Here’s a fun logic puzzle (see also here; originally found here). There’s a family resemblance to the Monty Hall problem, but the basic ideas are pretty distinct.
An eccentric benefactor holds two envelopes, and explains to you that they each contain money; one has two times as much cash as the other one. You are encouraged to open one, and you find $4,000 inside. Now your benefactor — who is a bit eccentric, remember — offers you a deal: you can either keep the $4,000, or you can trade for the other envelope. Which do you choose?
If you’re a tiny bit mathematically inclined, but don’t think too hard about it, it’s easy to jump to the conclusion that you should definitely switch. After all, there seems to be a 50% chance that the other envelope contains $2,000, and a 50% chance that it contains $8,000. So your expected value from switching is the average of what you will gain — ($2,000 + $8,000)/2 = $5,000 — minus the $4,000 you lose, for a net gain of $1,000. Pretty easy choice, right?
A moment’s reflection reveals a puzzle. The logic that convinces you to switch would have worked perfectly well no matter what had been in the first envelope you opened. But that original choice was complete arbitrary — you had an equal chance to choose either of the envelopes. So how could it always be right to switch after the choice was made, even though there is no Monty Hall figure who has given you new inside information?
Here’s where the non-normalizable measure comes in, as explained here and here. Think of it this way: imagine that we tweaked the setup by positing that one envelope had 100,000 times as much money as the other one. Then, upon opening the first one, you found $100,000 inside. Would you be tempted to switch?
I’m guessing you wouldn’t, for a simple reason: the two alternatives are that the other envelope contains $1 or $10,000,000,000, and they don’t seem equally likely. Eccentric or not, your benefactor is more likely to be risking one dollar as part of a crazy logic game than to be risking ten billion dollars. This seems like something of a extra-logical cop-out, but in fact it’s exactly the opposite; it takes the parameters of the problem very seriously.
The issue in this problem is that there couldn’t be a uniform distribution of probabilities for the amounts of money in the envelopes that stretches from zero to infinity. The total probability has to be normalized to one, which means that there can’t be an equal probability (no matter how small) for all possible initial values. Like it or not, you have to pick some initial probability distribution for how much money was in the envelopes — and if that distribution is finite (“normalizable”), you can extract yourself from the original puzzle.
We can make it more concrete. In the initial formulation of the problem, where one envelope has twice as much money as the other one, imagine that your assumed probability distribution is the following: it’s equally probable that the envelope with less money has any possible amount between $1 and $10,000. You see immediately that this changes the problem: namely, if you open the first envelope and find some amount between $10,001 and $20,000, you should absolutely not switch! Whereas, if you find $10,000 or less, there is a good argument for switching. But now it’s clear that you have indeed obtained new information by opening the first envelope; you can compare what was in that envelope to the assumed probability distribution. That particular probability distribution makes the point especially clear, but any well-defined choice will lead to a clear answer to the problem.
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Oh and there is also a bit of a cheat in this puzzle, the conceit being the implicit use of a scale (dollars). If you re-formulate the problem as:
A game master has two tokens marked A and B, and a sealed envelop which contains the game masters prior choice of the relative values of A and B. In the game you choose A or B, and then are allowed to choose the other, after which the envelop is opened.
In this context one can see the second choice is irrelevant. This also demonstrates the significant impact of the extra information of the choice of units in the puzzle. If you left out just the dollars part the puzzle is trivial.
So what this teaches us is not anything about how counter intuitive and strange probability is, but rather how tricky our implicit assumptions are.
@Aaron Sheldon
Re: Post 50. Tosh and piffle!
Re: Post 51. Excellent point, probably the most useful post so far. 🙂
@49 😉
But can anyone help with the meta-question @48?
Sean?
Beuller?
@53: The only way I have been able to personally rectify the debate is by adopting the following definition of probability that cleanly separates the role of empirical frequency:
Probability is a hypothesized inference on the frequency of occurrence of future events that must agree with the actual observed empirical frequency of events in the limit of a large number of repetitions of a trial.
There are similar statements in the literature, this is essentially an adaption of consistency. I have found this definition useful because it allows me to find a logical rectification between Bayesian and Frequentest interpretations. The reason why this works is because the statistical tests in either interpretation can be shown to be consistent estimators; the predicted probability converges to the empirical frequency. Both interpretations also suffer from the same flaws of not being particularly robust against violations of assumptions on the form of the sampling distribution.
Huh. Certainly doesn’t square with the ordinary meaning of probability. If there are 3 envelopes and 2 contain money, then the “probability” of choosing one with money is 2/3, no? Doesn’t require any prior trials. But if you’re distinguishing “odds” from “probability” — why? Your definition seems to mean we can’t deduce probabilities (2 of 3 envelopes) and can only get them by induction. Why?
Your example contains an implicit hypothesis, that 2 of 3 envelops contain money. You can only ever statistically test that hypothesis by repeating the trial a large number of times. One trial of opening a single envelop cannot meaningfully test the hypothesis that 2 of 3 envelopes contain money.
There other non-statistical method is to open all three envelops at once.
The main reason for cutting such a delicate surgery is to avoid the paradoxes involved in interpreting frequencies of events that have already occurred as probabilities, which leads to contradictory statements on the odds of observing events that have already occurred.
This is in part a method of expanding statements like:
The probability of getting a heads given the flip of a fair coin is 1/2
Into the more precise and analytically tractable:
Given the hypothesis that the coin and the toss is fair then the probability of flipping a heads is 1/2.
The former statement more clearly illustrates that it will take a large number of trials to statistically test the hypothesis.
*by paradox I mean real actual paradoxes of the same order that had to be overcome during the formulation of set theory.
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Aaron, good point about the hypothesis assumption. Couple of things, too:
There is no such thing as a fair coin. The fairness of a coin is a continuous chaotic function of time, in the real world, that changes even as the coin is in the air. And by relating ‘fairness’ of the experiment to the outcome, all statistics essentially makes a grand assumption that the probability is a frequency distribution with no underlying information. You are defining the probability of one out of two events to be one out of times frequency, if it is ‘fair’. The fairness is an illusion that can be violated very heavily depending on the nature of the chaotic function producing the outcome.
About the problem of the two envelopes, one containing twice as much money as the other, can we just say that the fact that at least one of two non-negative numbers is positive does not imply that the two numbers must be equal? Upon discovering that the envelope we have chosen contains $4000, all we really know is that at least one of the two a priori probabilities P(2000, 4000) and P(4000, 8000) is nonzero.
What happens if the money is a foreign currency that I don’t recognize and can’t value?
It seems to me that mathematically, switching should still be the correct move.
My friend, Hermione Granger, explained the problem to me this way. Call the amounts in the two envelopes that you are presented with Y and 2Y. If the envelope you chose contains Y dollars, then you will gain 2Y – Y = Y dollars by switching. If , on the equally likely other hand, you chose the envelope containing 2Y dollars, then you will lose 2Y – Y = Y dollars by switching. Hermione says there is no advantage in switching. I started to object, but I just can’t stand that patronbizing look of hers. Too bad we can’t ask Dumbledore.
I’d be like, “Woohoo, 4k in free money!”
Then think again and say to myself, “Well, 2k would be free too, and it wouldn’t take that much to earn the other 2k to get back up to my current 4k. Certainly easier than coming up with the 4k to get to the potential 8k.”
“Okay, gimme the other envelope.”
That’s right; I’m not a mathematician. 😉
I am disturbed by the contrast between “Hermione”‘s elegant solution (#61) and the gross failure of what seems to be a closely related algebraic approach i.e. why should calculating in dollar deltas work better than in dollar fractions ?
Hermione’s solution is simple and leads to the intuitively correct result, so why does the mathematical formulation of the solution in the original post fail ?
I.e. the description:
“After all, there seems to be a 50% chance that the other envelope contains $2,000, and a 50% chance that it contains $8,000. So your expected value from switching is the average of what you will gain ($2,000 + $8,000)/2 = $5,000 minus the $4,000 you lose, for a net gain of $1,000.”
would seem to be equivalent to:
The value of the current envelope is, say, X:
Vc = X
and the expected value of the other envelope is the average of 0.5X and 2X:
Vo = 0.5 * (0.5X + 2X) = 1.25X
Vo > Vc for positive X and so it is always worth switching.
Ironically, considering the original post is a non-physics problem in a physics blog, my fix is to introduce some units.
As given, the problem involves two sums of money e.g. $2000 and $4000. Call the smaller amount one “small” (S) and the larger amount one “large” (L) where the conversion rate is 1L = 2S.
Having chosen an envelope the value of the current envelope is either 1S or 1L, with 50% probability each:
Vc = 0.5 * (1S + 1L) = 0.5 * (1S + 2S) = 1.5S
When the current envelope holds an S then the other holds an L and vice versa. The expected value of the other envelope is then:
Vo = 0.5 * (1L + 1S) = 0.5 * (2S + 1S) = 1.5S
So, now Vc = Vo and there is no value in switching.
Introduction of units has forced us to consider the probabilistic nature of Vc (rather than just calling it X) and then allowed us to calculate Vc and Vo in a consistent manner i.e. in units of S.
Likewise Hermione’s solution implicitly uses a consistent set of units i.e. also my S units, where Y = 1S.
Note that efp (#41) gave effectively the same explanation as here but I was keen to clarify the contrasting success of Hermione’s solution and the original post’s approach.
The puzzle results from “stopping” in the middle and doing the expected value. That is, since the values in the envelopes don’t change, you have to calculate the expected value of the entire process of the initial choice and the decision to switch or not. So there are 4 states, pick envelope A and stick with A, pick A switch to B, pick B and stick with B, or pick B then switch to A. Let A have X dollars and B have 2X dollars. For the two states where you stick with your original choice the total value is 3X so the EV = 3X/2. For the two states where you switch, again the total value is 3X for an EV of 3X/2. So there is no value in switching.
Another way to put it is to only look at what you end up with at the end of your two choices, 2 states yield X, 2 states yield 2X all with equal probability. The “gain or loss” after making only one choice is an illusion, because it doesn’t matter whether you open the first envelope or not, it is still just as unknown (with respect to the value in the other envelope) as it was before. Since no real information is gained by opening your first envelope, your probability of having chosen the 2X envelope has not changed. Just as in the Monty Hall problem where Monty really doesn’t add any information so the probability that you chose the correct door initially remains 1/3.
@SteveM… Monty does add information because he opens up a goat door among the two doors not initially chosen. In the end, it tells you the probability of finding the car between the door you chose and the only other remaining door.
@Atider84, The reason Monty does not add any information is that you already know that there is at least one door with a goat. Monty showing you a goat does not add any information you didn’t already know. If Monty added any information then the probability of your door being the winner would have to change.