Your responses help me formulate my thoughts more clearly. Each time I think about how to reply to your comments, my thinking about the situation becomes clearer.
TEP = 2-envelope paradox
TEP logic = “half the time I trade I’ll get twice my amount and half the time I trade I’ll get half my amount”
Here’s the problem subtly built into the TEP logic that disqualifies it from being used to calculate the odds of trading every time in the 2-envelope game:
When combined with the rules of the 2-envelope game, the 2-envelope paradox logic has a positive feedback loop in it that makes it inapplicable to that game. Together, they demand that, in each game, both twice and half of every amount drawn are included in the set of numbers drawn. This means that every game must have an infinite number of numbers drawn. Since no game has an infinite number of numbers drawn, the 2-envelope paradox logic can’t be used for any of these games.
That feedback loop isn’t apparent when thinking about only one pair of envelopes. It seems like the TEP logic that is applied to one pair can be generalized to an entire game, but it can’t be generalized because of the feedback loop. This becomes apparent when trying to devise the multiple trials that must exist in every game.
Here is the problem stated another way:
Since every 2-envelope game must have a highest amount that doesn’t have twice its amount in the game, the TEP logic doesn’t work in the 2-envelope game.
The above analysis leads to this:
The TEP logic requires multiple trials by definition. The unresolvable difficulty with applying the TEP logic to the 2-envelope game is deciding what the multiple trials will look like.
I will explain and exemplify this problem in the following paragraphs. I’ll make my argument in 3 different ways: operationally, mathematically, and with language.
First, I’ll illustrate by operationalizing the logic:
First, I’ll play a typical game. I decide the first amount using a random number generator. Suppose it is 20D. I fill one envelope with 20D. Then I flip a coin. If it’s heads, I fill the other envelope with twice the amount and, if it’s tails, I fill the other envelope with half the amount. It’s tails, so I fill the other envelope with 10D. I offer the two envelopes to the player. She chooses one envelope and opens it. It contains 10D. She has decided to use the “trade every time” strategy, so she trades. She gets the 20D envelope and wins 10D. For the next trial, I randomly generate 54D, which I put into the first envelope. The coin flip is tails again, so I put 27D into the second envelope. She chooses the 54D envelope and trades. She loses 27D. For that game (2 trials), she loses 17D.
I want to figure out the odds that her strategy is working in the above game, so I repeat it. I decide not to tell her that I’m repeating the same game, so she continues the “trade every time” strategy. I offer her the 10D and 20D envelopes again. She chooses the 10D envelope again and wins 10D. Then I offer the 27D and 54D envelopes to her again. This time, she chooses the 27D envelope and wins 27D by trading. In the second game she wins 37D.
She wins 20D average for the first 2 trials.
If I repeat this game many times, her winnings will be close to zero. I’ll think that her strategy doesn’t work. Mathematically, I can prove that it doesn’t work.
Notice that the TEP logic doesn’t apply to this game because she’ll never be able to trade the 54D envelope for a 108D envelope.
Next, I’ll try to play the game in a way that I can use the TEP logic. I’ll start the same way. The first round will happen exactly the same that it did above. But I don’t know what to do about the second round. What amounts do I use? If I use the same amounts, she’ll never draw a 108D envelope. She needs to be able to draw a 108D envelope for the TEP logic to work, but there is no way to make that happen. I could fill 2 envelopes with 54D and 108D and 1) give her the 54D envelope, but that’s not allowed by the rules of the game, or 2) let her choose one of them. But if she chooses the 108D envelope, I’m back where I started. I could keep adding envelopes with different amounts, but then she’d never pick all of the amounts more than once, and might not even pick some of the amounts. I could make the game just one trial and keep giving her the same amount over and over again while I change the amount in the other envelope. At least then the TEP logic would work, but then I’d be playing a different game. In fact, the TEP logic will never work for the 2-envelope game using the proper rules.
Next, I’ll approach this mathematically:
In this section, I’ll apply statistics to a single envelope, a trial of 2 envelopes, and then multiple trials of 2 envelopes. The trick is to keep track of the variables in the problem. Once that is done correctly, I think that the paradox resolves. Let’s see if you agree.
Here’s a review: Odds are calculated by averaging an outcome over repeated events/trials. If I have only one variable, then I average the values of that variable over repeated events. The value repeats according to rules of the game. If I have multiple variables, I hold all but one constant and average the outcome of the variable amount over multiple trials. That average can then be treated as a constant (it doesn’t change assuming that the variables are independent, which they are in the TEP), and it can be averaged over some other variable. And so on.
Single envelope: In a single-envelope “game,” I’ll fill an envelope with a random amount and the player chooses it. How much will the player make on average? The only variable is the amount put into the envelope, which changes with each repetition, so I calculate the amount won by averaging the amounts put into the envelope. The result will depend on how the rules of the game dictate that I fill the envelope. Not much of a game, so let’s introduce a second envelope.
Two envelopes: Let’s look at the variables using 2 envelopes. They are: 1) The amount put into the first envelope, 2) the amount put into the second envelope, and 3) the amount (envelope) chosen. I must keep two of these variables fixed and vary the other one to apply statistics.
First, I’ll keep variable #1 constant by always putting the same amount X into the first envelope. That leaves 2 variables. I’ll hold one of those variables constant and vary the other one (A), and then I’ll switch the one I hold constant (B). This brings up two different scenarios: A) The amount put into the other envelope is fixed and the player can choose either envelope, or B) The amount put into the other envelope varies and the player chooses the same amount (envelope) each time, which must be the first envelope because it’s the only envelope with the same amount in the repeated trials.
The 2 scenarios above describe 2 different games: A) the same 2 envelopes with the same 2 amounts are repeatedly offered to a player who chooses one of them at random. B) A player is given a specific amount and then the other envelope is filled with twice that amount half the time and half that amount half the time. (Note that the TEP logic is consistent with game B but not consistent with game A.)
I can calculate the odds of winning by trading every time in each of the above scenarios like this: A) if the amounts are X and 2X, then I’ll win X when I draw X and lose X when I draw 2X. On average, I’ll win zero. B) In the second scenario, if I have X, I’ll win X half the time and lose 0.5X half the time, so on average I’ll win .5X.
Now that I’ve dealt with those 2 variables, I can vary the last variable - the amount put into the first envelope - in each scenario. I’ll let X be the variable amount put into the first envelope.
A) If 2X is put into the second envelope, the trial will consist of 2 envelopes containing X and 2X. If .5X is put into the second envelope, the trial will consist of 2 envelopes containing X and .5X. These constitute 2 separate, independent trials. Each trial will average out to zero gain by trading every time.
B) In each trial, the envelope chosen will contain X. It will be traded half the time for 2X and half the time for .5X. For every value X, I’ll make, on average, .5X extra by trading every time. The average amount won will be .5 times the average value of X.
Scenarios A and B define specific, different games with specific, different rules. Game A is the 2-envelope game. Game B is the “coin-toss” game I mentioned in the previous post. The TEP logic is consistent with game B, but is not consistent with game A.
Last, I’ll explain the paradox in words.
I’m glad you mentioned analyzing only the 2 envelopes on the table. Suppose I play one trial. I choose 20D. I trade it for the other envelope, which has 10D. I lose 10D. She declares that my strategy of always trading didn’t work. “Look,” she says, “you used your strategy and you lost!” At this point, I begin to talk about multiple trials.
I’ve included this tiny vignette to demonstrate why one trial with 2 envelopes alone cannot be used to figure out and/or decide whether or not a strategy works. Once further trials are discussed, the following logic applies:
The result in a single trial of the 2-envelope game cannot be generalized to multiple trials using the TEP logic. The TEP logic states that the other envelope can contain half my amount or twice my amount, so I’ll win more than I’ll lose on average by trading every time. “On average” and “every time” imply multiple trials. It seems obvious that multiple trials that fit the TEP logic must exist, but they don’t. That’s the trick. Try to figure out how to do the multiple trials necessary for the TEP logic to work and you’ll quickly see that 1) it can’t be done, or 2) to do it the rules of the game must change.
I draw an amount and say, “I’m going to trade every time because the other amount will be twice this amount half the time and half this amount half the time.” She says, “That can’t happen in this trial because there’s only one other envelope. When will that happen?” I say, “In future trials.” She says, “In future trials, you’ll draw a different amount.” I say, “But I’ll trade those different amounts for either twice or half their amounts, too.” She says, “Not in the trial in which you first pick them. You’ll have to wait for yet more future trials for that to happen.” I say, “Yes.” She says, “But in those future trials, you’ll choose yet different amounts.” I say, “Yes, but I’ll trade those different amounts for either twice or half their amounts, too.” She says, “Not in the trial in which you first pick them. You’ll have to wait for even more future trials for that to happen.” I say, “Yes.” She says, “But in those future trials, you’ll pick yet different numbers. You’re building up a lot of numbers that you have to choose again, but you’ll keep picking different numbers before you choose those first numbers again. You’ll get behind in choosing the same numbers multiple times, and you’ll never catch up. Even if you do this for all eternity, you’ll still never catch up. You’ll just keep choosing new numbers. Your strategy doesn’t work for this game.” She’s right.
The TEP logic cannot apply to the many possible 2-envelope games in which the same number never repeats. Therefore, the TEP logic cannot be used to determine the odds in the 2-envelope problem.
The TEP logic cannot be used in any game in which some amount doesn’t have twice its amount. Since the highest amount in every game will never have twice its amount in that game, the TEP logic doesn’t ever apply to any game.
The logic applied to a game must follow the rules of the game. If something doesn’t repeat, then logic that depends on that something repeating is likely to lead to a faulty conclusion. Or, if something repeats only a finite number of times, then logic that demands that it repeat an infinite number of times will also likely lead to a faulty conclusion. Specifically, the TEP logic implies that the same amount will be drawn repeatedly and the amounts in the other envelopes will be twice the starting amount half the time and half the starting amount half the time. If I choose 20D, the paradox logic implies that, during multiple trials of an actual game, I will, on average, definitely draw 20D multiple times and that the other envelope will definitely contain 40D half the time and 10D half the time on average. In addition, it demands either that I draw no other amounts or that the half/twice logic is true for every amount drawn. Since the rules of the 2-envelope game never allow all of these things to happen, this logic can’t be used for the 2-envelope game. Even with an infinite number of trials, these things won’t happen because I’ll keep drawing new numbers forever.
With multiple trials, 1) if I repeatedly use the same set of paired envelopes, that set will contain a highest amount that won’t have twice its amount available: the TEP won’t work. 2) If I instead repeated use new amounts to try to include both half and twice the value of every amount, then many values will only be drawn once. Instead, new value will be continually drawn. Once again, the TEP logic won’t work. There are no other ways of repeating the trials.
You also mentioned that the lowest amount isn’t ever paired with half its amount. It is true that the lowest amount will always win, just as the highest amount will always lose. That’s the symmetry in the logic. But the amount always lost (the highest amount) will always be more than the amount always won (the lowest amount). Then you might wonder why I don’t always lose when trading every time. That’s because some middle amounts will win on average. In any game repeated many times, the amounts won and the amounts lost will be equal.
Notice that the amounts can be drawn from any set of numbers: finite, infinite, bounded, or unbounded.