TheNightfly
Joined: May 21, 2010
• Posts: 480
August 14th, 2010 at 4:18:15 PM permalink
Hello all -

Here is the "solution" to this fun little conundrum. Keep in mind that when given your choice, you know that you will always end up with a number above zero as your final result. This is important to note because if your options were "double or nothing" then it would be much easier to make your decision. I'll use a few scenarios to show why this is so. I'm also going to begin the first 2 scenarios by having you hold an envelope (the one you first choose) that is worth \$100. We are dealing with finite numbers here (regardless of what some may suggest) and \$100 is a nice, round number. One more point that is crucial here is that your benefactor can afford to part with any amount of money. To say that the amount of the first envelope will give some indication of what may be in the second envelope is not so... at least according to the facts introduced in Dorothy's original post. (Sorry, I know it's long...)

SCENARIO #1 (which assumes that your options are "double or nothing". I know that this is not the case in this particular problem but follow me and it should help you see the actual problem in the right light.)

- When you choose your first envelope and BEFORE you look inside, your benefactor tells you that it is worth \$100 to you. He goes on to say that neither envelope contains any money but that each one has a slip of paper. Upon one piece of paper is the figure "0%" and on the other, "200%". You now have two options; take the \$100 or switch envelopes. It is easy to see that if you were to switch an even amount of times (whether you do it twice or one million times) your EV is precisely 100%. You also know that if you change envelopes that there is PRECISELY a 50% chance of the envelope being the "0%" option and PRECISELY a 50% chance of it being the "200%" option. If you select the "0%" you end up with nothing and if you select the "200%" you end up with \$200. Do this an even amount of times and your EV remains at 100% of your \$100. Therefore it makes no sense to change envelopes any amount of times as you already have made \$100 in the bargain. IF this were the problem at hand it is easy to see why you should not change your envelope UNLESS you are enough of a gambler to take the chance of doubling your money at the risk losing everything. (Sounds like a good ol' coin flip to me with a double or nothing outcome... an even money proposition - no more, no less.) The point here is that switching envelopes does not have a +EV therefore you should not switch... the best you could do in the long run is to break even and the worst you could do in one trial is to lose everything.

SCENARIO #2 (Now we get closer to the real problem at hand using the facts introduced by Dorothy in her original post; A switch will either double your amount or cost you 50% of that amount)

- When you choose your first envelope and BEFORE you look inside, your benefactor tells you that it is worth \$100 to you. He goes on to say that neither envelope contains any money but that each one has a slip of paper. Upon one piece of paper is the figure "50%" and on the other, "200%". You now have two options; take the \$100 or switch envelopes. It is easy to see that if you were to switch an even amount of times (whether you do it twice or one million times) your EV is precisely 125%. THIS IS BECAUSE ALTHOUGH YOUR ODDS OF CHOOSING THE LARGER AMOUNT ARE THE SAME AS CHOOSING THE SMALLER AMOUNT (precisely 50%), THE LARGER AMOUNT INCREASES YOUR TOTAL BY 200% WHERE THE SMALLER AMOUNT ONLY REDUCES YOUR TOTAL BY ONLY 50%.

***** Let's look at this by putting it into practice. If I am offered this opportunity once and I change envelopes then I will either end up with \$200 or \$50. The median amount is \$125. Therefore if I were to be offered this opportunity twice and I changed both times there's a good chance that I would select the smaller amount once and the larger amount once. I'd end up with \$50 one time and \$200 the next for a total of \$250. This would give me an average gain of \$125 between the two trials. Over an infinte number of trials I'd simply get closer to this +EV of 125%. If I had kept the \$100 both times I'd end up with \$200 for an average of a \$100 gain between the two trials. You can see that my EV for switching is 125% and for staying my EV is 100%. (Akin to the casino taking commission on a winning Pai Gow bet... when you lose they take 100% of your money and when you win they take 5% of your money. In this scenario when I choose the small amount I lose only 50% of my money but when I choose the larger amount I win 100% of my money) Now THERE'S a coin flip I'd take every time! *****

Do this any amount of times and your EV remains at 125% of your \$100. Therefore it makes sense to change envelopes ONCE as you have a +EV. As I said earlier, we are dealing with a finite amount and to suggest that swapping envelopes more than once would give you a +EV for every switch is a fallacy. IF YOU SWITCHED, AFTER YOU'VE SWICHED THE FIRST TIME, SWITCHING AGAIN CAN ONLY PRODUCE THE OPPOSITE RESULT AS THERE ARE ONLY 2 OPTIONS. The point here is that switching envelopes has a +EV therefore you should switch - but only once as switching again would put you back where you began. The odds of choosing the greater or lesser amount will always be 50/50 but half of the time you will choose the larger amount and INCREASE YOUR TOTAL BY TWICE AS MUCH as you'd lose if you chose the lesser amount.

SCENARIO #3 (And now we use Dorothy's actual problem with the facts as she presented them. We will use real amounts and find that again we are dealing with finite figures and options.)

- You are offered two envelopes. You are told that one envelope contains an amount that is precisely twice what is in the other. (It therefore stands to reason the one envelope contains an amount the is precisely 50% of what is in the other.) When you choose your first envelope and look inside, you see a \$100 bill. Your benefactor tells you that you may keep the \$100 or change envelopes. It is easy to see (by looking at SCENARIO #2) that if you were to switch an even amount of times (whether you do it twice or one million times) your EV is precisely 125%. You also know that if you change envelopes that there is PRECISELY a 50% chance of the new envelope having \$50 and PRECISELY a 50% chance of it containing \$200. At this point you realize that your benefactor is willing to part with up to \$100 OR up to \$200.

***** Let's look at this by putting it into practice. There are only two possibilities; the two envelopes contain either \$50 and \$100 or they contain \$100 and \$200... there are no other options. If I'm holding an envelope that contains \$100 then I know I have the larger or the smaller amount and there is an EQUAL CHANCE of either probability. I know that by changing I will either lose \$50 or gain \$100. In SCENARIO #2 we have already shown that I should change envelopes as my EV is 125%. It has also been shown that changing more than once is not of ANY benefit to me as I'd simply be back to where I began, with my original envelope with \$100. As we know what's in the original envelope we chose it's rather silly to think that by changing back to this original envelope after having changed envelopes once the amount will somehow have miraculously increased... nope, you just got the same original envelope with the same \$100 you originally had. *****

This is more a logic puzzle than a math problem. The math tells you that you have a +EV by changing envelopes only one time. Once you realize that there are finite options then the math becomes moot and logic has to dictate your actions. The "flaw in the logic" is that it is easy to look at the odds (50/50) and the EV (125%) and wonder how they can be different... and then spend hours trying to figure out why they are. It's obvious that they are different because they define two wholly separate facts. One (the odds) defines the chances of choosing a higher or lower amount and the the other (the EV) defines what the result will be when you've made your choice. I suggest that by changing an EVEN AMOUNT of times will give you a +EV but the fact is that you will always have a +EV. I simply wished to let you work out the numbers for yourself by using 2 (or any even amount) of trials as a simple proof. Dorothy's original formula holds true but as we are dealing with finite numbers and a finite number of possibilities then the EV is also finite at 125%.

THE END.
Happiness is underrated
DorothyGale
Joined: Nov 23, 2009
• Posts: 639
August 14th, 2010 at 4:37:38 PM permalink
Quote: TheNightfly

Hello all - Here is the "solution" to this fun little conundrum... [infinitely long explanation ... let's just say it's the first ordinal larger than countably infinite ] ... THE END.

Zounds!

Surely someone can help me scrape my exploded brain off the wall ...

--Dorothy
"Who would have thought a good little girl like you could destroy my beautiful wickedness!"
Wizard
Joined: Oct 14, 2009
• Posts: 23117
August 14th, 2010 at 9:07:49 PM permalink
Quote: TheNightfly

Here is the "solution" to this fun little conundrum...

That is a fancy explanation about why the EV=\$125 argument is allegedly correct. However, it doesn't make common sense that it is. Suppose Bill Gates is hosting the contest and the two envelops contain \$50 and \$100. He then offers the contest to thousands of people individually. By your argument everybody would switch. However, would switching benefit the group as a whole? The half that picked \$50 would switch to \$100, and the half that switched to \$100 would end up with \$50. Overall there would be an average increase in wealth of 25% per person of (ave(100%,-50%)), but the total amount of money won would stay the same. Sorry, but you feel into the trap.
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 14th, 2010 at 11:20:21 PM permalink
Quote: Wizard

That is a fancy explanation about why the EV=\$125 argument is allegedly correct. However, it doesn't make common sense that it is. Suppose Bill Gates is hosting the contest and the two envelops contain \$50 and \$100. He then offers the contest to thousands of people individually. By your argument everybody would would switch. However, would switching benefit the group as a whole? The half that picked \$50 would switch to \$100, and the half that switched to \$100 would end up with \$50. Overall there would be an average increase in wealth of 25% per person of (ave(100%,-50%)), but the total amount of money won would stay the same. Sorry, but you feel into the trap.

Well, I think this is a trap that most people fall into. After all, we keep stuffing politicians into office whose ideological shibboleth is that taking money from some people and giving it to other people ACTUALLY INCREASES THE TOTAL AMOUNT OF SAID MONEY. Taxation, like alchemy, creates something out of nothing! (I beg your pardon--not "taxation", "revenue creation". My bad.) But since we keep sending the same clowns back over and over, and in REALLY delusional times, such as recently, we send a whole gang of them, I think that is manifest proof that we as a society have no idea how the monetary system/the economy works.

P.S. I just put a dollar bill in one desk drawer and two dollar bills in another. I then proceeded to alternately open and close each drawer. To my total surprise, after only fifteen minutes of doing that, the top desk drawer contained a billion dollars! Maybe you're wrong after all, Wizard.

P.P.S. After I discovered the billion dollars, I got greedy, reasoning that the bottom drawer must now contain 1.25 billion dollars. So I closed the top drawer and opened the bottom--but that act caused the amount in the top drawer to increase by 312.5 million. The increased weight was too much: the desk crashed through the floor, and I fell into the resulting hole. I woke up in the emergency room, and when I got back to the wreckage of my house, I found that someone had taken all the money. Live and learn!
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
Doc
Joined: Feb 27, 2010
• Posts: 7086
August 15th, 2010 at 4:52:47 AM permalink
Quote: weaselman

Quote: Doc

Here I believe I have properly calculated expected values using the 50% factors as they should be used. The calculations show that for any value of X (and 2X), players of the game cannot reasonably expect to gain or lose on average by either swapping or not swapping, although if you DO swap, you will either gain or lose X.

This is the (yes, properly calculated) expected value BEFORE the envelope is opened. Once you know the amount in one of the envelopes, the expectations of swapping and not swapping change, but to find out the new values, we need to know the new probabilities.

I personally think, your previous explanation was perfect (and, way better than this one :)). #2 was right to the point.

weaselman: I missed this post of yours last night. Sorry. Didn't mean to ignore it. I think you and I are pretty much in agreement on this problem. I do disagree with some aspects of your comments on my post.

Things I did do:
(1) State a 50% probability of being lucky or unlucky in your first random guess of an envelope.
(2) Calculate expected values of the net effects of swapping and not swapping envelopes.

Things I did not do:
(1) State any probability for any particular amount of money being in the mystery envelope.
(2) Calculate an expected value of the contents of the mystery envelope.

I contend that the probabilities I stated are correct both before and after you open the envelope. After you open it, either you were or you were not lucky (you still don't know which), but the probability didn't change.

I also contend that the expected values that I calculated (effect of swapping or not swapping) do not change after you open the envelope. They both stay at zero. The amount of money that you might gain or lose by swapping changes once you have a dollar value to work with, but the expected value of swapping is still zero. So I guess we disagree on that point.

We both agree (I think) that the probabilities of an amount of money in the mystery envelope in unknown and unknowable without some inside information. Perhaps those probabilities do change when you open an envelope, but I made no claim about them.

As for which of my explanations (#1, #2, or #3) is best, I think that depends upon the initial viewpoint of the person I am explaining it to. That is why I tried to show it different ways.

Now, I am off to Mississipi for a week of practical experiments on a subject of interest to most of us. Hope you folks have a lot of fun on this forum while I am gone.
Wizard
Joined: Oct 14, 2009
• Posts: 23117
August 15th, 2010 at 5:17:03 AM permalink
Quote: mkl654321

After all, we keep stuffing politicians into office...

Your rant on politics is so significantly off topic that it should have been made into a separate thread. Please copy and paste it into a separate thread if you wish this to remain on the board.
It's not whether you win or lose; it's whether or not you had a good bet.
Joined: Nov 30, 2009
• Posts: 550
August 15th, 2010 at 9:02:47 AM permalink
Quote: Wizard

Suppose Bill Gates is hosting the contest and the two envelops contain \$50 and \$100. He then offers the contest to thousands of people individually. By your argument everybody would switch. However, would switching benefit the group as a whole? The half that picked \$50 would switch to \$100, and the half that switched to \$100 would end up with \$50. Overall there would be an average increase in wealth of 25% per person of (ave(100%,-50%)), but the total amount of money won would stay the same. Sorry, but you feel into the trap.

I agree with the Wizard here. Suppose I were Satan and offered you those three envelopes, but instead called the prizes: Hell, Purgatory, and Heaven. Then you pick the envelope for Purgatory. Aww shucks. I offer you the opportunity to choose one or the other envelopes. Do you really risk it because the expected value is slightly positive? Afterall, the expected value is a concept that goes hand in hand with the Central Limit Theorum, as you run through multiple iterations to approach an "average" value. In this instance, you only get 1 shot at heaven or hell.

Now what if I were really sneaky. What if I made this offer with only 2 envelopes. You choose 1 = Purgatory, and I tell you the second envelope contains either Hell or Heaven. You would assume your scenario is purely binary with a 50% probability. But I'm Satan, and I've made this offer many times to many souls. And I know, That for every 1 envelopes filled with Heaven, and 3 envelopes are filled with Hell. Your 50% probability isn't looking too hot right now. Bahahaha.
weaselman
Joined: Jul 11, 2010
• Posts: 2349
August 15th, 2010 at 9:07:26 AM permalink
Quote: Doc

I contend that the probabilities I stated are correct both before and after you open the envelope. After you open it, either you were or you were not lucky (you still don't know which), but the probability didn't change.

I also contend that the expected values that I calculated (effect of swapping or not swapping) do not change after you open the envelope. They both stay at zero.

This is where we disagree. Suppose, the person stuffing the envelopes, follows this rule - pick a uniformly distributed number between 0 and 1, multiply it by \$1000, and put the amount into the first envelope, then either double the amount or divide it by two with a 50/50 chance, and put the result into the second envelope.
Suppose, after opening the first envelope, you see \$1026 in it. What is the probability, that the second envelope has more money? It is 0. What if you see \$950? The probability of the other one having more is 1/3 now. What if it's \$400? Now it's 50%.
This illustrates, how the probabilities and expectations change depending on the first amount unveiled. I did pick an arbitrary distribution, but it is not important - the same kind of mechanics happens for any distribution. It's much like the blackjack player's strategy changing depending on the dealer's up cards.

Do you really risk it because the expected value is slightly positive?

That depends on the monetary values you chose to assign to your soul going to each of those places. I suspect, for most people, the expected value will not be positive. If it is positive however, then yes, mathematics says that you should do it, all the other, non-mathematical reasons some might state for not doing it, are irrelevant (not because they are unimportant, but because they are unique for each individual, and cannot be generalized or formalized). I personally think, that if you do decide to not swap even though the expectation is positive, it simply means that the numerical values you have assigned to each of the outcomes are incorrect in your case (provided, of course, that you understand well what exactly you are doing, and all the math behind it, and are not simply acting on an impulse).

Quote:

Now what if I were really sneaky. What if I made this offer with only 2 envelopes. You choose 1 = Purgatory, and I tell you the second envelope contains either Hell or Heaven. You would assume your scenario is purely binary with a 50% probability.

Just like seeing a dinosaurus on the street :) Either you see it, or you do not. 50% :)
"When two people always agree one of them is unnecessary"
DorothyGale
Joined: Nov 23, 2009
• Posts: 639
August 15th, 2010 at 9:32:19 AM permalink
Quote: weaselman

pick a uniformly distributed number between 0 and 1 ...

Off topic, but this is not possible. There is no pseudo-random generation algorithm for picking a "random number" from any continuum. Indeed, except for a very small number of values between 0 and 1, there is no way of even describing most of these numbers algorithmically. Moreover, it is impossible to pick an integer at random. Why? because, with countably infinite many exceptions, individual integers are too large to describe in any language.

Densely yours in NP world,

--Dorothy
"Who would have thought a good little girl like you could destroy my beautiful wickedness!"
weaselman
Joined: Jul 11, 2010
• Posts: 2349
August 15th, 2010 at 9:39:18 AM permalink
Quote: DorothyGale

Off topic, but this is not possible. There is no pseudo-random generation algorithm for picking a "random number" from any continuum.

Who said "continuum"? Or pseudo-random for that matter? :)
Quote:

Indeed, except for a very small number of values between 0 and 1, there is no way of even describing most of these numbers algorithmically.

This is true

Quote:

because, with countably infinite many exceptions, individual integers are too large to describe in any language.

This is not true. A set of all possible finite combinations of words is isomoprphic to the set of all integers (or rationals).
"When two people always agree one of them is unnecessary"