Comment by kazinator | Hacker News Reader

Comment by kazinator | Hacker News Reader

It is very simple.

Let's represent the poorer envelope by the value 0.5 and the more bountiful one by 1.0.

The expected value from the envelope draw is 0.75: 0.5 drawn with a 50% probability contributing 0.25 to the expected value, and 1.0 drawn with a 50% probability contributing another 0.5 to the expected value for a total of 0.75.

If you were to do a large number N of these draws, the value you will obtain will be from a distribution that is centered on 0.75 N. So far so good?

So the way to look at a draw is this: you're going to be getting the expected/average value of 0.75, together with either a 0.25 bonus if you pick the better envelope, or a else -0.25 penalty if you pick the poorer envelope.

So you see, those two situations are equal and opposite. By switching, you either turn a 0.25 bonus into a -0.25 penalty (- 0.5) or vice versa (+ 0.5) with equal probability.

If you need any more complex analysis than this, you need to be thwacked on the head.

The rhetoric about doubling versus halving the money is a distracting red herring. If you switch from the 1.0 envelope to the 0.5 envelope, you're down 0.5, and if you switch the other way, you're up 0.5. That's it.

Accounting is based on debits credits not halving and doubling. You would never update a ledger by crediting double some amount to one account, and debiting half the amount from another account. It's all purely additive.

You have to look at what you're potentially gaining or losing on its own not as a fraction of some guaranteed fixed portion that you're getting from either envelope.

>The puzzle is to find the flaw in the very compelling line of reasoning above. This includes determining exactly why and under what conditions that step is not correct, in order to be sure not to make this mistake in a more complicated situation where the misstep may not be so obvious. In short, the problem is to solve the paradox. Thus, in particular, the puzzle is not solved by the very simple task of finding another way to calculate the probabilities that does not lead to a contradiction.

I did solve the paradox: it emanates from falsely looking at the switch as doubling or halving the money.

If the envelopes contain, say, $50 and $100, it looks like a doubling/halving. But nothing materially changes if we add $1000 to each one to make $1050 and $1100. The probabilities don't depend on what is in the envelopes, and the win/loss is just a spread of the difference between their values.

Why would anyone, while solving some problem, worry about some unspecified more complicated situation which supposedly resembles that problem?

Give me the specific of that more complicated problem and I will knock it out of the ballpark using whatever reasoning fits that problem.

You are missing the point. The problem of whether you should actually switch is obvious and trivial (you should not). The challenge here is explaining why the plausible 'proof' that you should switch is wrong.

I read that in the Wikipedia page. There is some obvious argument involving repeatedly switching sides and whatnot, and it has some subtle logical flaw. Why bother with that at all.

OK, but let's look at it:

1. Denote by A the amount in the player's selected envelope.

2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.

3. The other envelope may contain either 2A or A/2.

OK, here is where this argument got seduced by the multiplicative thinking. The correct reasoning is: let's call the smaller base amount A and the larger one 2A. One envelope contains A, the other one 2A. That's it.

3. If A is the smaller amount, then the other envelope contains 2A.

4. If A is the larger amount, then the other envelope contains A/2.

Here, the argument is splitting into cases because of the IF, and in these two cases, A has a different meaning.

5. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.

6. So the expected value of the money in the other envelope is:

This is wonky. The overall two-envelope situation has an expected value, not the individual envelope.

The correct view is that if this is the richer envelope, it contains X more money. If it is the less endowed one, then it contains X less.

    1        1           5
    - 2A +   - (A/2) =   - A
    2        2           4

7. This is greater than A so, on average, the person reasons that they stand to gain by swapping.

That calculation is wrong though, by itself. That issue should be attacked directly.

Separate calculations have to be performed for the two cases: A is the smaller amount and A is the bigger amount. For each of these cases, the expected value from both envelopes has to be calculated.

The way the above formula is doing the expected value calculation is off because it's equivocating on A.

It ends up working with two unrelated quantities: 2A and A/2. 2A is four times bigger than A/2. A is simultaneously denoting the smaller amount and the larger one.

8. After the switch, denote that content by B and reason in exactly the same manner as above.

Yes, and if you do, you will end up with 5/4B. So that can't be right and the whole remaining argument about infinite switching being irrational compared to just opening an envelope is superfluous.

Because the variable A refers to an amount, has to have a consistent meaning across the two cases, and so it has to be:

1. We have A; the other envelope has 2A; or else

2. We have 2A; and the other envelope has A.

In case 1, if we switch we gain A.

In case 2, if we switch, we lose A.

We have no idea which case we are in.

We cannot use the variable A to denote our initially selected envelope in both cases, because then A simultaneously represents the smaller amount and the larger amount, and we are absurdly mixing the quantities A/2 and 2A in the same calculation. No quantities in the actual problem are related by a factor of four!

People might be seduced into the A equivocation perhaps because they are initially thinking of A as being a label for the envelope, which is arbitrary and innocent enough. We have some envelope A and it has some money in it. Oh, but the other envelope then has either A/2 or 2A. At that point, A is no longer an envelope name, but a name for an amount. Then it has to pin down a specific number: it has to denote either the smaller value, or the larger one, and that value is found in different envelopes in the different cases; A cannot both be "name of initially chosen envelope" and "amount in that envelope".

I'm not sure it's that simple.

let's imagine the first envelope contained between 1 andand 1000 dollars, the second double the first.

if you open your envelope and it contains a whole number of dollars between 1 and 1000, then you should swap, id take "250 or 750" over 500.

Only if each value is equally likely. If you see $1,000 but figure the envelope-filler is a lot more likely to have been willing to put $1,500 in than $3,000 then you should stick.

But the problem is explicitly about swapping before opening any envelope.

Reply on news.ycombinator.com