再谈两个信封问题

我在想这个问题。

http://en.wikipedia.org/wiki/Two_envelopes_problem

我相信该解决方案，并且我认为我理解它，但是如果采用以下方法，我将感到完全困惑。

问题1：

我将为您提供以下游戏。您付给我10 美元，我会掷一枚公平的硬币。头部给我5 美元，尾巴给我20 美元。

期望是12.5 美元，因此您将始终可以玩游戏。

问题2：

我会给你一个10 美元的信封，信封已经打开，你可以检查一下。那么我告诉你一个信封，关闭，这个时候，告诉你：这个信封或者具有$ 5或以相同的概率$ 20吧。您要交换吗？

我觉得这与问题1完全一样，您放弃了$ 10换了$ 5或$ 20，所以再次您总是会切换。

问题三：

我做与上述相同，但是请关闭信封。因此，您不知道有10美元，但有多少X。我告诉您另一个信封有两倍或一半。现在，如果您遵循相同的逻辑，则要切换。这就是信封悖论。

我打开信封后发生了什么变化？

编辑：

有些人认为问题3并不是信封问题，我将尝试通过分析每个人对游戏的看法，在下面提供我认为为什么的问题。而且，它为游戏提供了更好的设置。

为问题3提供一些澄清：

从组织游戏的人的角度来看：

我拿着两个信封。我合上了一张10 美元的货币，将其交给玩家。然后我告诉他，我还有一个信封，是我刚给你的信封的两倍或一半。您要切换吗？然后，我继续掷出一枚公平的硬币，往正面扔了5 美元，向尾摆放了 20 美元，然后递给他信封。然后我问他。您刚给我的信封是您所持信封的两倍或一半。您要切换吗？

从玩家的角度来看：

我得到一个信封，并告诉我还有另一个信封，信封的数量是其概率的两倍或一半。我要切换吗？我认为我有$X$，因此$\frac{1}{2}(\frac{1}{2}X + 2X) > X$所以我想切换。我得到了信封，突然之间我面临着完全相同的情况。我想再次切换，因为另一个信封的数量是原来的两倍或一半。

1.不必要的概率。

本注释的下两部分使用标准的决策理论工具（2）分析“猜大”和“两个包络”问题。这种方法虽然简单明了，但似乎是新的。特别是，它确定了两个信封问题的一组决策程序，这些决策程序明显优于“始终切换”或“永不切换”程序。

第2节介绍（标准）术语，概念和符号。它针对“猜测更大的问题”分析了所有可能的决策程序。熟悉本资料的读者可能会跳过本节。第三部分对两个包络线问题进行了类似的分析。第四节的结论总结了关键点。

关于这些难题的所有已发表的分析均假设存在控制自然界可能状态的概率分布。但是，此假设不是难题陈述的一部分。这些分析的关键思想是放弃这种（毫无根据的）假设可以简单地解决这些难题中的明显悖论。

2.“更大的猜测”问题。

实验人员被告知，不同的实数$x_1$在两张纸上写着和$x_2$。她在一个随机选择的单上看这个数字。仅根据这一观察结果，她必须确定这是两个数字中的较小还是较大。

诸如此类的关于概率的简单但开放性的问题因令人困惑和违反直觉而臭名昭著。特别是，至少有三种不同的方式使概率进入图像。为了澄清这一点，让我们采用正式的实验观点（2）。

首先指定损失函数。从下面定义的意义上说，我们的目标将是最大程度地减少其期望。一个不错的选择是当实验者正确猜测时使损失等于$1$，否则使损失等于$0$。该损失函数的期望是错误猜测的可能性。通常，通过为错误的猜测分配各种惩罚，损失函数可以正确捕获猜测的目的。可以肯定的是，采用损失函数就像在x 1上假设先验概率分布一样任意。 $x_1$ 和 $x_2$，但更为自然和根本。当我们面对决策时，我们自然会考虑对或错的后果。如果两种方法都没有后果，那为什么要关心呢？每当做出（合理的）决策时，我们都会隐含地考虑潜在损失，因此我们会从显式的损失考虑中受益，而使用概率来描述纸条上的可能值是不必要的，人为的，并且-我们将看到-可能阻止我们获得有用的解决方案。

决策理论对观测结果进行建模并对其进行分析。它使用了三个附加的数学对象：一个样本空间，一组“自然状态”和一个决策程序。

• 样本空间 $S$ 由所有可能的观察组成；在这里可以用R标识 $\mathbb{R}$ （实数集）。

• 自然状态$\Omega$ 是控制实验结果的可能概率分布。（这是我们可以谈论事件的“概率”的第一种感觉。）在“猜测更大”的问题中，这些是离散分布，具有分别 具有相等概率的不同实数 $x_1$ 和 $x_2$值共每个值 2个。Ω 可以由参数{ω=（X1，X2）∈[R×[R| x1>x2}$\frac{1}{2}$$\Omega$$\{\omega = (x_1, x_2) \in \mathbb{R}\times\mathbb{R}\ |\ x_1 \gt x_2\}.$

• 决策空间是二进制集 $\Delta = \{\text{smaller}, \text{larger}\}$可能决策。

用这些术语，损耗函数是定义为Ω ×的实值函数。它告诉我们一个决策（第二个论点）与现实（第一个论点）相比有多“糟糕”。$\Omega \times \Delta$

实验人员可以使用的最通用的决策程序 是一种随机决策程序：对于任何实验结果，其值都是Δ上的概率分布 。即，观察结果x时做出的决定 不一定是确定的，而是根据分布δ （x ）随机选择 。（这是涉及概率的第二种方法。）$\delta$$\Delta$$x$$\delta(x)$

当 只有两个元素时，任何随机过程都可以通过它分配给预定决策的概率来识别，具体来说，我们认为它是“更大”的。 $\Delta$

物理旋转器实现这样一个二进制随机过程：可自由旋转的指针会停在上部区域中，对应于一个决定，具有概率δ，否则将在左下区域停止与概率1 - δ （x ）。所说的纺丝器完全由指定的值确定δ （X ）∈ [ 0 ，1 ]。$\Delta$$\delta$$1-\delta(x)$$\delta(x)\in[0,1]$

因此，可以将决策程序视为函数

$$\delta^\prime:S\to[0,1],$$

哪里

$${\Pr}_{\delta(x)}(\text{larger}) = \delta^\prime(x)\ \text{ and }\ {\Pr}_{\delta(x)}(\text{smaller})=1-\delta^\prime(x).$$

相反，任何这样的函数 确定随机决定过程。随机决定包括在特殊情况确定性决定，其中的范围 δ '在于{ 0 ，1 }。$\delta^\prime$$\delta^\prime$$\{0,1\}$

我们说，结果 x的决策过程δ的成本 是δ （x ）的预期损失。期望是关于决策空间Δ上的 概率分布 δ （x ）。每个自然状态ω（回想起来，是样本空间S上的二项式概率分布 ）确定任何过程δ的预期成本 ；这是危险的 δ 为 ω，风险δ（ω ）$\delta$$x$$\delta(x)$$\delta(x)$$\Delta$$\omega$$S$$\delta$$\delta$$\omega$$\text{Risk}_\delta(\omega)$。在此，期望是关于自然状态 $\omega$。

比较决策程序的风险功能。 当自然状态真正未知时，和δ是两个过程，且所有 ω的风险ε（ω ）≥ 风险δ（ω ），则使用过程ε是没有意义的 ，因为过程 δ永远不会更糟并且在某些情况下可能会更好）。这样的程序 $\varepsilon$$\delta$$\text{Risk}_\varepsilon(\omega)\ge \text{Risk}_\delta(\omega)$$\omega$$\varepsilon$$\delta$ 是不允许的$\varepsilon$; 否则，它是可接受的。通常存在许多可接受的程序。我们认为它们中的任何一个都是“好”的，因为它们在任何其他程序中都无法始终胜过它们。

请注意，没有在上引入先验分布 （C的“混合策略$\Omega$$C$（（1）中的“ ”）。这是概率可能是问题设置一部分的第三种方法。使用它使当前分析比（1）及其参考文献的分析更笼统，同时更简单。

表1评估了当真实自然状态由给出时的风险 $\omega=(x_1, x_2).$ 回想一下 $x_1 \gt x_2.$

表格1。

$$\matrix { \text{Decision:}& & \text{Larger} & \text{Larger} & \text{Smaller} & \text{Smaller}\\ \text{Outcome} & \text{Probability} & \text{Probability} & \text{Loss} & \text{Probability} & \text{Loss} & \text{Cost} \\ x_1 & 1/2 & \delta^\prime(x_1) & 0 & 1 - \delta^\prime(x_1) & 1 & 1 - \delta^\prime(x_1) \\ x_2 & 1/2 & \delta^\prime(x_2) & 1 & 1 - \delta^\prime(x_2) & 0 & 1 - \delta^\prime(x_2) }$$

$$\text{Risk}(x_1,x_2):\ (1 - \delta^\prime(x_1) + \delta^\prime(x_2))/2.$$

用这些术语来说，“猜大”的问题变成了

鉴于你一无所知 和 X 2，不同之处在于它们是不同的，你能找到一个决策程序 δ为其风险 [ 1 - δ '（最大（X 1，X $x_1$$x_2$$\delta$ 肯定小于$[1 – \delta^\prime(\max(x_1, x_2)) + \delta^\prime(\min(x_1, x_2))]/2$？$\frac{1}{2}$

该陈述等效于要求 $\delta^\prime(x)\gt \delta^\prime(y)$每当 从那里，它是必要的和足够的实验者的判断程序，以由一些严格递增函数来指定 δ '：小号→ [ 0 ，1 ] 。 这组过程包括，但比，所有的“混合策略大Q的” 1。有很多$x \gt y.$$\delta^\prime: S\to [0, 1].$$Q$ 比任何非随机过程都更好的随机决策过程！

3.“两个信封”问题。

令人鼓舞的是，这种简单的分析揭示了针对“猜大”问题的大量解决方案，其中包括以前未发现的好的方案。 让我们看看用相同的方法可以解决摆在我们面前的另一个问题的问题，即“两个信封”问题（有时也称为“盒子问题”）。这涉及一种通过随机选择两个信封之一来玩的游戏，已知其中一个信封的钱是另一个信封的两倍。打开信封并观察数量 $x$ 玩家决定是否将钱存放在未打开的信封中（“切换”）还是将​​钱存放在打开的信封中。人们会认为切换和不切换是同样可以接受的策略，因为玩家同样不确定哪个信封包含更大的金额。矛盾的是，切换似乎是更好的选择，因为它在和 x / 2的收益之间提供了“等概率”的选择 ，其预期值 5 x / 4 超过了打开包络中的值。请注意，这两种策略都是确定性的和恒定的。$2x$$x/2,$$5x/4$

在这种情况下，我们可以正式写

$$\eqalign{ S &= \{ x\in \mathbb{R}\ |\ x \gt 0\}, \\ \Omega &= \{\text{Discrete distributions supported on }\{\omega, 2\omega\}\ |\ \omega \gt 0 \text{ and }\Pr(\omega) = \frac{1}{2}\}, \text{and}\\ \Delta &= \{\text{Switch}, \text{Do not switch}\}. }$$

如前，任何决定过程 可以被认为是从一个函数 小号 至[ 0 ，1 ] $\delta$$S$$[0, 1],$通过将其与不进行切换，这又可以写入的概率相关联此时。开关必须的当然的概率是互补值1 - δ '（X ）$\delta^\prime(x)$$1–\delta^\prime(x).$

如表2所示，损失是游戏收益的负数。这是自然的真实状态的函数，结果X（其可以是 ω 或 2 ω），和决定的，这依赖于结果。$\omega$$x$$\omega$$2\omega$

表2。

$$\matrix{ & \text{Loss}&\text{Loss} &\\ \text{Outcome}(x) & \text{Switch} & \text{Do not switch} & \text{Cost}\\ \omega & -2\omega & -\omega & -\omega[2(1-\delta^\prime(\omega)) + \delta^\prime(\omega)]\\ 2\omega & -\omega & -2\omega & -\omega[1 - \delta^\prime(2\omega) + 2\delta^\prime(2\omega)] }$$

表2除了显示损失函数外，还计算了任意决策程序的成本 。因为游戏产生两个结果的概率相等为$\delta$，ω 为自然状态时的风险 为$\frac{1}{2}$$\omega$

$$\eqalign{ \text{Risk}_\delta(\omega) &=-\omega[2(1-\delta^\prime(\omega)) + \delta^\prime(\omega)]/2 + -\omega[1 - \delta^\prime(2\omega) + 2\delta^\prime(2\omega)]/2 \\ &= (-\omega/2)[3 + \delta^\prime(2\omega) - \delta^\prime(\omega)]. }$$

恒定过程，该装置总是切换（）或始终站在PAT（δ '（X ）= 1），将具有风险 - 3 ω / 2。任何严格递增函数，或者更一般地，任何函数δ ' 与范围在[ 0 ，1 ]针对 δ '（2 X ）> δ '（X ） 对于所有正实 $\delta^\prime(x)=0$$\delta^\prime(x)=1$$-3\omega/2$$\delta^\prime$$[0, 1]$$\delta^\prime(2x) \gt \delta^\prime(x)$确定过程 δ 具有风险函数总是严格小于 - 3 ω / 2，因此优于任一恒定过程，无论自然的真实状态的 ω！ 因此，恒定程序是不可接受的，因为存在某些程序，无论其自然状态如何，其风险有时较低，而从来没有较高。$x,$$\delta$$-3\omega/2$$\omega$

将此与前面的“猜大了”问题的解决方案进行比较，可以看出两者之间的紧密联系。在这两种情况下，适当选择的随机程序显然要优于“显而易见的”恒定策略。

这些随机策略具有一些显着的特性：

• 随机策略不会有坏情况：无论如何选择总金额，从长远来看，这些策略都不会比恒定策略差。

• 没有与限制值随机化策略和1个占优势的任何其他人的：如果对于期望δ时（ω ，2 ω ）是在该包络线超过了期望ε，则存在与一些其它可能的状态（η ，2包络中的η ）和ε的期望值超过 δ的期望值。$0$$1$$\delta$$(\omega, 2\omega)$$\varepsilon$$(\eta, 2\eta)$$\varepsilon$$\delta$

• 该 策略包括，作为特殊情况，战略等同于许多贝叶斯策略。任何策略，指出“如果开关X小于某个阈Ť和留否则”对应于δ （X ）= 1时X ≥ Ť ，δ （X ）= 0否则。$\delta$$x$$T$$\delta(x)=1$$x \ge T, \delta(x) = 0$

那么，主张始终转换的谬论到底是什么呢？ 隐含的假设是，替代方案根本没有任何概率分布。具体而言， 在打开的信封中观察到 时，用于切换的直观论据基于条件概率Prob（未打开的信封中的量| x 被观察到），这些概率是在基础自然状态集上定义的概率。但是这些不能从数据中计算出来。决策理论框架不需要 为了解决问题而在Ω上分配概率 ，问题也没有指定一个。$x$$x$$\Omega$

此结果与（1）及其参考文献所获得的结果有细微但重要的区别。 其他解决方案都假设（即使无关紧要）在上存在先验概率分布 ，然后从本质上证明它在S上必须是均匀的 。 反过来，这是不可能的。但是，这里给出的二包络问题的解决方案并不是作为某些给定先验分布的最佳决策程序而出现的，因此，这种分析方法对其没有予以关注。在当前的处理中，是否可以存在先验概率分布根本不重要。我们可以将其描述为$\Omega$$S.$不确定信封的内容（如先前的分发所述）与完全不知道信封的内容（因此没有相关的先前分发）之间的对比。

4。结论。

在“猜大”的问题中，一个好的程序是随机决定观察值是两者中的较大者，并随着观察值的增加而增加。没有单一的最佳程序。在“两个信封”问题中，一个好的程序是再次随机确定观察到的货币量值得保留（也就是说，这是两者中的较大者），并且概率会随着观察值的增加而增加。同样，没有一个最佳过程。在这两种情况下，如果许多玩家使用这样的程序并针对给定的独立玩游戏，那么（无论ω的值如何）总体上，他们的获胜多于输，因为他们的决策程序倾向于选择更大的金额。$\omega$$\omega$

在这两个问题中，都不是问题的一部分做出另外的假设-自然状态的事先分布-会引起明显的悖论。通过专注于每个问题中指定的内容，可以完全避免这种假设（可能会做出诱人的尝试），从而使悖论消失并产生直接的解决方案。

参考资料

（1）D. Samet，I。Samet和D. Schmeidler，两个信封拼图背后的一个观察。 美国数学月刊111（2004年4月）347-351。

（2）J. Kiefer，《统计推断导论》。 纽约，Springer-Verlag，1987年。

The issue in general with the two envelope problem is that the problem as presented on wikipedia allows the size of the values in the envelopes to change after the first choice has been made. The problem has been formulized incorrectly.

However, a real world formulation of the problem is this: you have two identical envelopes: $A$ and $B$, where $B=2A$. You can pick either envelope and then are offered to swap.

Case 1: You've picked $A$. If you switch you gain $A$ dollars.

Case 2: You've picked $B$. If you switch you loose $A$ dollars.

This is where the flaw in the two-envelope paradox enters in. While you are looking at loosing half the value or doubling your money, you still don't know the original value of $A$ and the value of $A$ has been fixed. What you are looking at is either $+A$ or $-A$, not $2A$ or $\frac{1}{2}A$.

If we assume that the probability of selecting $A$ or $B$ at each step is equal,. the after the first offered swap, the results can be either:

Case 1: Picked $A$, No swap: Reward $A$

Case 2: Picked $A$, Swapped for $B$: Reward $2A$

Case 3: Picked $B$, No swap: Reward $2A$

Case 4: Picked $B$, Swapped for $A$: Reward $A$

The end result is that half the time you get $A$ and half the time you get $2A$. This will not change no matter how many times you are offered a swap, nor will it change based upon knowing what is in one envelope.

My interpretation of the question

I am assuming that the setting in problem 3 is as follows: the organizer first selects amount $X$ and puts $X$ in the first envelope. Then, the organizer flips a fair coin and based on that puts either $0.5X$ or $2X$ to the second envelope. The player knows all this, but not $X$ nor the result of the coin-flip. The organizer gives the player the first envelope (closed) and asks if the player wants to switch. The questioner argues 1. that the player wants to switch because the switching increases expectation (correct) and 2. that after switching, the same reasoning symmetrically holds and the player wants to switch back (incorrect). I also assume the player is a rational risk-neutral Bayesian agent that puts a probability distribution over $X$ and maximizes expected amount of money earned.

Note that if the we player did not know about the coin-flip procedure, there might be no reason in the first place to argue that the probabilities are 0.5 for the second envelope to be higher/lower.

Why there is no paradox

Your problem 3 (as interpreted in my answer) is not the envelope paradox. Let the $Z$ be a Bernoulli random variable with $P(Z=1)=0.5$. Define the amount $Y$ in the 2nd envelope so that $Z=1$ implies $Y=2X$ and $Z=0$ implies $Y=0.5X$. In the scenario here, $X$ is selected without knowledge of the result of the coin-flip and thus $Z$ and $X$ are independent, which implies $E(Y\mid X) = 1.25X$.

$$\begin{equation} E(Y) = E(E(Y\mid X)) = E(1.25X) = 1.25E(X) \end{equation}$$ Thus, if if X>0 (or at least $E(X)>0$), the player will prefer to switch to envelope 2. However, there is nothing paradoxical about the fact that if you offer me a good deal (envelope 1) and an opportunity to switch to a better deal (envelope 2), I will want to switch to the better deal.

To invoke the paradox, you would have to make the situation symmetric, so that you could argue that I also want to switch from envelope 2 to envelope 1. Only this would be the paradox: that I would want to keep switching forever. In the question, you argue that the situation indeed is symmetric, however, there is no justification provided. The situation is not symmetric: the second envelope contains the amount that was picked as a function of a coin-flip and the amount in the first envelope, while the amount in the first envelope was not picked as a function of a coin-flip and the amount in the second envelope. Hence, the argument for switching back from the second envelope is not valid.

Example with small number of possibilities

Let us assume that (the player's belief is that) $X=10$ or $X=40$ with equal probabilities, and work out the computations case by case. In this case, the possibilities for $(X,Y)$ are $\{(10,5),(10,20),(40,20),(40,80)\}$, each of which has probability $1/4$. First, we look at the player's reasoning when holding the first envelope.

1. If my envelope contains $10$, the second envelope contains either $5$ or $20$ with equal probabilities, thus by switching I gain on average $0.5\times(-5) + 0.5\times10 = 2.5$.
2. If my envelope contains $40$, the second envelope contains either $20$ or $80$ with equal probabilities, thus by switching I gain on average $0.5\times(-20) + 0.5\times(40) = 10$.

Taking the average over these, the expected gain of switching is $0.5\times2.5 + 0.5\times10 = 6.25$, so the player switches. Now, let us make similar case-by-case analysis of switching back:

1. If my envelope contains $5$, the old envelope with probability 1 contains $10$, and I gain $5$ by switching.
2. If my envelope contains $20$, the old envelope contains $10$ or $40$ with equal probabilities, and by switching I gain $0.5\times(-10) + 0.5\times20 = 5$.
3. If my envelope contains $80$, the old envelope with probability 1 contains $40$ and I lose $40$ by switching.

Now, the expected value, i.e. probability-weighted average, of gain by switching back is $0.25\times5+0.5\times5+0.25\times(-40) = -6.25$. So, switching back exactly cancels the expected utility gain.

Another example with a continuum of possibilities

You might object to my previous example by claiming that I maybe cleverly selected the distribution over $X$ so that in the $Y=80$ case the player knows that he is losing. Let us now consider a case where $X$ has a continuous unbounded distribution: $X \sim \textrm{Exp}(1)$, $Z$ independent of $X$ as previously, and $Y$ as a function of $X$ and $Z$ as previously. The expected gain of switching from $X$ to $Y$ is again $E(0.25X) = 0.25E(X) = 0.25$. For the back-switch, we first compute the conditional probability $P(X=0.5Y \mid Y=y)$ using Bayes' theorem:

$$\begin{equation} P(X=0.5Y \mid Y=y) = P(Z=1 \mid Y=y) = \frac{p(Y=y\mid Z=1)P(Z=1)}{p(Y=y)} = \frac{p(2X=y)P(Z=1)}{p(Y=y)} = \frac{0.25e^{-0.5y}}{p(Y=y)} \end{equation}$$ and similarly $P(X=2Y \mid Y=y) = \frac{e^{-2y}}{p(Y=y)}$, wherefore the conditional expected gain of switching back to the first envelope is $$\begin{equation} E(X-Y \mid Y=y) = \frac{-0.125y e^{-0.5y} + ye^{-2y}}{p(Y=y)}, \end{equation}$$ and taking the expectation over $Y$, this becomes $$\begin{equation} E(X-Y) = \int_0^\infty \frac{-0.125y e^{-0.5y} + ye^{-2y}}{p(Y=y)}p(Y=y) dy = -0.25, \end{equation}$$ which cancels out the expected gain of the first switch.

General solution

The situation seen in the two examples must always occur: you cannot construct a probability distribution for $X,Z,Y$ with these conditions: $X$ is not a.s. 0, $Z$ is Bernoulli with $P(Z=1)=0.5$, $Z$ is independent of $X$, $Y=2X$ when $Z=1$ and $0.5X$ otherwise and also $Y,Z$ are independent. This is explained in the Wikipedia article under heading 'Proposed resolutions to the alternative interpretation': such a condition would imply that the probability that the smaller envelope has amount between $2^n,2^{n+1}$ ($P(2^n<=\min(X,Y)<2^{n+1})$ with my notation) would be a constant over all natural numbers $n$, which is impossible for a proper probability distribution.

Note that there is another version of the paradox where the probabilities need not be 0.5, but the expectation of other envelope conditional on the amount in this envelope is still always higher. Probability distributions satisfying this type of condition exist (e.g., let the amounts in the envelopes be independent half-Cauchy), but as the Wikipedia article explains, they require infinite mean. I think this part is rather unrelated to your question, but for completeness wanted to mention this.

Problem 1: Agreed, play the game. The key here is that you know the actual probabilities of winning 5 vs 20 since the outcome is dependent upon the flip of a fair coin.

Problem 2: The problem is the same as problem 1 because you are told that there is an equal probability that either 5 or 20 is in the other envelope.

Problem 3: The difference in problem 3 is that telling me the other envelope has either $X/2$ or $2X$ in it does not mean that I should assume that the two possibilities are equally likely for all possible values of $X$. Doing so implies an improper prior on the possible values of $X$. See the Bayesian resolution to the paradox.

This is a potential explanation that I have. I think it is wrong but I'm not sure. I will post it to be voted on and commented on. Hopefully someone will offer a better explanation.

So the only thing that changed between problem 2 and problem 3 is that the amount became in the envelope you hold became random. If you allow that amount to be negative so there might be a bill there instead of money then it makes perfect sense. The extra information you get when you open the envelope is whether it's a bill or money hence you care to switch in one case while in the other you don't.

If however you are told the bill is not a possibility then the problem remains. (of course do you assign a probability that they lie?)

Problem 2A: 100 note cards are in an opaque jar. "$10" is written on one side of each card; the opposite side has either "$5" or "$20" written on it. You get to pick a card and look at one side only. You then get to choose one side (the revealed, or the hidden), and you win the amount on that side.

If you see "$5," you know you should choose the hidden side and will win $10. If you see "$20," you know you should choose the revealed side and will win $20. But if you see "$10," I have not given you enough information calculate an expectation for the hidden side. Had I said there were an equal number of {$5,$10} cards as {$10,$20} cards, the expectation would be $12.50. But you can't find the expectation from $only$ the fact - which was still true - that you had equal chances to reveal the higher, or lower, value on the card. You need to know how many of each kind of card there were.

Problem 3A: The same jar is used, but this time the cards all have different, and unknown, values written on them. The only thing that is the same, is that on each card one side is twice the value of the other.

Pick a card, and a side, but don't look at it. There is a 50% chance that it is the higher side, or the lower side. One possible solution is that the card is either {X/2,X} or {X,2X} with 50% probability, where X is your side. But we saw above that the the probability of choosing high or low is not the same thing as these two $different$ cards being equally likely to be in the jar.

What changed between your Problem 2 and Problem 3, is that you made these two probabilities the same in Problem 2 by saying "This envelope either has $5 or $20 in it with equal probability." With unknown values, that can't be true in Problem 3.

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 problem here 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.