Two envelope (sub)-problem / related problem [closed]

Question

This old thread began along the lines I had been thinking about recently:

Two envelope problem revisited

Now, I realise this is old hat to many, but please bear with / be gentle if possible... Essentially my experience of playing with the two envelope puzzle is that it generates sub-puzzles which sometimes (or even often) loop back to the main one. So, with that in mind, here's a sub puzzle or perhaps a related puzzle which I think is of interest in itself and might shed some light on aspects of the two envelope puzzle.

In the standard presentation of the two envelope puzzle, the way in which the amounts are selected and the order in which the envelopes are filled is not specified. In the sub/related puzzle I present below (Scenario 3), the the way in which the amounts are selected, the order in which the envelopes are filled and the fact that one envelope is opened, are specified. What is more, in addition to getting to open an envelope, the player is told how they were filled and which one was filled first.

But consider first Scenarios 1 and 2. In Scenario 1 the player is given 10 dollars. He is then told that he can keep that or he can have 20 or 5 depending on the toss of a coin. We presume he's willing to play the odds and accordingly he takes up the offer as there's an expectation of gain. In Scenario 2 it is as in Scenario 1 except that we tell him that we have already performed the coin flip and put the resulting amount in a sealed envelope. No tricks, so it is 20 or 5 again. The player goes for it again, and rightly so.

Now, Scenario 3, the main event. In this case he is given an envelope and told it contains an unspecified amount of money in the form of an IOU. That amount is in there already. It was chosen by the master of ceremonies who literally just 'thought of a number. The player is then told that another envelope we are prepared to offer him instead contains an IOU for an amount created by tossing a coin to double or halve the value in the first envelope. Which one should he take home with him? Following the logic of Scenario 2, he should swap. There is a definite amount in the first one one, and because of the way the amount in the second one was created it has an expected value of 1.25 times the first one.

The player then opens the second envelope. He sees an amount of money on the IOU slip. It occurs to him that the other one, the one he started with, must contain either half or double the amount he is looking at. It also occurs to him that if the odds of this are 50-50 he seems to have chosen the wrong envelope. But he can't have because the logic of Scenarios 1, 2 and 3 lead him to the firm conclusion that the second envelope was the better one.

So, he reasons, the odds can't be 50-50. In fact, he reasons, the chance that the first envelope contains twice as much as the second envelope must be less than 50%, by quite a way. He does a quick calculation and reckons that for there to be an expectation that the second envelope would contain 1.25 times the amount in the first, the likelihood that the first one would contain twice as much as the second one must be a mere 20%, with an 80% chance it contain half as much (suppose he sees 250 on opening. The logic of the way the envelopes were created seems to mean he should think that the expected value of the other envelope is 200. It can't be 200 of course, it can only be 125 or 500. So he supposes that there is a 4/5 chance that the original sum was 125 and a 1/5 chance that it was 500).

Is the player wrong in thinking that his interpretation of Scenario 3 follows from Scenarios 1 and 2? And / or is he wrong in his calculation of the likelihood of the original values? Or is something else going on?

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Comment

The attraction of this sub problem for me is that the way the envelopes are filled constrains some of the possibilities while at the same time amplifying some of the (apparent?) paradoxes of the two envelope puzzle: if mathematical calculations (distributions) are to be used to resolve the matter they have an even bigger job to do than in the general two envelope puzzle in the sense that in this sub puzzle there is a clear (and valid?) expectation of gain one way (swapping), which only seems to work if larger values for the original amount are a lot less likely than smaller ones.

Answer 1

The two envelope paradox, and the above variations, play on the semantic difficulties of the English language. The paradox, as typically stated, has a hidden condition that is not apparent using common English. The statement, “You can either double or half your money” makes it sound like both of those choices currently exist. In fact, once you’ve chosen an envelope, one of those choices goes away. You just don’t know which one it was. More complex language is needed to convey this information. The following examples may clarify this for you:

First, an example in which the language is accurate: Suppose I have 3 envelopes, one with some amount (n), another with twice that amount (2n), and the third with half that amount (0.5n). If I give the middle amount (n) envelope to the person, I can truthfully say that they have the choice to either double or half their money by choosing one of the other envelopes because, in fact, both choices exist.

However, if I fill 2 envelopes with 2 amounts, one amount double the other (as in the typical envelope paradox), and give one to the person, I must truthfully say that, by changing envelopes, they may double their money or half their money, but not both. Which will happen is determined by the somewhat hidden condition of which envelope they picked. The following language exposes the hidden condition: “You can double your money if you picked the envelope with the smaller amount (there’s the hidden condition), or you can half your money if you picked the envelope with the larger amount (the condition once again).” More concisely, I’d say, “You can half the larger amount or double the smaller amount.” This makes it clearer that changing envelopes does not increase the chance of getting more money.

Hope that helps.

Answer 2

The paradox is that, since “when I see an amount, the other envelope will contain twice my amount half the time and half my amount half the time” is true, therefore trading every time will win on average, when, in fact, trading every time won’t win on average. The paradox is this contradiction.

To solve the paradox, I must show that the belief that “when I see an amount, the other envelope will contain twice my amount half the time and half my amount half the time” is false. That way, the conclusion that trading every time will win on average cannot be drawn, the contradiction is removed, and the paradox resolves.

The belief that “half the time I’ll lose half the amount and half the time I’ll win twice the amount” implies that that outcome is always true for every amount chosen. If the quoted logic fails for any amount, the logic is deemed false, which makes trading every time suspect, which removes the contradiction and, therefore, the paradox.

In fact, that outcome is never true for every amount chosen because it is never true for the largest amount put into whatever set of envelopes is chosen.

I think that the above paragraphs successfully resolve the paradox.

In your paired example, the logic always fails for the largest amount ([2^k]n at k-max).

For a moment I considered that the only amount in your sequential pairs example that loses money on average is the largest amount. That means that the odds of picking an amount that wins money is greater than the odds of picking an amount that loses money, and the paradox still holds. While the logic is correct as far as it goes, it misses a necessary piece of information: the amount lost in that sole losing choice is equal to the winning amount for all the other amounts combined. Thus, the (weighted) average of losing and winning is still zero, and the paradox doesn’t hold.

If someone says, “I’ll win half the amount half the time and twice the amount the other half of the time,” I’ll say, “That’s not true with every amount that you can choose.” That stops the logic from concluding that, “I’ll win on average by trading every time,” which stops the paradox. For example, if she sees 20D and says, “half the time I’ll win 20D and half the time I’ll lose 10D,” I’ll say, “That’s not true if the person has decided to never use 40D.” She might say, “But I think he’s likely to use 40D, so I’ll trade every time with this amount.” Her word “likely” clearly demonstrates her understanding that trading every time might lose (but she thinks it is unlikely) which then stops the logic from being definitely concluded, which resolves the paradox. She is demonstrating a different strategy that I have previously described.

Your wrote, “Some philosophers say that the solution to the puzzle is getting the player to take this viewpoint and ignore everything else as illusory.” I agree. For me, the problem with this approach is that it doesn’t address the reason that the paradox is wrong. It just shows that the conclusion is wrong, but not why. For me, this is no small issue. At different times during my education, I approached professors with questions about a problem. I’d explain my logic and they’d explain the correct logic. Sometimes, the correct logic didn’t explain the problem with my logic. Their approach taught me how to approach the problem correctly, but I didn’t learn what was wrong with my logic. I might ask, but they’d often just defer to the correct answer. When I was able use the correct answer to eventually understand why my logic was incorrect, I then understood the problem and the correct logic much better, and in the future, with different but similar problems that were more complex, I’d avoid the same faulty logic. In fairness, explaining the correct logic is much easier and more time efficient, and trying to understand the twisted logic that sometimes arises when people are learning new and complex math can be very difficult and time consuming. So, I understand why I sometimes was told the right answer without addressing my logic. Still, I think that effort put into explaining faulty logic (and not just giving the correct answer) can have significant benefits.

Answer 3

A game consists of a finite number of paired envelopes. In every game, some envelope will contain an amount that doesn’t ever, in that game, have a paired envelope with twice its amount. Therefore, the idea that “twice the amount will exist half the time” with any amount chosen is false, and the paradox is resolved.

For instance, if I open an envelope and see 20D, I cannot correctly assume that the other envelope will contain 40D half the time. If 20D is the highest amount in that game, then a 40D envelope will never exist in that game. This logic applies to any amount seen.

An envelope such as I just described will be present in every game. The player will never know when she will pick it.

In every game, the amount lost when choosing the largest amounts (that have no paired envelope with twice their amounts) exactly equals the amount won on average when trading the lower amount envelopes that do have paired envelopes with twice their amounts. Therefore, in any game, the player who trades every amount seen will win, on average, exactly as much as she loses.

That, I think, is the heart of the faulty logic that leads to the paradox.

Answer 4

The logic in your statement, “It is indeed all about doubling the smaller amount or halving the larger amount” is not apparent from the original language, and that’s why the paradox arises. Specifically, the original language includes the “doubling” or “halving,” but doesn’t include the “smaller” or “larger” amount language. In that way, the original language gives incomplete information about the problem to be solved, and solving the problem with only that incomplete information leads to the wrong conclusion. We say that this is a paradox because it seems clear that trading envelopes won’t increase the odds of getting more money, but logic that is based on the incomplete language suggests that trading will increase the odds. Original language: “I can either double or halve my amount by trading.” Correct, inclusive language,” I can either double my amount -- if I have the smaller amount, or I can halve my amount -- if I have the larger amount, by trading.” The original ambiguously unclear language leads to the paradox. The accurate, inclusive language makes the decision obvious, as your statement implies. That’s why I think that this is really a problem of English semantics and not math. Once English is correctly applied to the problem, the math, and then the correct answer, is obvious indeed.

The tricky language for me is the idea that, with one envelope in hand, I have both the choice to double or halve my money. This seems intuitively correct. However, it also seems intuitively correct to say that both of those choices don’t exist for the amount in that particular envelope. “Can I double or halve my money by trading?” “Sure. Well, it depends. For that envelope, one of those choices doesn’t exist. I don’t know which choice doesn’t exist for that envelope. I’d have to look inside to find that out. So, I don’t know which choice doesn’t exist, but I know that, for that envelope, one of those choices doesn’t exist.” “But it could double or halve, couldn’t it?” “Well, seems like it. But…”

Aaarrrgghhhhh!!!!!

Addendum 1:

I still think that the paradox arises from ambiguity in the language. Here’s how this happens:

Odds for any experiment can be empirically estimated using repeated trials. I want to set up the actual experiment so we know what is repeated. You’ll see that after looking at one envelope’s amount, there are 2 different things that can be repeated that lead to different outcomes. Since the same language seems to be able to be interpreted in both ways, the same language can seem to result in 2 outcomes that contradict.

The problem begins the same for each case. You are given the choice of one of 2 envelopes. You know that one envelope has twice the amount of money as the other envelope. You choose one and see that it has 20D inside. It’s what happens next that matters:

1: Repeatedly give the same 2 envelopes. Then, on the next try, you get to choose from the same 2 envelopes. Maybe you get the 20D again, or maybe you get the 40D. You repeat this process multiple times, always employing some strategy to see if you can increase your odds of ending up with the 40D envelope. The strategy of changing every time doesn’t work.

That’s the way this problem is typically conceived by people who show that changing envelopes after seeing 20D doesn’t increase the odds.

2: Repeatedly see the same amount of money. Then, on all subsequent tries the envelope you choose has 20D. In this case, what changes is the amount in the other envelope: it has either 10D or 40D. If the chance of the other envelope having 10D or 40D is the same (which will depend on the exact method for filling them), then always trading will increase the odds of making more money.

That’s the way that wins when always trading.

Addendum 2

I think that the above must be taken into consideration with my prior discussion that both choices don’t exist. Here’s that discussion after seeing the amount in the chosen envelope and also using the above ideas.

Suppose I fill the envelopes with 20D and 40D. You pick one, see 20D, and then state, “the other envelope has either 10D or 40D.” I’d correctly say, “No, it doesn’t,” because it can’t have 10D because neither envelope had 10D to begin with. Just because you don’t know that doesn’t give you the choice of 10D. In other words, you are wrong if you think the other envelope can have 10D in it. You might say to yourself, “I think it can have either 10D or 40D, but it can’t. One of those choices is not available to me. I just don’t know which one it is, but I could find out by doing this a few times. Until I do that, there’s no way for me to know whether or not trading will make me more money.” It’s the “doing it a few times” part that helps me see that both choices don’t actually exist.

This is similar to my argument that the amount in the other envelope depends on whether I chose the larger or smaller amount (the hidden conditional), and until I know that, I don’t know enough to decide whether or not to trade. “I see 20D but I don’t know if this is the smaller or larger amount. Until I know whether I chose the larger or smaller amount, I can’t know what the choice is for the other envelope, so I can’t know if trading will help me or not.”

This is both fun and frustrating to try and understand and then explain. ☺

Addendum 3

Even better: After you get the envelope with 20D, ask how the envelopes are being filled. The answer will determine whether or not you can improve your odds beyond 50-50. So, I ask, “How are you deciding what amounts to put into these envelopes?” You say, 1) “I’m not telling you,” or don’t answer. I assert: There is no way to know what percent of the time there will be 10D or 40D in the other envelope, so there is no way to determine when to trade, 2) “I put 20D in one and then half the time I put 10D in the other and the other half of the time I put 40D in the other,” or “I put 20D in one and then I put in either half the amount or twice that amount randomly into the other envelope.” I assert that trading every time will increase the odds of winning more money, 3) “I put 20D into one and then some percent of the time I’ll put 10D into the other and some other percent of the time I’ll put 40D into the other. I’m not telling you what the percent’s are.” I assert that there is no way to determine when to trade.

The typical assumption from your language is that there will be half the amount half the time and twice the amount half the time, in which case trading will increase the average amount of money acquired. But that assumption depends on the method by which the envelopes are being filled. If the method isn’t known, then I can’t make that assumption. If I know the method, then I can form a strategy or know that there is no strategy as I discuss above.

Answer 5

The setup: Fill one envelope with some amount, then pair that envelope with each of two other envelopes, one having half and the other twice the amount of the first envelope. I’ll call the amount put into the first envelope the “starting amount.” Using this method, there are 2 ways to proceed:

1) Pick starting amounts from a finite range. In this case, a possible starting amount X will be separately paired with 2X and .5X. For each pair of envelopes, the player will have an equal chance of choosing either the larger or smaller amount. That means that, on average, the player will choose the larger amount half the time and the smaller amount half the time from every combination of amounts. Consequently, by trading every time, the player will win, on average, the same amount that is lost for each pair. So, trading every time doesn’t improve the odds of increasing your money.

This seems counterintuitive when thinking about an actual amount like 20D because it will likely be traded half the time for 10D and half the time for 40D, so on average the player will make money when she sees 20D. In fact, the player will never lose money, and will usually win money, by trading any amount equal to or less than the highest starting amount. That’s because, on average, the player will make twice the amount once but often lose only half the amount once, and at most twice. However, for amounts that are greater than the starting amounts, the player will always lose when trading every time. For example, suppose I use starting amounts of 10, 20, 30, and 40 dollars. Some envelopes will contain 5 (half of 10), 15 (half of 30), 60 (twice 30), and 80 (twice 40) dollars. If I always trade, I will always lose when I choose a 60 or 80-dollar envelope because neither is paired with twice its amount, and I will always win when choosing a 5, 10, or 15-dollar envelope because none of these is paired with half their amounts. For the 20D and 40D amounts, I will win twice the amount half the time and lose half the time half the time for an overall gain. For the 30D amount, I’ll trade for 60D once and for 30D twice on average and will, therefore, break even. On average, the total losses and total gains are equal, which must be true because of the logic of the preceding paragraph. Thus, trading every time confers no advantage. Notice that the logic that leads to the conclusion that trading every time is beneficial, which is, “I’ll win twice the amount half the time and lose only half the amount half the time if I always trade” is incorrect because I’ll never win twice the amount of any amount larger than the largest starting amount.

Just for fun, I did this on a spreadsheet and saw the results for starting amounts 10-50 by 10D increments. Each envelope contained one of these amounts: 5, 10, 15, 20, 25, 30, 40, 50, 60, 80, and 100 dollars. The 60, 80, and 100-dollar envelopes lost money when traded, as expected. All the rest except for 40D made money on average. The 40D amount broke even. Of course, the overall amount gained by always trading was zero.

This logic must hold true for any combination of amounts chosen because of the logic of the paragraph 1) above. Here are several somewhat random pairings of amounts: 10-20, 10-20, 10-20, 20-40, 30-60, 50-100. I will randomly pick each listed amount once and always trade. That is, I’ll pick 10 3x’s and earn 30, pick 20 4x’s and lose 10, 30 1x and earn 30, 40 1x and lose 20, 50 1x and win 50, 60 1x and lose 30, and 100 1x and lose 50. That totals 30-10+30-20+50-30-50=0.

Notice that if you correctly guess the range of starting values used, you can then correctly decide whether or not to trade. For instance, you might correctly guess that the person doing the game will use values up to 20D or more. If you then see 4D, then each possible pair of amounts [(2D and 4D) or (4D and 8D)] will be equally likely, in which case you should trade. I think that this fact contributes to the confusion of the paradox. Upon seeing 20D, it is natural to think that it is in the range of starting amounts, in which case always trading will increase the chance of winning. As I mentioned above, this thinking leads directly to the faulty reasoning that 20D will always be in the range of starting amounts, and thus to the faulty conclusion that trading every time will increase the chance of winning more money.

Conversely, if you think that the chance of twice the amount you see being included in the game is less than 50%, then you should not trade. In other words, if you do this with a person you know, and you choose an envelope with 40D, you might think to yourself, “she would not likely put 80D into an envelope for this game, so I won’t trade.” If you’re right more than half the time, you’ll win money. This logic counteracts the logic of the last paragraph that suggests that trading every time will be beneficial.

2) Pick starting amounts from an infinite range. Using this method, it is difficult to know exactly how the game is actually played. How is a number randomly chosen from an infinite set of numbers? I don’t know. How likely is it that some particular number is chosen? That probability is zero. These difficulties make the problem impossible for me to fully conceptualize. However, when I try to conceptualize the problem, it goes something like this: I choose an envelope and it has 2^50D. I trade and get half the amount. I’m losing. Suppose the next envelope I choose has 2^100D. I trade and get the larger amount. Now I’m winning. The next envelope contains 2^150D. I trade and get the smaller amount. I’m losing again. That process goes on indefinitely, and I can never know whether I’ll win or lose because each subsequent envelope can decide my entire fate.

Answer 6

The player chooses one envelope and looks inside. She sees 20D. She is told that the other envelope has either half or twice that amount, determined randomly (e.g. by a coin toss). The logic that half the time she’ll make 10D and the other half of the time she’ll lose only 5D is based on the faulty idea that this scenario, and only this scenario, will be replayed multiple times and she’ll win the average amount over those repeated trials. But that’s not how the game is played. If I repeat that scenario multiple times, which the phrase “the other half of the time” implies must happen in order to use her logic, I’m then requiring that she always choose the 20D envelope out of 2 envelopes while I vary the amount of the other envelope between 10D and 40D. If instead I’m thinking about some random future trial in which she chooses a 20D envelope and the other envelope contains an amount different than it contained on the first trial (e.g. the first time she chose 20D the other envelope had 10D, but some time in the future when she chose the 20D envelope, the other envelope contained 40D), then I must also consider what happens with all the intervening trials to calculate my odds. In some of those intervening trials, if she chooses the larger amount, then double that amount will never happen. In fact, after seeing the 20D, the idea that half the time the other envelope will contain 40D may be faulty. That’s because for amounts larger than the largest original starting amount are never doubled. Whether or not 20D is larger than the largest starting amount is not known, so whether or not the other envelope could contain 40D is also not known.

My logic in a previous post seems fitting here. Two envelopes are filled with double/half their amounts. When one is drawn, and it is 20D, it is incomplete and therefore incorrect to think that the other envelope could contain either 10D or 40D because that logic presumes the amount in the other envelope is random. It isn’t. Within the context of the whole game, the amount in the other envelope is constrained by 2 conditions: 1) which envelope is chosen, the larger or smaller amount, and 2) the largest starting amount used. The correct logic sounds like this: “The other envelope contains 10D if I’ve chosen the larger amount and 40D if I’ve chosen that smaller amount.” But each of those choices may not occur in the course of the game, depending on what the highest starting amount is, and therefore the odds of winning cannot generally be determined by averaging the outcomes as though those choices were equally likely. When the odds of winning is determined as though those choices are equally likely, the paradox arises.

Here’s another problem with the language sometimes used: Suppose one envelope is filled with some amount of money X and then the other envelope is filled with either half or twice that amount, determined by a coin toss. If the player chooses the envelope with X dollars, it is fair to say that the other envelope was filled with either twice or half that amount randomly by a coin toss. But if the player chooses either other envelope, it is not correct to say that the “other envelope” was randomly filled with either twice or half that amount. The correct language is this: “If you choose one of the envelopes, the other envelope was filled with either twice or half that amount randomly, but if you choose the other envelope, then the other envelope contains twice your amount if you chose the smaller amount and half your amount if you chose the larger amount.” Notice that if the player chooses the envelope without X in it, then the amount in the other envelope is no longer random.

It will help me if, when setting up scenarios, you give me an example using envelopes with specific amounts that exemplify the language. You can tell me who knows what, and then what decisions are made according to what logic. It will help to include several rounds of choosing envelopes so I can see the odds playing out. This seems ok to do because in every actual case there must be multiple rounds (to demonstrate the odds) with exact amounts in the envelopes offered in each round. For example, you might say, “I told her that I put double/half the amount into two envelopes. (One had 20D, the other 40D, but she doesn’t know this.) She chose the envelope with 20D. She reasoned thus and so and then decided to trade/not trade and then made/lost money. The next time I offered 2 envelopes with these 2 amounts and told her…and so on.

For example, the following language confused me: “a chain of pairs in which every amount appears twice, once as the smaller amount in the pair, once as the larger amount in the pair… If the player is looking at, say, £100 and is told it is an extreme, what should he do? 50% chance it is the smallest, 50% chance it is the largest. We're back to the single envelope case.” If you give me a small example as I described in the previous paragraph, I think I’ll better understand what you mean.

You wrote, “It seems to me that some mathematical arguments about prior probabilities effectively say that on opening the first envelope the amount inside allows the player to make a judgment about the prior probabilities.” I’m not sure what you mean by “prior probabilities.”

You wrote, “So thinking about finite collections reduces to thinking about a chain which in turn reduces to thinking about a single pair.” Yes, and if a single pair of envelopes is repeatedly given as the envelopes from which to choose, it will quickly become apparent that the logic of, “when I see an amount, the other envelope will have half the amount half the time and twice the amount half the time,” is not true. That logic seems true until you try to play the game using multiple trials, which is necessary for that logic to hold.

I think I have addressed your sub-problem above. If it isn’t addressed, it will help me if you restate it and give me an example as I’ve described above.

Answer 7

I was wrong! The set from which the starting amounts are drawn is irrelevant. All that matters is the starting amounts used in any particular game.

The logic I suggested in a previous post that, for example, the highest amount included in any game will never be doubled, along with the fact that every game must end, is all that is needed to resolve the paradox, as I discuss below.

After all of these posts, I think that the following is a complete summary of all of the relevant and correct ideas already discussed:

I’ll define a trial as each time 2 envelopes are filled and one is chosen. I’ll define a game as any finite number of trials. This assumes that every game ends. No game goes on indefinitely.

The setup: An envelope is filled with some amount of money. Another envelope is then filled with either half or twice that amount, randomly. The player chooses one of those envelopes and then decides to keep the chosen envelope or trade for the other envelope. This continues with different pairs of envelopes until the game ends.

The 2 envelope paradox, which falsely claims that trading every time will increase a players winnings, is based on this logic: “When I see an amount, I can trade half the time for twice the amount and half the time for half the amount. Therefore, I should trade every time since I’ll make more on average than I’ll lose.”

In order to draw that conclusion, that logic must be true for every amount seen. However, in any actual game, the envelope that contains the highest amount used in that game cannot be traded for twice its amount. Therefore, the logic fails, and so does the conclusion that trading every time will increase the chance of winning.

In fact, in any game, 1) some of the lowest amounts will never be paired with half their amount, 2) some amounts will be paired with both half and twice their amount, and 3) some of the largest amounts will never be paired with twice their amount. When the envelopes in 1) are chosen, the player always wins by trading. When the envelopes in 2) are chosen, on average, the player wins or loses by trading according to what percent of the time the other amount is higher or lower. When the envelopes in 3) are chosen, the player always loses when trading. In any game in which the player either never trades or always trades, the sum of the amounts won or lost in each of the 3 possibilities described above equals zero.

For example, if the envelope I choose has 20D, I don’t know if any other envelope will ever have 40D in the game I’m playing. Maybe the first envelope was filled with 10D, and then, randomly, the other envelope was filled with 20D, and I picked the 20D envelope. In that case, then next time the 10D envelope is filled first, the other envelope may have 5D. The first envelope filled may never contain 20D in that game, and the 40D envelope may never appear. If a 40D envelope does exist, then I will draw it some percent of the time, and I’ll repeat the same logic – maybe an 80D envelope will never exist in this game. This logic applies to any amount drawn.

If someone says, “I see 20D. The other envelope will have 10D half the time and 40D the other half of the time.” I’ll say, “That’s not necessarily true. Maybe the 10D envelope was filled first and, randomly, the other envelope, which is the one you chose, was filled with 20D. If so, then the next time the 10D envelope is filled first, the other envelope may be filled with 5D. A 40D envelope may never exist in this game.” She may say, "But the other envelope could have 40D." I'll say, "Yes, but in the games that don't have a 40D envelope, you'll lose money every time you trade the 20D envelope. Those games are examples of when your strategy of always trading won't work." The same logic applies to any amount drawn or even considered.

I think that this exposes the illogic that leads to the paradox.

Both of your statements below were generated from my mistake. Still, I’ll address them individually.

You wrote, “What troubles me still is this: while I think it is the case that it is impossible to work with the idea of 'choosing' from infinite possibilities, this does not imply that the finite set of possibilities is fixed in the sense that there is a given finite set of pairs from which the pair the player plays is selected.”

I was wrong about this, and I agree with what you state above. I realize that the starting amounts can be drawn from an infinite set and my logic still holds as long as the game ends. It’s not the set from which the initial starting amounts were drawn that matters; it’s the actual amounts drawn that matter.

You asked, “if I say I create a pair by 'thinking of a number', then flipping a coin to put double or half into the other envelope', what is the finite set from which this pair was selected from?”

My answer is: Any set (finite or infinite) that contains whatever starting amounts were used in that particular game.

Answer 8

Overview

I believe that they way you have broken out the problem is completely correct. You need to distinguish the "Coin Flip" scenario, from the situation where the money is added to the envelope before the envelope is chosen

Not distinguishing those scenarios lies at the root of many people's confusion.

Problem 1

If you are flipping a coin to decide if either double your money or lose half, always play the game. Instead of double or nothing, it is double or lose some.

Problem 2

This is exactly the same as the coin flip scenario. The only difference is that the person picking the envelope flipped before giving you the first envelope. Note You Did Not Choose an Envelope!!!! You were given one envelope, and then given the choice to switch This is a subtle but important difference over problem 3, which affects the distribution of the priors

Problem 3

This is the classical setup to the two envelope problem. Here you are given the choice between the two envelopes. The most important points to realize are

• There is a maximum amount of money that can be in the any envelope. Because the person running the game has finite resources, or a finite amount they are willing to invest
• If you call the maximum money that could be in the envelope M, you are not equally likely to get any number between 0 and M. If you assume a random amount of money between 0 and M was put in the first envelope, and half of that for the second (or double, the math still works) If you open an envelope, you are 3 times as likely to see something less than M/2 than above M/2. (This is because half the time both envelopes will have less than M/2, and the other half the time 1 envelope will)
• Since there is not an even distribution, the 50% of the time you double, 50% of the time you cut in half doesn't apply
• When you work out the actual probabilities, you find the expected value of the first envelope is M/2, and the EV of the second envelope, switching or not is also M/2

Interestingly, if you can make some guess as to what the maximum money in the envelope can be, or if you can play the game multiple times, then you can benefit by switching, whenever you open an envelope less than M/2. I have simulated this two envelope scenario and find that if you have this outside information, on average you can do 1.25 as well as just always switching or never switching.

Answer 9

I appreciate your replies.

I realize that I have not been explaining the paradox, but instead simply applying different logic to the problem to come to the correct solution, despite me specifically trying to avoid just that.

The paradox arises in part due to thinking about how much is won or lost, and not focusing on the actual goal of winning or losing, regardless of the amount.

The goal is to develop a strategy to win more money than you lose. Winning more money than is lost is the strategy’s only goal. The amount of money won is irrelevant.

Single trial: In any single trial with only 2 envelopes, the odds of winning are 50%. The strategy of trading every time doesn’t improve those odds. The fact that more is won than lost on average by repeatedly choosing the same amount doesn’t change the odds of winning a single trial.

Multiple trials: In a multiple trial game, each individual trial is won 50% of the time. However, if the player trades every time, the player will lose 100% of the time that she chooses the largest amount in that game. On average, that and other losses will exactly equal the amounts won if she trades every time. That’s why the strategy of trading every time doesn’t increase the chances of winning.

Consider a game with only one trial of 2 envelopes. The odds of winning and losing are both 50%. Trading doesn’t increase the chance of winning for any single set of envelopes. It doesn’t matter how much is won or lost for a single set of envelopes because there aren’t more trials (in a single trial game) to either win back or lose whatever is lost or won in that trial. There is only that one trial. You either win or lose, 50-50.

Of course, she will say, “but if I lose this time, maybe I’ll win the money back in further trials with my strategy.” Let’s do some more trials and see what happens.

Suppose the envelopes in the second trial contain higher amounts than in the first trial. Then, for both of the trials combined, all that matters is what happens in the second trial because the amounts are bigger. But for that second trial, regardless of whether or not she trades, the odds of winning are once again 50-50. That means that the overall odds of winning are also 50-50. Her strategy doesn’t work.

Suppose the envelopes in the second trial contain the same amounts as in the first trial. On average, she’ll chose the other amount that she didn’t choose in the first trial and will either win or lose depending on whether or not she lost or won, respectively, in the first trial. Therefore, on average, she’ll exactly break even on those 2 trials. Her strategy doesn’t work.

Suppose the envelopes in the second trial contain smaller amounts than in the first trial. Then they don’t matter to the overall winning or losing. Only the first trial matters, and the odds of winning on the first trial are still 50-50. The strategy still doesn’t increase the chance of winning.

This logic continues for any number of envelopes.

The logic that leads to the paradox goes awry in two places. First is in understanding that the amount won or lost in a single trial is irrelevant to the odds of winning or losing that trial. It seems liking winning more is better than losing less, but for a single trial, it isn’t. We either win or lose, and that’s it. 50-50. Second, the prospect of winning more in future identical trials than we lost in the first trial entices us into thinking about a game in which the same amount is repeatedly drawn, and then generalizing the outcome of that repeated event to include any number drawn. While a game in which the same number is repeatedly drawn seems intuitively possible, it rarely happens on average. Consequently, generalizing that (rare) result leads to an incorrect conclusion that spawns the paradox.

Answer 10

I appreciate your reply. I’m interested to know if the following explanation helps.

This paradox is an example of the problems that occur when infinity is subtly introduced into a problem. It also exemplifies the difficulties that may occur when defining an “event” for statistical purposes.

Since both twice and half of any number exist, it is tempting to think that both twice and half of any amount drawn in any game also can, on average, exist in that game, or in some combination of games. But both twice and half of every possible amount drawn never exist for every amount in any game or any finite combination of games because in any finite number of games, a highest amount will exist, and twice that amount won’t exist. Only a game with an infinite number of paired envelopes, or an infinite number of games, can have both half and double of every amount in that (those) game(s). Either not realizing that only an infinite set satisfies the logic or the difficulty conceptualizing infinite sets leads to the faulty logic that results in the paradox.

In the 2-envelope problem, any particular game can be won, lost, or tied using the “trade every time” strategy. A successful strategy doesn’t assure a win every time a particular game is played, it assures a win on average. But, what is being averaged? A successful strategy must assure a win, on average, in any particular game that is played repeatedly.

In the 2-envelope problem, every envelope with amount X can be paired with another envelope that contains either 2X or .5X. The paired envelope might contain 2X in one game, and the paired envelope might contain .5X in a different game, but both of the paired 2X and .5X envelopes typically won’t appear in the same game. If I choose an envelope containing X and I average the outcome of trading for 2X with the outcome of trading for .5X, I’m (typically) averaging the results from two different games. This does not help determine whether or not the strategy employed is effective in either game.

Instead, what must be averaged is what happens when the same game is played multiple times -- that is, when the same n pairs of envelopes are presented multiple times. In that case, it’s easy to see that, in any particular game, each envelope will be selected the same number of times on average. (It’s also easy to see that, for the envelopes in that particular game, some envelope will contain the largest amount that cannot be doubled by being traded.) If each envelope is traded every time, the loss will equal the gain in each pair of envelopes on average. Therefore, trading every time will not increase the odds of winning in any game.

Typical mathematical approaches that average the outcomes of 2X and .5X are attempting to determine the effect of the strategy on the outcome of an infinite number of games simultaneously. That’s because 2X and .5X define different games, and the 2X and .5X amounts needed for those different games will be in yet other games, and so on. Introducing infinity in this way into the statistics calculations leads to difficulties.

For statistical purposes, a single, repeatable trial is defined by two filled envelopes being presented to a player and her picking one of them. A single, repeatable trial is not defined by filling the second envelope randomly. A trial with envelopes containing 20D and 10D is different from a trial with envelopes containing 20D and 40D. And yet the logic that leads to the paradox mistakenly considers both of these trials simultaneously. That leads to problems. When repeating a trial, only 2 amounts appear. What changes in each repetition is the amount that is chosen.

By contrast, here’s a game for which the incorrect 2-envelope logic applies correctly. Suppose I start with an envelope that contains some random amount. I toss a coin and fill the other envelope with twice the starting amount when the coin reads heads and half the starting amount when the coin reads tails. In this game, the trial starts when I have the starting amount in hand, before the other envelope is filled. In this game, I will be able to choose twice the amount half the time and half the amount half the time. I can then correctly average the amount won or lost over multiple repetitions and determine that trading every time is better than never trading. In this game, when repeating a trial, 3 amounts appear. What changes in each repetition is the amount in the other envelope.

When calculating odds for a card game, the player’s hand remains constant and the opposing hands change. When the 2-envelope game is played, the envelope amounts stay the same and the chosen envelope changes. The amount drawn is not what stays the same, so calculating the odds of winning that game based on choosing the same amount repeatedly won’t work.

One way to counter the faulty logic of generalizing the possible results (double or half the amount traded) of trading any particular amount is to find language that ties those proposed results of trading to an actual game. The fact that those 2 results are (typically) in different games – and therefore cannot be used to determine the outcome of any particular game -- must become more apparent in order break the seemingly irrefutable logic that leads to the paradox. The following paragraph contains my attempt to use such language.

I say, “I see 20D. I should always trade because half the time the other envelope will contain 10D and the other half of the time the other envelope will contain 40D.” She says, “While both of those amounts general can exist, they may not exist in this game. If 20D is the highest amount in the game you’re playing, you’ll always lose when trading it.” I say, “True, but in the next game, the 40D envelope might exist, so between the 2 games, I’ll still win on average when always trading 20D envelope. I’ll just average those 2 games together to calculate my odds.” She says, “That’s true for the 20D amount, but in that next game, that 40D envelope might be the highest amount. An 80D envelope might not exist in that next game. When you play that 2nd game, you’ll eventually draw the 40D envelope. When you do, you’ll always lose when you trade it. That means that you’ll lose in the second game more than you can win in the first game by trading the 20D envelope. So, if you average those 2 games together, you still won’t win overall by trading every time.” I say, “That’s true, too, but the 3rd game I play might have the 80D envelope, so I’ll still win money when trading the 40D envelope if I average the 3 games together.” She says, “That’s true for the 40D envelope, but the 80D envelope in the 3rd game might be the biggest amount in that game. You’ll eventually draw it, and when you do, you’ll always lose when you trade it. In fact, you can add as many games as you desire, but no matter how many games you add, another game will still be necessary for the logic to be true. You can’t play an infinite number of games, although you will need to for your logic to work. That’s why your logic doesn’t work.”

Answer 11

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.

Answer 12

I’ve been coming at this from several points of view, so I’ll post what I have and will await your response. I hope this all makes sense.

I want to convince you that only one trial of anything is useless in determining odds/outcomes. You must always have multiple trials. In every case that odds are determined, multiple trials are involved. Those trials must be defined precisely. In the course of those multiple trials, something must repeat. That something also must be defined precisely. If something doesn’t repeat, then no odds can be calculated on it. The problem with the 2-envelope game is that it always contains an amount that doesn’t repeat. It seems like this isn’t true, but it is. That’s why it’s confusing.

You wrote, “every trial offers a chance to go up or down a fixed amount (fixed for the trial).” It doesn’t matter that an amount can go up or down if it doesn’t ever go up or down in that game no matter how many times it is played. In all of the 2-envelope games every played or conceived, some amounts have never gone up and down! In fact, if you consider that every amount put into an envelope is a part of the game because it can be drawn, then you can play a billion trials, a billion games, or a billion combinations of those, and some amount still won’t have gone up and down. You can play an infinite number of trials, games, or combinations thereof and there will still be an amount that didn’t go up and down. What happens to the TEP logic for those amounts? What difference does it make that an amount could go up or down when it never does?

Here’s another way to respond to your statement above: so you must average the number of times it goes up with the number of times it goes down. How do you do that when, in theory, every amount doesn’t go both up and down, even though, in fact, both of those amounts exist?

You might say that every amount drawn has gone both up and down, but to consider your odds, but that’s simply luck that that number was drawn twice before twice its amount was drawn once. That luck can’t be used to calculate odds. You must consider the odds for every amount put into an envelope during any game because there is some non-zero change that every envelope will be chosen.

Below, I’ll explain in a different way, but I think that having you actually play a game and show me how the logic applies to it will be the most instructive.

Odds are always based on multiple trials with something repeating and something (the outcome) varying. It is always evident what constitutes a trial and what repeats. In fact, these must be precisely determined in order to calculate odds.

Here’s an example: I want to calculate the odds of a coin coming up heads when I toss it. I don’t know if the coin is fair. So, I toss the coin repeatedly (that’s what repeats) and average the outcomes (that’s what varies). The average of the outcomes will get closer to the answer with repeated trials.

Suppose I shuffle a deck of cards and then remove the top 10 cards. I want to figure out the odds that two pair will win against every other possible hand. I can do this by playing many hands (that’s what repeats) and recording the number of times 2 pairs wins (that’s what varies). Once again, the average of the outcomes will get closer to the answer with repeated trials.

This is always how odds are calculated. Sometimes, math can aid, but sometimes it can’t. For example, in the coin problem above, no math will suggest an answer. Only doing the experiment or analyzing the physical makeup of the coin will suggest an answer.

If an experiment can’t be described in a similar way, then odds cannot be calculated.

So, I’d like you to play the 2-envelope game several times for me (i.e. write out what you did and tell me the outcome) and then show me how you applied the TEP logic to the game and then calculated the odds. You can just play 3 or 4 trials and then show my how the odds apply.