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Topic: HARD: ENVELOPE GAMBLE (Read 9811 times) |
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Grimbal
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Re: HARD: ENVELOPE GAMBLE
« Reply #50 on: Jun 1st, 2004, 4:15pm » |
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Let's be practical.
There are some reasons, whether valid or not, that tell us that we should switch. There are some other reasons, valid or not, that tell us it doesn't matter. So, just in case, we should switch. Then get the money.
But in my opinion, there is no distribution such that for every X, P(X=high|picked=X) = P(X=low|picked=X), which is implied by the problem.
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rmsgrey
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Re: HARD: ENVELOPE GAMBLE
« Reply #51 on: Jun 2nd, 2004, 6:38am » |
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on Jun 1st, 2004, 4:15pm, Grimbal wrote:Let's be practical.
There are some reasons, whether valid or not, that tell us that we should switch. There are some other reasons, valid or not, that tell us it doesn't matter. So, just in case, we should switch. Then get the money.
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Very practical. As long as no-one comes up with a reason, whether valid or not, to say we lose by switching...
As for the variation with 1-2 and 2-4, take the 2 or switch, it isn't obviously equivalent to the original problem because in the original, the amounts in the two envelopes is determined first, then you pick an envelope. In the variation, you pick an envelope, and that determines the value.
Obviously, for any finite set of possible values for X, where the envelopes contain values X and 2X, the expected return for always switching without seeing the contents of the envelopes (or equivalently with no knowledge of the set of values X can take) must be the same as that for always keeping your first pick. Looking at a uniform distribution for X over the first n powers of 2, (2,4,...,2n) the expected return for never switching is 3(2n-1)/n, while the expected return from always switching after seeing the contents of the envelope except when opening 2n is (7*2n-6)/2n - a gain of 2n-1/n. On the other hand, I'm sure Icarus would have harsh words for me if I then tried taking the limit as n goes to infinity  - after all, a uniform distribution across an infinite number of possibilities has probability 0 of any given event, which means you're looking for some sort of integral rather than a summation, and the profit lies in the strange effects for the maximum value, which means the value of the integral is determined by a discontinuity, which gets even harder to cope with, and oh no, Three Hands has gone cross-eyed...
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Three Hands
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Re: HARD: ENVELOPE GAMBLE
« Reply #52 on: Jun 2nd, 2004, 10:38am » |
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on Jun 2nd, 2004, 6:38am, rmsgrey wrote:
after all, a uniform distribution across an infinite number of possibilities has probability 0 of any given event, which means you're looking for some sort of integral rather than a summation, and the profit lies in the strange effects for the maximum value, which means the value of the integral is determined by a discontinuity, which gets even harder to cope with, and oh no, Three Hands has gone cross-eyed... |
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Dammit - and done in such a way that I was struggling to keep them uncrossed to begin with 
After all, I'm a philosopher, not a mathemawhatsit... 
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pedronunezmd
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Re: HARD: ENVELOPE GAMBLE
« Reply #53 on: Jun 3rd, 2004, 9:42pm » |
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First of all, I do not think this is a distribution problem. The fact of the matter is that, in the riddle, the gamemaster has already chosen the x and 2x that go into each envelope, and it does not state this was done randomly.
The problem with calculating the expected payoff using the rationale stated in the riddle is as follows:
The envelopes are already set when the game begins. If you choose an envelope and call it's value X, then when you calculate the expected payoff by switching, you are calculating the average amount you will expect per game if: you are given these same 2 envelopes already chosen, and you are handed the same envelope each time which is the one you have now called X, and you switch to the other each time.
The problem is you are now saying that there is a P(A) = chance you are playing the (x/2,x) game, and P(B) = chance you are playing the (x,2x) game, and that P(A) = P(b) = 50%.
Therefore, payoff = P(A)(x/2) + P(B)(2x) = 1.25x
But this is not correct. P(A) does not equal P(B), in fact, since the envelopes are already set, either P(A) = 1 or P(B) = 1, you just don't know which. Just because you don't know, does not mean you call them 50% chance each. You have already playing either the A game or the B game, you just don't know which it is.
So actually the payoff = either x/2 or 2x, but you don't know which, when choosing to do the calculations by calling your envelope X.
Please consider this analogy: I have a coin, I flip it, then hide the results from you. Now I offer the following: if it is heads, you win $2, if it is tails, you win nothing. If the orientation of the coin never changes, the expected payoff is either $0 each time you play, or $2 each time you play. If I charged you $1 to play the game forever, you would either win an infinite amount of money, or lose an infinite amount of money. However, this is different from the game where, each time you play, I flip the coin anew, in which case the expected payoff is $1, and if you play an infinite number of times, you would in the long run break even having paid $1 per game.
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pedronunezmd
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Re: HARD: ENVELOPE GAMBLE
« Reply #54 on: Jun 3rd, 2004, 9:51pm » |
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I must also note that what the riddle is asking you to do is calculate the expected payoff this scenario.
Suppose the gamemaster has a set of an infinite number of envelopes laying on the table, and for any envelope x, he always remembers which other envelope is the corresponding "2x" envelope. He always picks 2 envelopes such that one contains twice the other.
On the first round, he hands you 2 envelopes. Before opening them, you label one envelope as "X" literally, with a marker or pen and give both envelopes back. You then agree to play the game an infinite number of times.
Each time he hands you envelopes and neither one is marked with your X, you reject playing the game and ask for a new pair. He is picking the pairs of envelope at random. For now, assume he can do this.
However, when you get 2 envelopes, and one has an "X" on it, you agree to play the game, and now you always switch to the other envelope!
In this version of the game, the chance that you are now playing the (x/2,x) versus the (x,2x) game is equal and set to 50%, and the expected payoff is as calculated in the riddle.
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towr
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Re: HARD: ENVELOPE GAMBLE
« Reply #55 on: Jun 4th, 2004, 3:37am » |
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on Jun 3rd, 2004, 9:42pm, pedronunezmd wrote:| But this is not correct. P(A) does not equal P(B) |
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Depends on how you chose which envelope you pick, if you label the envolopes A and B, then flip a coin and choise envelope A if it's heads and B if it's tails, then P(A) = P(B).
Choosing the 2x envelope is equally likely as choosing the x envelope.
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Grimbal
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Re: HARD: ENVELOPE GAMBLE
« Reply #56 on: Jun 4th, 2004, 8:23am » |
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on Jun 3rd, 2004, 9:42pm, pedronunezmd wrote:| First of all, I do not think this is a distribution problem. The fact of the matter is that, in the riddle, the gamemaster has already chosen the x and 2x that go into each envelope, and it does not state this was done randomly. |
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But you don't know the numbers. So you have to consider the game master as a random process. If you don't, you can not analyze the problem probabilistically.
Suppose the game master gives you one envelope with some amount int it, what is the probability that it is over $1?
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| « Last Edit: Jun 4th, 2004, 8:26am by Grimbal » |
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pedronunezmd
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Re: HARD: ENVELOPE GAMBLE
« Reply #57 on: Jun 4th, 2004, 10:03am » |
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Quote:Depends on how you chose which envelope you pick, if you label the envolopes A and B, then flip a coin and choise envelope A if it's heads and B if it's tails, then P(A) = P(B).
Choosing the 2x envelope is equally likely as choosing the x envelope. |
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Actually, I defined "A" as "you are playing the (x/2,x) game" and "B" as "you are playing the (x,2x) game", not as labels for each of the two envelopes. Whichever envelope you choose, you are stating the amount inside it is "x" per the riddles suggestion. You know that if the envelope you have has contains x dollars, the other envelope has either x/2 dollars, if it turns out you are playing the A game, and 2x dollars, if it turns out you are playing the B game. But the P(you are playing the "A" game) = P(A) is either 1 or 0, because the envelopes have already been set. It is incorrect, given the way envelopes' amounts have already been set, to state that P(A) = .5
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rmsgrey
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Re: HARD: ENVELOPE GAMBLE
« Reply #58 on: Jun 4th, 2004, 2:14pm » |
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But which game you're playing isn't determined in advance - you determine which game you're playing by your choice of envelope. The contents of the envelopes are set as, say, M and N with N=2M, but which you label as X depends on which you pick (with even chance each way) so you have an equal chance of being in the X,2X game or the X,X/2 game. The catch is that X in the first case is M, and in the second N, so the latter is worth twice the former.
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pedronunezmd
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Re: HARD: ENVELOPE GAMBLE
« Reply #59 on: Jun 4th, 2004, 4:49pm » |
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Quote:| But which game you're playing isn't determined in advance - you determine which game you're playing by your choice of envelope. |
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I'm not saying that which game you're playing is determined in advance. However, if you choose to determine the expected payoff of the game, once you have the 2 envelopes in front of you, by calling one "x", then "which game you are playing" is now fixed, as far as for the calculations of your expected payoff are concerned. You have specified and labeled the one envelope as "x", and you have either picked the x such that you are playing the A game or the B game, for purposes of calculating the expected payoff. One of the two games is now "the actual game" and the other is irrelevant. You don't know which, but you can't now assume that P(A) = P(B) at this point in the calculations, which is why the reasoning stated in the riddle is wrong.
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rmsgrey
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Re: HARD: ENVELOPE GAMBLE
« Reply #60 on: Jun 5th, 2004, 7:14am » |
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But I have no information about whether I'm playing game A or game B. Assuming that my choice of which envelope to label 'X' is independent of which has more money, then if I played the game a million times, about 500,000 of them would be game A and the remainder game B.
At the point where you face the choice of switching or not, then you are in one of two universes, one where P(A)=1, and one where P(B)=1, but you have no way of knowing which universe you are in, so, from the information you have, P(P(A)=1,P(B)=0) = P(P(A)=0,P(B)=1) = 0.5, in other words, P(A)=P(B)=0.5
Quote:| Please consider this analogy: I have a coin, I flip it, then hide the results from you. Now I offer the following: if it is heads, you win $2, if it is tails, you win nothing. If the orientation of the coin never changes, the expected payoff is either $0 each time you play, or $2 each time you play. If I charged you $1 to play the game forever, you would either win an infinite amount of money, or lose an infinite amount of money. However, this is different from the game where, each time you play, I flip the coin anew, in which case the expected payoff is $1, and if you play an infinite number of times, you would in the long run break even having paid $1 per game. |
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While the range of possible outcomes is different between the two coin toss games, in both of them, the expected return before the coin is revealed is 0. After the first round, the expected return per round for the first game is fixed at one of +1 or -1, but until you know which, the two are equally likely to be the case, so the expected return is still 0.
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asterix
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Re: HARD: ENVELOPE GAMBLE
« Reply #61 on: Jun 5th, 2004, 11:40am » |
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Suppose the two numbers in the envelope were 2 and 4, and you played the game a million times, but with short term memory loss, so you never figured out that the numbers never changed. Half the time you'd draw the 2, figure the other envelope must be 1 or 4, so you'd switch, and always win. The rest of the time you'd draw the 4, figure the other contains either 2 or 8, so you'd switch, and always lose. Final total is simply the average of 2 and 4, and there is no advantage whatsoever to switching.
Now go back to the original game where the numbers are random from an infinite distribution, and play the game an infinite number of times. There will be an infinite number of games where there is a 2 and a 4. And you'll draw each an equal number of times, and have the same results as I described, a simple average with no advantage to switching. There will also be an infinite number of times when you'll get every other possible pair, and in every case there will be no advantage to switching (unless you get a number that cannot be divided by 2 and you can determine logically that it's the low number).
So instead of looking at a 4 and saying, the other number is either a 2 or an 8; the average is 5, so I should switch. You should look at that 4 and say, "half the time this 4 is part of a 2,4 pair, and I'll always switch, so I'll average 3 and gain no advantage; the rest of the time this 4 is part of a 4,8 pair, and I'll always switch so I'll average 6 and gain no advantage."
In the long run there's no advantage to switching. But in the short run (playing the game just once) the average is not possible; you either gain or lose. And since this short run does not allow you to take into consideration the times when you draw 1/2 the actual number and win a little by switching or the times when you draw 2x the actual number and lose a lot by switching, a single game is going to give the impression of gaining by the switch (and I'd still switch).
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pedronunezmd
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Re: HARD: ENVELOPE GAMBLE
« Reply #62 on: Jun 6th, 2004, 7:38am » |
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In response to Rmsgrey's recent post:
Let me address the coin flip game first. There is a fundamental difference between calculating the expected outcome before the coin has been flipped and after the coin has been flipped.
Prior to the coin being flipped, the P(head) = P(tails) = .5 and you can use that in calculating the expected payoff. If you play the game one million times, starting before the coin flip each time, then approx 50% of the time you win with heads, and the other 50% of the time you lose with tails.
After the coin is flipped, the either P(heads) = 1 or P(tails) = 1, you just don't know which. If you play the game one million times, once the coin has been flipped, the either you are going to win one million times in a row, or you are going to lose one million times in a row. The expected outcome of the game is more accurately stated as "either you win $1 each time or you lose $1 each time", not "you break even".
Second, to address the first part of your post:
Quote:| But I have no information about whether I'm playing game A or game B. Assuming that my choice of which envelope to label 'X' is independent of which has more money, then if I played the game a million times, about 500,000 of them would be game A and the remainder game B. |
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When you choose to calculate your expected return by labeling one envelope as "x", you are calculating the average return if, given the same 2 envelopes, you always pick out the envelope that you labelled "x". If you played the game 1 million times, then 1 million times you are getting the envelope labeled "x".
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pedronunezmd
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Re: HARD: ENVELOPE GAMBLE
« Reply #63 on: Jun 6th, 2004, 7:49am » |
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In response to Asterix's recent post:
What you have posted, at least in the first 80% of your post, is another way to restate the solution first proposed by S.Owen in the very first post of this thread, which is the "correct" way to figure out the expected payoff of the game.
I hope that everyone agrees that the expected gain by switching is zero, as stated by S.Owen. If not, my offer still stands to challenge anyone who disagrees with recreating the game as a computer program.
The challenge of this riddle, however, is not to state what the "correct" way to calculate the gain from switching is, but to show why the method proposed in the riddle is "not correct". For example, if I showed you a proof that 0=1, then it is not complete to just say "wrong! I know that 0 does not equal 1", but it would be complete to say "here is the flaw in your proof that 0=1".
This is what I'm trying to do. The reasoning in the riddle is wrong. To classify this riddle as "solved", we will need to show why the reasoning is wrong.
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towr
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Re: HARD: ENVELOPE GAMBLE
« Reply #64 on: Jun 6th, 2004, 11:40pm » |
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on Jun 6th, 2004, 7:38am, pedronunezmd wrote:| The expected outcome of the game is more accurately stated as "either you win $1 each time or you lose $1 each time", not "you break even". |
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This is not true, because 'expected outcome' is one single value, not a statement. It's [sum] p(X=a)*a
If P(X=x)=1, then the expected outcome is x, if P(X=y)=1, then the expected outcome is. If X is either x or y, then you have to look at the chance of either state being the case, and here P(X=x|x or y) = 1/2 and P(X=y|x or y) = 1/2 (as said before), and therefor the expected outcome is (x+y)/2
Quote:| This is what I'm trying to do. The reasoning in the riddle is wrong. To classify this riddle as "solved", we will need to show why the reasoning is wrong. |
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It has allready been shown several times why it is wrong, and what mistake is made.
The X from (X/2, X) isn't the same X as in (X, 2X), because it's a given that the two pairs are the same (because it's the same envelope)
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carpao
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Re: HARD: ENVELOPE GAMBLE
« Reply #65 on: Jun 7th, 2004, 9:13am » |
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on Jun 6th, 2004, 7:49am, pedronunezmd wrote: If not, my offer still stands to challenge anyone who disagrees with recreating the game as a computer program.
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If someone is interested ... avery somple Java Program...
Code:
public class gambler {
static public void main (String[] args)
{
int i;
int[] j;
int x;
int switchTotal = 0;
int noSwitchTotal = 0;
int c0=0;
int c1=1;
j = new int[2];
for (i=0;i<10000;i++)
{
//extract a number j between
j[0]= (int) (StrictMath.random()*10000);
j[1]= j[0]*2;
//choose randomly one envelope
x= (int)(StrictMath.random()+0.5);
if (x==0)
c0++;
else
c1++;
//what about switching?
switchTotal += j[(x+1)%2];
//what about not switching?
noSwitchTotal += j[x];
}
System.out.println("NoSwitch:"+noSwitchTotal);
System.out.println("Switch:"+switchTotal);
System.out.println("X = 0:"+c0);
System.out.println("X = 1:"+c1);
}
}
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JocK
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Re: HARD: ENVELOPE GAMBLE
« Reply #67 on: Jul 23rd, 2004, 4:20pm » |
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Before opening the envelope, the expectation value for the payout is infinity (it has to be, or else P(X=high|picked=X) = P(X=low|picked=X) can not be true for all X). So, the other envelope contains twice infinity or half infinity. Swapping brings no advantage because infinity is infinity, regardless whether you multiply by 2 or by 1/2.
The paradox disappears when using an a-priori distribution with a finite expectation value. (E.g. n dollars is put in the first envelope with probability (1/2)^n, (n=1, 2, ...), and double the amount of the money in the first envelope is put in the second envelope.)
J CK
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| « Last Edit: Jul 24th, 2004, 10:57am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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JocK
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Re: HARD: ENVELOPE GAMBLE
« Reply #68 on: Jul 23rd, 2004, 4:45pm » |
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on Jul 23rd, 2004, 4:20pm, JocK wrote:| The paradox disappears when using an a-priori distribution with a finite expectation value. |
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This brings me to a variant of the two-envelope paradox that nicely stresses that switching paradoxes only occur when the expected pay-out for both envelopes is infinite.:
The St. Petersburg Two-Envelope Paradox
You are presented with two envelopes. You are told that each of them contains an amount determined by the following procedure, performed separately for each envelope: starting with 1$ a coin was flipped until it came up heads. Each time a tail showed the amount was doubled. So, if the first nth trials showed tail, but head came up in the (n+1)th trial, an amount of 2^n was put into the envelope. This procedure is performed separately for each envelope. You randomly select an envelope, and are offered the options of keeping this envelope or switching to the other. What should you do?
You might reason as follows:
- Before opening the envelopes, the expected value in each is infinite.
- For any x, if you knew that your envelope contained x, then the expected value in the other would still be infinite.
- So for all x, if you knew that yours contained x, you would have an expected gain in switching to the other envelope.
- So you should switch. But this seems clearly wrong, as your information about both envelopes is symmetrical and following this line of thought you should keep switching between both envelopes.
JCK
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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mattian
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Re: HARD: ENVELOPE GAMBLE
« Reply #69 on: Jul 26th, 2004, 11:29am » |
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There is no strategic benefit to switching - there is a 50%chance of receiving the higher of the two sums of money - regardless of your decision.
Proof:
Show me an argument that illustrates the benefit in switching, and I'll show you the same argument that illustrates the benefit in not switching.
For example: Suppose by switching, you stand to gain double the amount that's in your hand, with the risk of losing half of what's in your hand. Clearly this must be a benefit because the prospective gain is higher than the potential loss. Right?
Wrong. If we reword it and say that after selecting an envelope, you have the option of keeping the envelope that you have or defaulting to the other envelope, we could make the same argument: By keeping the envelope we have, we stand to walk away with twice the amount that's in the other envelope, and we stand to miss out on half the amount that's in the other envelope. This time it is clearly beneficial to keep what we have.
Since these two arguments - though contradictory - are balanced, the odds of either choice HAS TO BE 50% and therefore there must be a flaw in the analysis which shows any bias, one way or the other.
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Padzok
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Re: HARD: ENVELOPE GAMBLE
« Reply #70 on: Sep 5th, 2004, 3:42pm » |
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Obviously, I agree with S Owens first post.
I have 3 points about why there is an apparent paradox.
I think my points are so similar to what has already been said, that I would not bother were it not for the fact that some posters are saying that the paradox has not been properly explained. So I will give it a go.
Point 1
Let's say the 2 envelopes are A and B, and A is the one you hold.
If you wish to analyse the problem by defining that one of the envelopes contains x, then you have the following possibilities.
1. A=x, B=0.5x.
2. A=x, B=2x.
3. A=0.5x, B=x
4. A=2x, B=x
So there are 4 possibilities, all equally possible.
If you swap, your gain, G, is
G = 0.25 (-0.5x) + 0.25 (x) + 0.25 (0.5x) +0.25 (-x) =0
The probability equation in the riddle ignores possibilities 3 and 4.
Point 2
If it were true (which it isnt) that swapping produced an increase in your chances of winning, then it would be true that swapping back would also increase your chances etc.
That the original riddle has its definitions wrong is easliy shown (and has been already in this thread).
Following the riddle's analysis you could say to yourself, I have Y in this envelope, the other envelope has 50% chance of having 0.5Y and 50% of having 2Y and so I will swap.
But if you remembered your last argument, pre-swap, you would think the other envelope has X. My envelope therefore has 50% chance of having 0.5X and 50% of having 2X and so I will not swap.
We do not know what X and Y are and so the analysis is meaningless.
(I believe, Mattian, this is the point you have just made).
Point 3
It will always be worth swapping after opening the envelope (if we ignore real word considerations) because your potential gain is twice your potential loss.
The reason point 1 no longer applies is that now we can ascribe a particular value to x.
Say the envelope contains $100.
We can have x=$100, and use possibilities 1 and 2 from above and ignore 3 and 4.
The reaon point 2 no longer applies is obvious. Once we have swapped, that is the end of the game. We know for certain if we have got $50 or $200. The question of whether it is better to swap again does not arise.
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BNC
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Re: HARD: ENVELOPE GAMBLE
« Reply #71 on: Sep 5th, 2004, 5:54pm » |
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on Sep 5th, 2004, 3:42pm, Padzok wrote:
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Point 3
It will always be worth swapping after opening the envelope (if we ignore real word considerations) because your potential gain is twice your potential loss.
The reason point 1 no longer applies is that now we can ascribe a particular value to x.
Say the envelope contains $100.
We can have x=$100, and use possibilities 1 and 2 from above and ignore 3 and 4.
The reaon point 2 no longer applies is obvious. Once we have swapped, that is the end of the game. We know for certain if we have got $50 or $200. The question of whether it is better to swap again does not arise.
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But don't you see the problem? Opening the envelope did not give any "important" (or relevant) information. So you found $100... your decesion would not have changed if it were $50, $200, or any other value. The sum of money in the envelope does not change the outcome, thus (IMHO) does not matter.
In other words, instead of opening the envelope, why not guess a sum (say $100) -- as it wouldn't matter anyway -- and go on from there?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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rmsgrey
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Re: HARD: ENVELOPE GAMBLE
« Reply #72 on: Sep 6th, 2004, 7:31am » |
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on Sep 5th, 2004, 5:54pm, BNC wrote:| Opening the envelope did not give any "important" (or relevant) information. |
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It depends what information you consider "important". To borrow language from Quantum Mechanics, before opening either envelope, both exist in a (symmetric) superposition of states. Once you open one envelope, you collapse the waveforms of both envelopes: the opened one to a single, definite value; the other to a superposition of two values.
If you prefer, you can look at it this way. Before opening the envelope, you have two absolutely symmetrical envelopes. By opening the envelope, you destroy the symmetry - one envelope is open, and the other closed. The open envelope has an absolutely known quantity of money inside, while the other has a chance of being worth more (dependent upon the distribution of possible values for the envelope contents).
As (yet) another way of looking at it: if you consider finite uniform distributions, you get a maximum possible value, for which you won't switch, but whatever else you open, you will switch. As the distribution gets wider, the chance of opening the value that causes you not to switch heads to 0, so the chance of not switching also heads that way, but, for any finite range of values, there's still the possibility of opening the counter-example.
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Padzok
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Re: HARD: ENVELOPE GAMBLE
« Reply #73 on: Sep 6th, 2004, 12:17pm » |
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on Sep 5th, 2004, 5:54pm, BNC wrote:
Opening the envelope did not give any "important" (or relevant) information.
It gives you 2 pieces of information.
1. Is that one of the envelopes contains $100
2. Is that the envelope you hold contains $100
If all you knew was that one of the envelopes contained $100, then the possibilities would be:
You have $100, other is $50
You have $100, other is $200
You have $50, other is $100
You have $200, other is $100
You are right, BNC, to say that there is nothing special about $100. You could write out the 4 possibilities for any positive integer.
The general case is as I had in last post,
1. A=x, B=0.5x.
2. A=x, B=2x.
3. A=0.5x, B=x
4. A=2x, B=x
where A is the envelope you hold.
Alternatively you can swap and then hold envelope B (without opening) and the possibilities remain unchanged.
The formulation given for working out the probabilities in the original post ignores the symmetry, and the problems which arise from that are as described in my last post (and by mattian in the post before that).
As rmsgrey has said, opening one envelope destroys the symmetry.
What is an acceptable solution to the riddle, bearing in mind that the riddle is apparently to show the flaw in the logic of the original argument?
I think several different explanations have been offered. There is no reason that more than one might not be correct.
Has anybody shown that ALL the arguments explaining the flaw in the riddle's logic are wrong? If not, then has the apparent paradox been satisfactorily explained?
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towr
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Re: HARD: ENVELOPE GAMBLE
« Reply #74 on: Sep 7th, 2004, 3:09am » |
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on Sep 5th, 2004, 3:42pm, Padzok wrote:Point 3
It will always be worth swapping after opening the envelope (if we ignore real word considerations) because your potential gain is twice your potential loss. |
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Let's say one of the envelopes contains $50 dollar, and the other a $100. And let's say you don't know that, but only know that one envelope contains twice as much as the other.
If you pick the envelope with $50 there is no actual chance the other contains $25 , nor if you pick the one with $100 is there any actual chance the other contains $200.
Your potential gain, and your potential loss are both $50 dollar. So opening the envelope doesn't make switching a good option all of a sudden. It's just as irrelevant as before.
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| « Last Edit: Sep 7th, 2004, 3:13am by towr » |
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