Пересмотр двух конвертов


16

Я думал об этой проблеме.

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

Я верю в решение и думаю, что понимаю его, но если я выберу следующий подход, я полностью запутался.

Проблема 1:

Я предложу вам следующую игру. Вы платите мне 10 долларов, а я подброшу честную монету. Головы я даю вам 5 долларов, а хвосты - 20 долларов .

Ожидание составляет $ 12,5, поэтому вы всегда будете играть в игру.

Проблема 2:

Я дам вам конверт с 10 долларами , конверт открыт, и вы можете проверить. Затем я покажу вам другой конверт, закрыл этот раз и сказать вам: Этот конверт имеет либо $ 5 или $ 20 в ней с равной вероятностью. Вы хотите поменяться?

Я считаю , что это точно так же , как и задачи 1, то отказываться от $ 10 для $ 5 или $ 20, так что снова вы всегда будете переключаться.

Проблема 3:

Я делаю то же, что и выше, но закрываю конверты. Таким образом, вы не знаете, что есть 10 долларов, но некоторая сумма X. Я говорю вам, что у другого конверта есть двойной или половина. Теперь, если вы следуете той же логике, которую хотите переключить. Это парадокс конверта.

Что изменилось, когда я закрыл конверт ??

РЕДАКТИРОВАТЬ:

Некоторые утверждают, что проблема 3 не является проблемой конверта, и я постараюсь изложить ниже, почему, по моему мнению, это так, анализируя взгляд каждого на игру. Кроме того, это дает лучшую настройку для игры.

Предоставление некоторых разъяснений по проблеме 3:

С точки зрения человека, организующего игру:

Я держу 2 конверта. В одном я положил 10 долларов, чтобы закрыть его и дать игроку. Затем я говорю ему, что у меня есть еще один конверт, в котором вдвое больше или вдвое меньше конверта, который я только что дал вам. Вы хотите переключиться? Затем я продолжаю подбрасывать честную монету, и головы, в которые я положил 5 долларов, и хвосты, которые я положил, 20 долларов, и вручить ему конверт. Я тогда спрашиваю его. Конверт, который вы только что дали мне, содержит вдвое или половину количества конверта, который вы держите. Вы хотите переключиться?

С точки зрения игрока:

Мне дают конверт и говорят, что есть другой конверт, который имеет удвоенную или половинную сумму с равной вероятностью. Хочу ли я переключиться. Я думаю, что у меня есть , поэтому поэтому я хочу переключиться. Я получаю конверт, и внезапно я сталкиваюсь с точно такой же ситуацией. Я хочу переключиться снова, поскольку другой конверт имеет либо двойную, либо половину суммы.1X12(12X+2X)>X


2
По крайней мере, для меня ключевое понимание заключается в том, что я не могу просто сказать: «У меня есть Х, следовательно (1/2 * Х + 2Х) / 2> Х» - общий средний шанс составляет 50/50, но для любого конкретного Х ожидаемые шансы больше не 50/50; и чем больше X, тем меньше вероятность наличия 2 * X в другой оболочке (для положительных конечных распределений); поэтому интегрирование по возможным X ' sum(p(X) * (1/2X*f(X) + 2X(1-f(X)) ) = X, где f (X) - это вероятность того, что первый конверт будет больше, учитывая любой конкретный X.
Петерис

1
В формулировке парадокса нет ничего, что говорит о том, что экспериментатор выбирает количество X, и затем экспериментатор случайным образом решает поместить X или X/2 в другую оболочку. Тот факт , что вы держите приравнивая ситуацию, созданную с помощью двух конвертов парадокса вы не понимаете , почему это неправильно для игрока считает , что есть шанс 50/50 другой конверт либо X/2 или 2X . В реальной проблеме двух конвертов вероятность того, что 2X находится в другом конверте, равна либо 0, либо 1.
jsk

вы правы. Я не понимаю :( отсюда вопрос. Я пытаюсь понять разницу между поставленной мною проблемой 3 и парадоксом конверта. Я понимаю, что в парадоксе есть два конверта, так что X и 2X и сделано, но я не не вижу, как это отличается от того, чтобы дать кому-то конверт, а затем подбросить монетку, чтобы принять решение ввести другую сумму.
evan54

1
Уловка к этому - ошибочное предположение, что или X/2 или 2X результаты одинаково вероятны. Если 2X находится в другом конверте, то ожидаемый выигрыш от переключения составляет 2XX=X . Если X/2 находится в другой огибающей, то ожидаемый выигрыш от переключения составляет X/2X=X/2 . Игрок не знает, в какой из этих ситуаций он находится, но это не значит, что он должен верить, что есть шанс 50/50.
Jsk

1
Давайте предположим , что конверты содержат X и 2X . Если вы в конечном итоге с X , то вероятность того, что 2X находится в другой конверт 1 и вероятность того, что X/2 находится в другой оболочке 0. Если вы в конечном итоге с 2X , то вероятность того, что 2(2X)=4X в другой конверте равно 0, а вероятность того, что 2X/2=X в другой конверте, равна 1.
jsk

Ответы:


23

1. НЕОБХОДИМЫЕ ВОЗМОЖНОСТИ.

В следующих двух разделах этой заметки анализируются проблемы «угадать, что больше» и «две оболочки» с использованием стандартных инструментов теории принятия решений (2). Этот подход, хотя и простой, кажется новым. В частности, он идентифицирует набор процедур принятия решения для двух проблем конверта, которые явно превосходят процедуры «всегда переключаться» или «никогда не переключаться».

Раздел 2 вводит (стандартную) терминологию, понятия и обозначения. Он анализирует все возможные процедуры принятия решения для «угадать, что является большой проблемой». Читатели, знакомые с этим материалом, могут пропустить этот раздел. Раздел 3 применяет аналогичный анализ к проблеме двух конвертов. Раздел 4, выводы, суммирует ключевые моменты.

Все опубликованные анализы этих головоломок предполагают, что существует распределение вероятностей, определяющее возможные состояния природы. Это предположение, однако, не является частью утверждений головоломки. Основная идея этих анализов заключается в том, что отбрасывание этого (необоснованного) предположения приводит к простому разрешению очевидных парадоксов в этих головоломках.

2. Проблема «Угадай, что больше».

Экспериментатору говорят, что на двух листках бумаги написаны разные действительные числа и x 2 . Она смотрит на номер на случайно выбранном бланке. Основываясь только на этом одном наблюдении, она должна решить, является ли оно меньшим или большим из двух чисел.x1x2

Простые, но открытые проблемы, такие как вероятность, известны своей запутанностью и нелогичностью. В частности, есть по крайней мере три различных способа, которыми вероятность входит в картину. Чтобы прояснить это, давайте примем формальную экспериментальную точку зрения (2).

Начните с определения функции потерь . Наша цель будет сводить к минимуму его ожидания, в смысле, который будет определен ниже. Хороший выбор - сделать потерю равной если экспериментатор угадал правильно, и 0 в противном случае. Ожидание этой функции потерь - вероятность неправильного угадывания. В общем случае, назначая различные штрафы неправильным догадкам, функция потерь фиксирует цель правильного угадывания. Чтобы быть уверенным, принятие функции потерь столь же произвольно, как и допущение предварительного распределения вероятностей по x 1 и x 210x1x2, но это более естественно и фундаментально. Когда мы сталкиваемся с принятием решения, мы, естественно, учитываем последствия правильности или неправильности. Если в любом случае нет никаких последствий, тогда зачем заботиться? Мы неявно принимаем во внимание возможные потери всякий раз, когда принимаем (рациональное) решение, и поэтому мы извлекаем выгоду из явного рассмотрения потери, в то время как использование вероятности для описания возможных значений на листочках бумаги излишне, искусственно и, поскольку мы увидим - может помешать нам получить полезные решения.

Теория решений моделирует результаты наблюдений и их анализ. Он использует три дополнительных математических объекта: образец пространства, набор «состояний природы» и процедуру принятия решения.

  • Пространство выборки состоит из всех возможных наблюдений; здесь его можно отождествить с R (множество действительных чисел). SR

  • Состояния природы являются возможными распределениями вероятностей, определяющими результаты эксперимента. (Это первый смысл, в котором мы можем говорить о «вероятности» события.) В задаче «угадай, что больше» это дискретные распределения, принимающие значения при различных действительных числах x 1 и x 2 с равными вероятностями из 1Ωx1x2 на каждое значение. Ω можно параметризировать с помощью{ω=(x1,x2)R×R| х1>х2}.12Ω{ω=(x1,x2)R×R | x1>x2}.

  • Пространство решений - это двоичный набор возможных решений.Δ={smaller,larger}

В этих терминах функция потерь - это действительная функция, определенная на . Он говорит нам, насколько «плохое» решение (второй аргумент) по сравнению с реальностью (первый аргумент).Ω×Δ

Наиболее общая процедура принятия решения доступны для экспериментатора является случайным образом один: его значение для любого экспериментального результата является распределение вероятностей на А . То есть решение принимать после наблюдения результата x не обязательно определено, а должно выбираться случайным образом в соответствии с распределением δ ( x ) . (Это второй способ, которым вероятность может быть задействована.)δΔxδ(x)

Когда имеет только два элемента, любая рандомизированная процедура может быть идентифицирована по вероятности, которую она назначает заранее определенному решению, которое, как конкретное, мы считаем «большим». Δ

волчок

Физический счетчик реализует такую ​​двоичную рандомизированную процедуру: свободно вращающийся указатель остановится в верхней области, соответствующей одному решению по , с вероятностью δ , а в противном случае остановится в нижней левой области с вероятностью 1 - δ ( х ) . Спиннер полностью определяется указанием значения δ ( x ) [ 0 , 1 ] .Δδ1δ(x)δ(x)[0,1]

Таким образом, процедура принятия решения может рассматриваться как функция

δ:S[0,1],

где

Prδ(x)(larger)=δ(x)  and  Prδ(x)(smaller)=1δ(x).

Наоборот, любая такая функция определяет процедуру рандомизированного решения. В рандомизированных решения включают детерминированные решения в частном случае , когда диапазон & delta ; ' лежит в { 0 , 1 } .δδ{0,1}

Скажем, что стоимость процедуры принятия решения для результата x - это ожидаемая потеря δ ( x ) . Ожидание относится к распределению вероятности δ ( x ) в пространстве решений Δ . Каждое состояние природы ω (которое, напомним, является биномиальным распределением вероятностей на выборочном пространстве S ) определяет ожидаемую стоимость любой процедуры δ ; это риск от б для ш , риск δ ( ω )δxδ(x)δ(x)ΔωSδδωRiskδ(ω), Здесь ожидание берется относительно естественного состояния .ω

Процедуры принятия решений сравниваются с точки зрения их функций риска. Когда естественное состояние действительно неизвестно, и δ - две процедуры, и Риск ε ( ω ) Риск δ ( ω ) для всех ω , то нет смысла использовать процедуру ε , поскольку процедура δ никогда не бывает хуже ( и может быть лучше в некоторых случаях). Такая процедура ε является недопустимымεδRiskε(ω)Riskδ(ω)ωεδε; в противном случае это допустимо. Часто существует много допустимых процедур. Мы будем считать любого из них «хорошим», потому что ни один из них не может быть последовательно обойден какой-либо другой процедурой.

Обратите внимание, что на вводится предварительное распределение («смешанная стратегия для C » в терминологии (1)). Это третий способ, которым вероятность может быть частью постановки задачи. Его использование делает настоящий анализ более общим, чем анализ (1) и его ссылок, но при этом более простым.ΩC

В таблице 1 оценивается риск, когда истинное состояние природы определяется как Напомним, что x 1 > x 2 .ω=(x1,x2).x1>x2.

Таблица 1.

Decision:LargerLargerSmallerSmallerOutcomeProbabilityProbabilityLossProbabilityLossCostx11/2δ(x1)01δ(x1)11δ(x1)x21/2δ(x2)11δ(x2)01δ(x2)

Risk(x1,x2): (1δ(x1)+δ(x2))/2.

In these terms the “guess which is larger” problem becomes

Given you know nothing about x1 and x2, except that they are distinct, can you find a decision procedure δ for which the risk [1δ(max(x1,x2))+δ(min(x1,x2))]/2 is surely less than 12?

This statement is equivalent to requiring δ(x)>δ(y) whenever x>y. Whence, it is necessary and sufficient for the experimenter's decision procedure to be specified by some strictly increasing function δ:S[0,1]. This set of procedures includes, but is larger than, all the “mixed strategies Q” of 1. There are lots of randomized decision procedures that are better than any unrandomized procedure!

3. THE “TWO ENVELOPE” PROBLEM.

It is encouraging that this straightforward analysis disclosed a large set of solutions to the “guess which is larger” problem, including good ones that have not been identified before. Let us see what the same approach can reveal about the other problem before us, the “two envelope” problem (or “box problem,” as it is sometimes called). This concerns a game played by randomly selecting one of two envelopes, one of which is known to have twice as much money in it as the other. After opening the envelope and observing the amount x of money in it, the player decides whether to keep the money in the unopened envelope (to “switch”) or to keep the money in the opened envelope. One would think that switching and not switching would be equally acceptable strategies, because the player is equally uncertain as to which envelope contains the larger amount. The paradox is that switching seems to be the superior option, because it offers “equally probable” alternatives between payoffs of 2x and x/2, whose expected value of 5x/4 exceeds the value in the opened envelope. Note that both these strategies are deterministic and constant.

In this situation, we may formally write

S={xR | x>0},Ω={Discrete distributions supported on {ω,2ω} | ω>0 and Pr(ω)=12},andΔ={Switch,Do not switch}.

As before, any decision procedure δ can be considered a function from S to [0,1], this time by associating it with the probability of not switching, which again can be written δ(x). The probability of switching must of course be the complementary value 1δ(x).

The loss, shown in Table 2, is the negative of the game's payoff. It is a function of the true state of nature ω, the outcome x (which can be either ω or 2ω), and the decision, which depends on the outcome.

Table 2.

LossLossOutcome(x)SwitchDo not switchCostω2ωωω[2(1δ(ω))+δ(ω)]2ωω2ωω[1δ(2ω)+2δ(2ω)]

In addition to displaying the loss function, Table 2 also computes the cost of an arbitrary decision procedure δ. Because the game produces the two outcomes with equal probabilities of 12, the risk when ω is the true state of nature is

Riskδ(ω)=ω[2(1δ(ω))+δ(ω)]/2+ω[1δ(2ω)+2δ(2ω)]/2=(ω/2)[3+δ(2ω)δ(ω)].

A constant procedure, which means always switching (δ(x)=0) or always standing pat (δ(x)=1), will have risk 3ω/2. Any strictly increasing function, or more generally, any function δ with range in [0,1] for which δ(2x)>δ(x) for all positive real x, determines a procedure δ having a risk function that is always strictly less than 3ω/2 and thus is superior to either constant procedure, regardless of the true state of nature ω! The constant procedures therefore are inadmissible because there exist procedures with risks that are sometimes lower, and never higher, regardless of the state of nature.

Strategy

Comparing this to the preceding solution of the “guess which is larger” problem shows the close connection between the two. In both cases, an appropriately chosen randomized procedure is demonstrably superior to the “obvious” constant strategies.

These randomized strategies have some notable properties:

  • There are no bad situations for the randomized strategies: no matter how the amount of money in the envelope is chosen, in the long run these strategies will be no worse than a constant strategy.

  • No randomized strategy with limiting values of 0 and 1 dominates any of the others: if the expectation for δ when (ω,2ω) is in the envelopes exceeds the expectation for ε, then there exists some other possible state with (η,2η) in the envelopes and the expectation of ε exceeds that of δ .

  • The δ strategies include, as special cases, strategies equivalent to many of the Bayesian strategies. Any strategy that says “switch if x is less than some threshold T and stay otherwise” corresponds to δ(x)=1 when xT,δ(x)=0 otherwise.

What, then, is the fallacy in the argument that favors always switching? It lies in the implicit assumption that there is any probability distribution at all for the alternatives. Specifically, having observed x in the opened envelope, the intuitive argument for switching is based on the conditional probabilities Prob(Amount in unopened envelope | x was observed), which are probabilities defined on the set of underlying states of nature. But these are not computable from the data. The decision-theoretic framework does not require a probability distribution on Ω in order to solve the problem, nor does the problem specify one.

This result differs from the ones obtained by (1) and its references in a subtle but important way. The other solutions all assume (even though it is irrelevant) there is a prior probability distribution on Ω and then show, essentially, that it must be uniform over S. That, in turn, is impossible. However, the solutions to the two-envelope problem given here do not arise as the best decision procedures for some given prior distribution and thereby are overlooked by such an analysis. In the present treatment, it simply does not matter whether a prior probability distribution can exist or not. We might characterize this as a contrast between being uncertain what the envelopes contain (as described by a prior distribution) and being completely ignorant of their contents (so that no prior distribution is relevant).

4. CONCLUSIONS.

In the “guess which is larger” problem, a good procedure is to decide randomly that the observed value is the larger of the two, with a probability that increases as the observed value increases. There is no single best procedure. In the “two envelope” problem, a good procedure is again to decide randomly that the observed amount of money is worth keeping (that is, that it is the larger of the two), with a probability that increases as the observed value increases. Again there is no single best procedure. In both cases, if many players used such a procedure and independently played games for a given ω, then (regardless of the value of ω) on the whole they would win more than they lose, because their decision procedures favor selecting the larger amounts.

In both problems, making an additional assumption-—a prior distribution on the states of nature—-that is not part of the problem gives rise to an apparent paradox. By focusing on what is specified in each problem, this assumption is altogether avoided (tempting as it may be to make), allowing the paradoxes to disappear and straightforward solutions to emerge.

REFERENCES

(1) D. Samet, I. Samet, and D. Schmeidler, One Observation behind Two-Envelope Puzzles. American Mathematical Monthly 111 (April 2004) 347-351.

(2) J. Kiefer, Introduction to Statistical Inference. Springer-Verlag, New York, 1987.


8
This is a short article I wrote ten years ago but never published. (The new editor of the AMM saw no mathematical interest in it.) I have given talks in which I played the two-envelope game with the audience, using substantial amounts of real money.
whuber

1
Very nice write up! Joe Blitzstein talked about the two evelope problem in a Harvard Stat 110 lecture which is available free on youtube if anyone is interested btw.
Benjamin Lindqvist

@whuber Consider this variant. Suppose I choose two amounts of money such that one is twice as much as the other. Then I flip a fair coin to decide which amount goes in which envelope. Now you pick an envelope at random, and imagine the amount inside it, calling it x (if this step is questionable, consider the case of opening up the envelope and looking at the actual amount - since the reasoning applies no matter what value you see inside, it should apply with a general x). Then calculate the expected value of the money in the other envelope as E=(1/2)(x/2)+(1/2)(2x)=1.25x>x...
Zubin Mukerjee

I guess I don't understand where in that reasoning I "assumed a prior distribution on the states of nature". Did I? Clearly the reasoning cannot be correct, because I cannot justify switching to the other envelope by merely thinking about the first envelope (since the same logic would apply to the second, once I switch once).
Zubin Mukerjee

2
@Zubin There is a basic (but interesting) mistake in that analysis. Let θ be the smaller amount in the two envelopes. Given an observation of x, you know that either θ=x or θ=x/2 and that the likelihood of this observation in either case is 1/2. In the former case the amount Y in the other envelope is 2x and in the latter case it is x/2, but in order to assign a valid expectation to Y you must assume there is some probability distribution for θ. Equal likelihood is not equivalent to equal probability.
whuber

7

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 12A.

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.


IMO, the problem says that you cannot lose A no matter what. So, your +A vs -A cannot be appropriate. You either win A or 2A.
Little Alien

7

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(YX)=1.25X.

E(Y)=E(E(YX))=E(1.25X)=1.25E(X)
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×(5)+0.5×10=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×(20)+0.5×(40)=10.

Taking the average over these, the expected gain of switching is 0.5×2.5+0.5×10=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×(10)+0.5×20=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 0.25×5+0.5×5+0.25×(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: XExp(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=0.5YY=y) using Bayes' theorem:

P(X=0.5YY=y)=P(Z=1Y=y)=p(Y=yZ=1)P(Z=1)p(Y=y)=p(2X=y)P(Z=1)p(Y=y)=0.25e0.5yp(Y=y)
and similarly P(X=2YY=y)=e2yp(Y=y), wherefore the conditional expected gain of switching back to the first envelope is
E(XYY=y)=0.125ye0.5y+ye2yp(Y=y),
and taking the expectation over Y, this becomes
E(XY)=00.125ye0.5y+ye2yp(Y=y)p(Y=y)dy=0.25,
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 2n,2n+1 (P(2n<=min(X,Y)<2n+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.


I edited my question trying to explain why I think it is similar to the envelope paradox and you would want to switch forever.
evan54

@evan54 I rewrote my answer to contain my interpretation of the setting problem 3, more explanation about why the situation is not symmetric, examples etc.
Juho Kokkala

I think I'm close to getting it. I think that once there is a coin flip and envelope 2 contains half/double the amount in your hand you are basically in the situation of the envelope paradox BUT the way you got there guarantees you that you are better off switching. Does that make sense?
evan54

also, if it does, is there a way to make it more formal? I may ponder on it more..
evan54

1
@evan54 Not sure. The whole point of the paradox is that it is a situation in which there is no advantage to switching. Thus, anything you change to the setup of the problem that results in it being advantageous to switch, at least initially, must therefore not be equivalent to the setup of the two envelope paradox. Note that in your setup, it only makes sense to switch the very first time. After you switch the first time, you expect to lose by switching back. The flawed logic in the paradox comes into play if you attempt to argue that you should switch back.
jsk

4

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.


I see we interpret problem 3 slightly differently. I assumed OP specifically constructs the setting in problem 3 so that the 2nd envelope has probabilities 0.5/0.5. This is clearly possible without improper distributions, but then the possibilities for envelope 1 are not equally likely given the amount in the second envelope.
Juho Kokkala

Agreed, if OP meant that you are told that the other envelope either has X/2 or 2X with equal probabilities, then problem 3 would not be equivalent to the 2 envelope paradox.
jsk

yes that was my thinking, that in problem 3 there is equal probability between X/2 and 2X. So you hold 3 envelopes give him the 10 and then flip a coin to see if you give him the 20 or 5 (they are closed) if he decides to switch
evan54

1
@evan54 - if you make the random flip after you choose which envelope to give me, then it's equivalent to problem 1; if you choose both amounts of money, and then make a random flip on which envelope you give me, then it's the situation described above; they're different situations.
Peteris

1
@evan54 - the optimal player's decision depends on how you made those envelopes. If you don't tell the player how you did that (only that 50/50 sentence), then the optimal strategy depends on player's assumptions on how likely you are to do it one way or another - the first envelope you prepared is less valuable than the second envelope you prepared; if they were fairly shuffled (and unopened) then it doesn't matter what the player chooses; if the player thinks that you likely (>50%) initially gave him the first envelope, then player should switch and stick with that.
Peteris

1

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?)


Introducing the possibility of negative amounts is an interesting observation, but not needed for resolving the issue in your question. See my answer.
Juho Kokkala

It is not necessary to assume the amount in the envelope is random: it suffices that it is unknown. Assuming randomness adduces information--however little it might be--that was not given in the problem!
whuber

1
The biggest difference between 2 and 3 is that being told the other amount is either X/2 or 2X is not the same as being told that the two possibilities are equally likely. Assuming the two amounts are equally likely is not the same as being told the two amounts are equally likely.
jsk

1

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.


0

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.

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