TheNightfly
Joined: May 21, 2010
• Posts: 480
August 15th, 2010 at 4:07:40 PM permalink
Hmmm...

For those who may have read through my novelette from a couple of pages ago, I profoundly apologize. The Great and Powerful Wiz was right (of course) and yes, I fell into the trap. As a matter of fact, I pushed myself into it by beginning my analysis with SCENARIO #1 in which I imagined an example where the first envelope has no monetary amount and the two envelopes have a slip of paper that say "50%" and "200%" respectively. If that were the case (which it is not) then yes, there would be a positive EV however, since it is not the case I simply painted myself into a corner by believing my own flawed logic.

As there are only 2 envelopes there is a 50/50 chance of selecting the large one and a 50/50 chance of selecting the small one. If the two envelopes contain \$50 and \$100 respectively then once you've chosen the \$100 envelope there is no possibility of this being the smaller of the two. Even though you still don't know which one you've chosen it doesn't change the fact that you've chosen the larger of the two. You may think that you have a chance to increase your total by trading envelopes but this is simply not the case. Once you've chosen the larger of the two a trade will only reduce your total by 50%. For the time when you select the smaller of the two a trade will increase your total by 100%.

The flawed logic presumes that once you've chosen an envelope that the other envelope could be a higher OR lower value than the one you hold. This cannot be the case as it must only be one or the other... there is not the possibility or potential for it to be both.

The Bill Gates example is a good way to show this and it has brought me to my "A-HA" moment.

Assuming that the two envelopes do hold \$50 and \$100, the median amount is \$75 yet there is no \$75 envelope. Selecting the \$50 and switching will result mean that you will end up with \$100. Selecting the \$100 envelope and switching will mean that you end up with \$50. Regardless of what the percentage gain or loss may be in relation to the original amount, the actual monetary gain or loss will be the same in both cases - \$50. Although you might believe that once you've looked in your envelope and found \$100 that you may have selected the smaller amount and have a chance of increasing your amount by \$100, this is not the case. You can only go down and never up, even thought you don't know this fact. Therefore, to believe that you have a chance of increasing your amount by 100% doesn't make it so.... you cannot as the \$200 envelope you are hoping to find does not exist.

Looking at the problem from the perspective of the person who is selecting the envelopes, you may convince yourself that whichever envelope you've chosen will always have a 50/50 chance of containing the smaller amount but you would be wrong. Only 50% of the time will you have a chance of increasing the amount you hold and that will ONLY happen when you have chosen the envelope containing the smaller amount. So, in the long run you will go up \$50 half of the time and go down \$50 half of the time. As \$50 is 100% of \$50, the increase or decrease will always be the same amount and the EV BETWEEN the two envelopes will be \$75 (the total of \$150 divided by two, the number of envelopes). In the long run, the person who is handing out the envelopes will lose an average of \$75 per envelope but never at \$75 per envelope but only in \$50 and \$100 increments.

The fallacy in logic is when you (or I) suggest that as this \$50 gain is 100% of \$50 and the \$50 loss is only 50% of \$100 that there is a percentage gain in the combination of the two of 50% (which divided by the two choices gives us an average increase of 25% per choice) or and EV of 125%. In actual fact the gain or loss will always be exactly \$50 (which is 100% of \$50) and in the long run you will end up making an average of \$75 per selection, which is exactly \$25 more than \$50 and \$25 less than \$100. In other words, don't change the envelope because the first choice you made is just as likely to hold the larger amount as it is the smaller and changing will not make it any more likely that you will then be holding the larger amount.

I know I've gone on much too long once again but I tend to use 10 words where 5 will do and this is the result.

Thanks to Dorothy for providing this puzzle and thanks Wiz for gently pointing out my error and leading me to the truth.
Happiness is underrated
konceptum
Joined: Mar 25, 2010
• Posts: 790
August 15th, 2010 at 8:40:28 PM permalink
I love being wrong. Thanks Wiz. Makes me actually do some thinking.

What if you think of it like this. There are two envelopes. Envelope A contains \$x, and envelope B contains \$2x.

Once you have picked an envelope, you are able to take that amount. The question is whether or not you should switch. So let's look at it from the point of view of what you gain or what you lose.

If you originally picked A, and don't switch, your gain is \$0.
If you originally picked A, and do switch, your gain is \$x.
If you originally picked B, and don't switch, your gain is \$0.
If you originally picked B, and do switch, your gain is -\$x, or a loss of \$x.

If you consider that all four of those combinations are possible, then the expected gain is:
(1/4)*0 + (1/4)*x + (1/4)*0 + (1/4)*(-x) = 0.

If you say that the person should always switch envelopes, then the expected gain still comes out to \$0. So, I guess what I'm saying is that the reason the problem doesn't seem to make sense is that we're thinking in terms of the amount that is originally found in the first envelope, when maybe all we should be looking at is what is the possible gain by switching.
Wizard

Joined: Oct 14, 2009
• Posts: 23117
August 15th, 2010 at 9:01:01 PM permalink
Quote: konceptum

...maybe all we should be looking at is what is the possible gain by switching.

You're welcome on the help. As they say, the best arguments are the ones you know you lost, because at least you learned something.

Your argument in favor of no gain by switching is completely correct. Doc argued the same thing earlier. However, how would you explain where the logical flaw is in averaging a gain of 100% and a loss of 50%, and coming up with an expected profit of 25% by switching? THAT is the puzzle.
It's not whether you win or lose; it's whether or not you had a good bet.
mkl654321
Joined: Aug 8, 2010
• Posts: 3412
August 15th, 2010 at 9:15:35 PM permalink
Quote: Wizard

You're welcome on the help. As they say, the best arguments are the ones you know you lost, because at least you learned something.

Your argument in favor of no gain by switching is completely correct. Doc argued the same thing earlier. However, how would you explain where the logical flaw is in averaging a gain of 100% and a loss of 50%, and coming up with an expected profit of 25% by switching? THAT is the puzzle.

I may have what you're driving at:

The logical flaw is in attempting to average those two numbers at all. Percentages of variables are NOT integers--they are unknowns (or, more precisely, derivatives of unknowns). I think I was skirting the concept when I said that a drop of 50% in the stock market followed by a gain of 50% does NOT restore prices to their former levels--in other words, the percentages are not equivalent, even though numerically equal, because they refer to different quantities. In the envelope game, a gain of 100% over the SMALLER amount is exactly equivalent (as an absolute number) to a loss of 50% from the GREATER amount. To average the numeric values of the percentages is to create GIGO.
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
Kelmo
Joined: Aug 15, 2010
• Posts: 85
August 15th, 2010 at 10:13:05 PM permalink
I would like to thank everyone for this discussion. I wrote the quote in the initial thread, which was posted by a avery smart friend of mine. He is helping me to get my head around this, but I think the rocks keep getting in the way. The way I though about this problem was to consider that the inital equation 0.5*x/2 + 0.5*2x actualy belong to two seperate possibilites. Whoever sets up the envelopes can choose and combination of values, as long as one is twice the amount as the other. Once we choose the envelope and know its contents, we narrow the possiblities down to two sets:
The total value placed in the envelopes is either 3x or 3x/2. Since we do not know which set we are in at this point, we should weight the expected values accordingly. The total value in both scenarios is 4.5x, 3x represents 2/3 and 3x/2 represents 1/3.
Now, there is a 50% chance that we are in the realm of 3x and the gain is x2-x, weighted by 2/3.
50%(x2-x)/0.666... = 0.75x
There is a 50% chance that we are in the realm of 3x/2, which is a gain of x/2-x, weighted by 1/3.
50%(x/2-x)/0.333... = -0.75x
The fact that we do not know how much money is placed in the envelopes at the outset creates the disparity.

Not sure if this makes sense, but would appreciate the feedback.

K
konceptum
Joined: Mar 25, 2010
• Posts: 790
August 16th, 2010 at 6:00:59 AM permalink
Quote: Wizard

You're welcome on the help. As they say, the best arguments are the ones you know you lost, because at least you learned something.

Your argument in favor of no gain by switching is completely correct. Doc argued the same thing earlier. However, how would you explain where the logical flaw is in averaging a gain of 100% and a loss of 50%, and coming up with an expected profit of 25% by switching? THAT is the puzzle.

I know that a major part of my problem in understanding the logic behind this puzzle is that I would always choose to switch envelopes. My reasoning for this is based loosely upon Pascal's Wager, and the fact that there is nothing to lose.

Without looking at the contents of the envelope, it would seem obvious that the envelope you choose makes no difference, and switching the envelopes makes no difference either. The expected value of the envelope you chose is (3/2)x. The expected value of the other envelope is also (3/2)x. This makes sense since (3/2)x + (3/2)x = 3x, which is the total value of the monies placed in the envelope. Since each envelope has the same expected value, then it doesn't matter which one you choose, and switching them doesn't make sense.

The logical flaw, I would think, comes about in the false concept that choosing the other envelope can result in "doubling" or "halving" your money. The fact remains that you do not yet have the money, until you decide whether to keep the envelope you have, or take the other envelope. And until you make the decision, the expected value of the envelope in hand is (3/2)x and the expected value of the envelope you could switch to is also (3/2)x.

There is an analogy to this in the stock market or housing market. People will say that they have lost a lot of money in their stocks, but until they actually sell those stocks, they haven't lost anything. The paper value of those stocks is meaningless, until it is actually turned into cash. The same is true with the envelopes. Until you turn one envelope in and cash it in, it has no value, other than an expected value of (3/2)x.

Even if you know the value of the envelope, it doesn't change. Let's say you buy a stock at \$40 per share and the stock drops to \$20 per share. You can't claim that you have lost money, because you haven't turned that stock in for cash yet. Thus, you haven't lost anything until you do so. Similarly for the envelope. If you open it and it contains \$100, you can't say that you have a 50% chance of doubling your money or a 50% of halving your money, because it isn't yet your money. It won't be your money unless you decide to keep it, in which case you will neither double nor halve it. Further, if you switch envelopes, you will get whatever is in that envelope. Since you never had the \$100 to begin with, you neither doubled nor halved by switching.

All that being said, I would still utilize Pascal's Wager and switch no matter what.
DorothyGale
Joined: Nov 23, 2009
• Posts: 639
August 16th, 2010 at 7:03:14 AM permalink
Quote: Kelmo

I wrote the quote in the initial thread ...

Welcome to the board Kelmo. Glad to have you. As you can tell, your question has raised a lot of very interesting discussion.

As for smart, well, at least I can sing ...

--Dorothy
"Who would have thought a good little girl like you could destroy my beautiful wickedness!"
Kelmo
Joined: Aug 15, 2010
• Posts: 85
August 16th, 2010 at 8:26:23 AM permalink
Quote: DorothyGale

Welcome to the board Kelmo. Glad to have you. As you can tell, your question has raised a lot of very interesting discussion.

As for smart, well, at least I can sing ...

--Dorothy

Thanks for introducing me to this website. It's gold!

K
Nareed
Joined: Nov 11, 2009
• Posts: 11413
August 16th, 2010 at 9:59:31 AM permalink
Here's a stupid and simple math problem to occupy three seconds of your day:

Using the numerals 1,1,1,1 (once each) and any mathematical symbols (any number of symbols any number of times), what's the largest number you can express?

Examples:

1+1+1+1=4

1+1*11=22

11*11=121

11^11 (eleven to the eleventh power)
Donald Trump is a fucking criminal
Wizard

Joined: Oct 14, 2009