# Logic question

I played the game a million times, just to see what would happen, and also because it took about 3 minutes to write the script, and it runs near instantly.

My rules are as described in the OP. You are handed the first envelope. It contains \$1 or \$2, but you don’t know which. The other envelope contains the complementary value. The statements on the envelopes hold true, because of the or in the statement.

Thinking about it this way, at least to me, makes it extremely obvious that across many runs the mean of the first set of envelopes equals the mean of the second set of envelopes. Staying or switching does not matter, because 1 and 2 are assigned randomly in each trial.

In this case, the gamemaster does not add any additional information to change the odds of staying or switching.

``````# Set number of envelope pairs
n <- 1000000
# Assign a value to the first envelope
envelopes <- data.frame(
env1 = sample(c(1, 2), n, replace = TRUE),
env2 = NA
)
# Assign the complementary value to the second envelope
envelopes\$env2 <- ifelse(envelopes\$env1 == 1, 2, 1)
# Result of always taking the first envelope
mean(envelopes\$env1) # 1.500479
# Result of always switching
mean(envelopes\$env2) # 1.499521
``````

I find it interesting that the Wikipedia article says that “no proposed solution is widely accepted as definitive”. IMO everything has been answered in this thread.

I think that’s just a result of different formulations of the puzzle. For any single proposed solution anyone expounds, you can always come up with some variant of the puzzle for which that solution doesn’t apply.

As well, wiki argues that the puzzle is not to correctly compute the expected value of switching / not switching, but to correctly formulate why the naïve interpretation is wrong.

There are lots of ways to argue why something is wrong. With lots of reasonable overlap to be expected.

But yeah, if you want to publish a new answer that no one has ever given, it’s sure helpful to create a new variation on the question being asked. Lots of “publish or perish” papers work on exactly that basis. Call it the paradox of infinite academic subdivision.

The question of what constitutes a satisfactory explanation or resolution is an interesting one.

Where there are two competing answers, both of which appear to be logically supported, merely stating or demonstrating that one of the answers is the correct one is not sufficient.

In this aspect it reminds me of the Missing dollar riddle - Wikipedia.

Or it could be that what constitutes a “solution” to this kind of informal paradox differs from person to person.

So, in a manner of speaking, the real paradox is inside the listener?

Well, maybe. It might only be half a paradox or two paradoces inside the listener.

Reminds me of another classical “logic puzzle”, which in order to solve you have to decide precisely how you are going to interpret the question:

Suppose a factory has produced tanks with consecutive serial numbers 1, 2, 3, …, up to some unknown number N, and you randomly encounter one tank whose number turns out to be, say, 42. What is the best estimate for N?

I’m going with 42.

Yeah, that’s a fun and interestingly historically relevant problem; always worth a mention. To be sure, though, note that there’s nothing particularly special about the German tank problem regarding interpretation decisions. The same interpretation and statistical framework decisions have to be made up front for any statistical inference problem.

Skimming the thread, I think this has probably been more thoroughly covered, but in my humble opinion the answer below is the correct one.

It all comes down to how the master decided to run the game, and unlike the similar Monte Haul problem, it is left entirely unspecified in this case.

Consider for example these three scenarios all starting with the same two requirements.

1. A way to generate a value M, I don’t care if its fixed, or drawn from a distribution or whatever, just write it down and put that much money into an envelope.
2. An envelope swapping machine which can randomly pick one of two envelopes.

The master then does one of three things.
A) Put \$M in the first envelope and \$2*M in the second envelope. Put them in a random mixer and hand one to the contestant leaving the other as the alternative.

B) Make three envelopes, with values \$M \$2M and \$M/2. Hand the first one to the contestant, and then send the other two to the mixer, to act as the second option they can switch to.

C) Make three envelopes, with values \$M \$2M and \$M/2. Set the first envelope aside as an alternate choice and and then send the other two to the mixer to be given to the contestant as a starting point.

In all three cases it is true that the first envelope has a 50% chance of having either half or twice the value of the second, and vis-versa, so all three scenarios fit the OP.

But in scenario A it is makes no difference which you choose, in scenario B it is better from an expected value point of view to choose the second, and in scenario C it is better from an expected value point of view to keep the first envelope rather than switch.

We have no way of knowing which scenario is actually the way the game was run, or if it was some combination with equal or unequal probabilities or if an entirely different mechanism chosen. Since the different scenarios give different optimal strategies there is no “correct one”.

Honestly not seeing your point. In the hypothetical, and your newly created ones, the player gets an envelope and then is offered a second envelope that has either twice or half as much dough.

I don’t see what changed. Perhaps you can show your math?

In reference to the last two posts:

Say M is \$10,

Let X be the amount in the participant’s initial envelope and Y be the amount in the alternative envelope.
E(X) is the expected value for staying
E(Y) is the expected value for switching
In all scenarios there are two equally likely possibilities

Scenario A: make two envelopes and pick a random one to give the contestant.
Case 1 X=\$10 and Y=\$20
Case 2 X=\$20 and Y=\$10
E(X) =\$15 E(Y) =\$15 No difference

Scenario B: Fill the participant’s envelope and then fill the alternative with either twice or half of that.
Case 1 X=\$10 and Y=\$5
Case 2 X=\$10 and Y=\$20
E(X) = \$10 E(Y) = \$12.5 better to switch

Scenario C: Fill the alternative envelope and give the participant either half or double that.
Case 1 X=\$5 and Y=\$10
Case 2 X=\$20 and Y=\$10
E(X) = \$12.5 E(Y)=\$10 better to stay.

The details changed. The statement “the other envelope has either half or double this amount” is still true in all of the cases, but knowing that isn’t everything you need to know.

Following on @Chronos excellent albeit brief point.

You are implicitly assuming that this:

then is offered a second envelope that has either twice or half as much dough.

actually says

then is offered a second envelope that has either twice or half as much dough with an equal probability for double or half.

But that’s not what is says. The statement as written

then is offered a second envelope that has either twice or half as much dough.

would still be true if there was a 100% chance the second was double and a 0% chance the second was half. The statement would still be true. What it would also be is “incomplete”.

@Buck_Godot’s excellent examples above are three very tightly distilled scenarios where the game operator can force the second envelope to have a 50/50 shot at double/half, a 100/0 shot at double/half, or a 0/100 shot at double/half.

And any of those three options completely complies with the actual problem statement.

Why is that? Because that problem statement is deliberately misleadingly incomplete. And you’ve inadvertently unknowingly substituted my added words onto your mental model of what the problem statement means.

But those words are yours, not the problem’s. The problem writers very carefully set that ambiguity trap and baited it with a big tasty blob of “common sense”. And you walked right into it and ate the bait. As do the vast majority of non-statistics types who see this problem. Including me wayy upthread.

That’s the actual stated goal of this problem in it’s “official form”. To notice the missing words people assume are there but aren’t really.

Now once folks understand that, it’s time to move on to Chapter 2: stay or switch?

In the actual problem, the game runner could have pre-chosen any of

a 50/50 shot at double/half, a 100/0 shot at double/half, or a 0/100 shot at double/half.

as their envelope filling and selection method without telling you which they chose. Your optimal strategy changes depending on which they chose. The ideal guaranteed winning strategy under scenario 2 is the exact opposite: a guaranteed losing strategy strategy under scenario 3. And vice versa.

Faced with three competing incompatible strategies and no way to pick, the rest of the meta-math says “It does not matter in the long run.” And it doesn’t. Just pick an envelope (or be given one) and be happy.

This is just like the naïve approach to the Monte Hall problem. There’s two choices so they must be 50-50, right? That’s like saying your lottery ticket will either win or not so the odds of winning are 50-50.

No not quite the same.

In the Monte Hall problem, the rules were effectively fully specified. The goat is behind one of three doors, effectively at random, and after you pick a door, Monte opens a door that you didn’t pick and which also didn’t contain a goat, effectively giving you partial information that makes it advantageous to switch.

In this case the rules are not fully specified. All that is specified is that the master put money in envelopes in such a way that one envelope contained twice or half the amount of the other with equal probability.

There are many ways that this could happen. I specified three possible scenarios which fit the bill. By the construction of the envelope swapping machine all of them result in two equally likely possibilities. And further by construction they all satisfied the specified requirement of the game.

Was I arguing that one of these scenarios was how the game was actually run? Definitely not. In fact I was arguing the opposite. I was arguing that we have no idea how the game was run, and pointed out three possible ways it could have been run that have different optimal strategies. Since we have no way to decide what way was actually used, there is no more optimal strategy to choose the envelope than there is an optimal strategy to playing rock, paper, scissors.

ETA: I just realized the in my original writeup I said,

An envelope swapping machine which can randomly pick one of two envelopes

I should have further specified that it would choose them with equal probability, rather than leaving it as an assumption.

Too late to add a TLDR:

The unlike the lottery, the two equally like possibilities were enforced by the way I wrote the scenarios, which in turn was made so as to satisfy the following requirement in the OP.

(quote edited to fix typo and add bolding)

To me, I read the following as meaning that there are only two envelopes. Not three, not two plus a cache of hidden ones, but just two. (I am reading that “change” is a typo for “chance”. If that’s not the case, then I’m not sure what’s happening here.)

I think if we take “equal chan[c]e” literally, then the envelopes are being handed out randomly, otherwise the chance would not be equal between half and double.

If we can accept that “two” or “equal” is a lie, then we can’t trust any aspect of the game, and can’t draw any conclusions, because it’s all rigged. The ball is always under one of the cups you did not pick.