around. It can represent 2 different values depending on which box is chosen.
There are two problems, in p3 and p4, both dealing with the definition of the variable X. The variable X is poorly defined all around. It can represent 2 different values depending on which box is chosen.
QUOTE (Mr. Revenge+ Dec 1 2005, 02:43 AM)
p3) Assuming you never switch, you will always end up with X money in each case (x being whatever you get)
Here in p3 the variable X changes with each trial. Since a box is randomly chosen each time, and the chooser does not know whether he has chosen the large quantity or the small one, X itself has an equal chance of being either the large quantity or the small one. X is never just X. Sometimes it is small and sometimes big. The probability of each scenario is "built into" the variable, but never accounted for.
Here in p3 the variable X changes with each trial. Since a box is randomly chosen each time, and the chooser does not know whether he has chosen the large quantity or the small one, X itself has an equal chance of being either the large quantity or the small one. X is never just X. Sometimes it is small and sometimes big. The probability of each scenario is "built into" the variable, but never accounted for.
QUOTE (Mr. Revenge+ Dec 1 2005, 02:43 AM)
p4) Assuming you always switch, you will end up with 2X half the time, and 1/2X half the time.
In p4, the variable X still changes each time, just as in p3. There is even more confusion, however, because the usage of 2X and 1/2X gives the illusion that the probability of either scenario is in fact being taken into account. But X is still poorly defined. If the large box is chosen, then X refers to the small quantity. If the small box is chosen, then X refers to the large quantity.
The correct approach is to assign a separate variable to the two different values. X = the small value. Y = the large value. We also include the definition that Y = 2X. No more ambiguity. Although X and Y can change from one scenario to the other, we can always be certain of their relationship, mathematically. Now we can say one box always contains X and one always contains Y. The probability of getting either X or Y when choosing one box is .5. Once a box has been chosen, we still have no idea whether we have gotten X or Y. Which means the other box is still unknown, and if we switch, there remains a .5 probability that it is smaller, and .5 that it is larger.
So if you always stay with the first box, you will make an average amount M, where M = .5X + .5Y = .5X + .5*2*X = 1.5X.
But if you throw that choice away and go with the second box, your odds remain exactly the same that the second box will be either X or Y. So again M = .5X + .5Y = .5X + .5*2*X = 1.5X.
The only way the odds change for the second box is if you know the exact value of X and Y when you make the choice. This way, choosing a box reveals what is in the other, and no "chance" is left, because the probability is exactly 1 that the second box contains X if the first contains Y, and vice versa. But this should go without saying.
So.... Hope that clears everything up. It really is simpler than a lot of people here made it seem.
In p4, the variable X still changes each time, just as in p3. There is even more confusion, however, because the usage of 2X and 1/2X gives the illusion that the probability of either scenario is in fact being taken into account. But X is still poorly defined. If the large box is chosen, then X refers to the small quantity. If the small box is chosen, then X refers to the large quantity.
The correct approach is to assign a separate variable to the two different values. X = the small value. Y = the large value. We also include the definition that Y = 2X. No more ambiguity. Although X and Y can change from one scenario to the other, we can always be certain of their relationship, mathematically. Now we can say one box always contains X and one always contains Y. The probability of getting either X or Y when choosing one box is .5. Once a box has been chosen, we still have no idea whether we have gotten X or Y. Which means the other box is still unknown, and if we switch, there remains a .5 probability that it is smaller, and .5 that it is larger.
So if you always stay with the first box, you will make an average amount M, where M = .5X + .5Y = .5X + .5*2*X = 1.5X.
But if you throw that choice away and go with the second box, your odds remain exactly the same that the second box will be either X or Y. So again M = .5X + .5Y = .5X + .5*2*X = 1.5X.
The only way the odds change for the second box is if you know the exact value of X and Y when you make the choice. This way, choosing a box reveals what is in the other, and no "chance" is left, because the probability is exactly 1 that the second box contains X if the first contains Y, and vice versa. But this should go without saying.
So.... Hope that clears everything up. It really is simpler than a lot of people here made it seem.
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