No.

The value of X is fixed. The two possibilities you consider must treat X as the same. You don't know what X is; but you do know that whatever it is, it does not change across the two scenarios you are considering. Your fallacious reasoning assumes that it does. — Snakes Alive

The amount in my envelope is also fixed. So as the amount in my envelope is fixed to £10 and if X is fixed to 10 then it would be false to say that there's a 50% chance of switching into X and a 50% chance of switching into 2X. We can only say that there's a 100% chance of switching into 2X (or if X is fixed to 5 then a 100% chance of switching into X).

But this isn't a model that we as participants can use. We don't know the value of X. The only information available to us is:

1. There's a 50% chance that my envelope is envelope X, and
2. My envelope contains £10.

You've accepted these premises. And from these we must infer:

3. There's a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20.

And you've accepted this as a valid inference. So it is as I said: a 2:1 payout with an even chance of winning. That's a good bet to make.

The amount in my envelope is also fixed. So as the amount in my envelope is fixed to £10 and if X is fixed to 10 then it would be false to say that there's a 50% chance of switching into X and a 50% chance of switching into 2X. We can only say that there's a 100% chance of switching into 2X (or if X is 5 then a 100% chance of switching into X). — Michael

No.

We're talking about epistemic possibilities. There are two, equipossible: first, that you have drawn X, and second, that you have drawn 2X. While you know how much you have drawn, you don't know whether this is X or 2X.

Hence, there is a 50% chance of your having drawn each, and a 50% chance of switching into the other, if you decide to switch. This is the same opportunity that not switching gives you.

3. There's a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20. — Michael

This conclusion does not show what you want it to. From this, it does not follow that switching is advantageous. For if the other envelope contains 5, you have drawn 2X, and if it contains 20, you have drawn X. In either case, your average output as a proportion of X is 1.5. This is the same as if you had not switched.

While you know how much you have drawn, you don't know whether this is X or 2X.

Hence, there is a 50% chance of your having drawn each, and a 50% chance of switching into the other, if you decide to switch. — Snakes Alive

That's exactly what I've said.

There's a 50% chance that my £10 envelope is envelope 2X, so a 50% chance of a switch getting me £5.

There's a 50% chance that my £10 envelope is envelope X, so a 50% chance of a switch getting me £20.

This is the same opportunity that not switching gives you.

In terms of probability, sure, but not earnings. The average of the above is £12.50, but the average of not switching is £10.00.

In terms of probability, sure, but not earnings. The average of the above is £12.50, but the average of not switching is £10.00. — Michael

I demonstrated that this is fallacious in my longer post above. It falsely assumes that the amount drawn is a variable independent of X. It is not.

There's a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20. — Michael

This conclusion does not show what you want it to. From this, it does not follow that switching is advantageous. — Snakes Alive

Yes it does. It shows that there's a 2:1 payout with an even chance of winning. How is that not advantageous?

Yes it does. It shows that there's a 2:1 payout with an even chance of winning. How is that not advantageous? — Michael

The payout is on average 1.5X. This is the same payout that not switching affords.

When betting the odds, all variables are independent. That is not the case here.

The payout is on average 1.5X. This is the same payout that not switching affords. — Snakes Alive

How do you get that?

If we play that game 100 times and switch then 50 times I win £20 and 50 times I win £5. That's £1,250.

If we play that game 100 times and I don't switch then 100 times I win £10. That's £1,000.

In order to reason this way, we must hold X constant for each case. If we played 100 times, where X = 10 in each game, then switching 50 times will net us on average 15 per game, thus 1500. This is the same payout as it we didn't switch at all, switched 75 times, or any other combination of switching or not switching.

The fallacy is that instead of holding constant X, you are holding constant the amount drawn, and acting as if this variable is independent of X.

I’m talking about the specific situation where I have £10 and there’s a 50% chance that the other envelope contains £5 and a 50% chance that the other envelope contains £20. This offers a 2:1 payout with an even chance of winning, and so is an advantageous bet.

The fact that a second bet with the same envelopes but a different initial pick would offset any winnings doesn’t change this. As a single bet that first bet is worth making.

The number of bets made is irrelevant.

Then forget about averages and just consider a single game. There’s a 50% chance of winning £10 and a 50% chance of losing £5. Is it advantageous to bet? I say yes.

That's not the game being played. There is no bet of an independent amount that returns half or double at equal rates. The amount seen first is itself already a function of X.

It is the game. I know that there’s £10 in my envelope. There’s a 50% chance that the other envelope contains £5 and a 50% chance that it contains £20. You’ve agreed with this. So switching has a 50% chance of winning me £10 and a 50% chance of losing me £5.

That doesn't model the case we're considering, which is that we know we have £10. — Michael

It models it precisely. Your knowing what's in one envelope does not change what's in the other envelope.

Here's an example.

Suppose for a given trial, X = 12. There's an envelope with 12 in it and another with 24.

Suppose you pick the one with 12 and look at it. Y = 12. We'll call the amount in the other envelope Z. You reason, quite correctly, that

(1) Y = 12 → . Z = 6 ⊻ Z = 24

This is absolutely true because Z = 24.

Suppose you picked the 24 envelope. Y = 24. You reason, quite correctly, that

(2) Y = 24 → . Z = 12 ⊻ Z = 48

This is absolutely true because Z = 12.

What you have to accept is that for each trial there are only 2 values in the space. In the example above, Z = 6 = X/2 is not actually a possibility; Z = 48 = 4X is not actually a possibility. You can choose between 12 and 24 as many times as you like and you will never, ever get either 6 or 48. Your thinking they are possibilities does not make them so.

It models it precisely. Your knowing what's in one envelope does not change what's in the other envelope.

Here's an example.

Suppose for a given trial, X = 12. There's an envelope with 12 in it and another with 24.

Suppose you pick the one with 12 and look at it. Y = 12. We'll call the amount in the other envelope Z. You reason, quite correctly, that

(1) Y = 12 → . Z = 6 ⊻ Z = 24

This is absolutely true because Z = 24.

Suppose you picked the 24 envelope. Y = 24. You reason, quite correctly, that

(2) Y = 24 → . Z = 12 ⊻ Z = 48

This is absolutely true because Z = 12.

What you have to accept is that for each trial there are only 2 values in the space. In the example above, Z = 6 = X/2 is not actually a possibility; Z = 48 = 4X is not actually a possibility. You can choose between 12 and 24 as many times as you like and you will never, ever get either 6 or 48. Your thinking they are possibilities does not make them so. — Srap Tasmaner

So let's say I flip a coin and it lands heads, but I hide the result from you. How would you model the probability that you will win with a guess of tails? You might say 50%, but then I counter with: "But suppose it's heads. Then you'll never win, no matter how many times you guess tails."

That's just a ridiculous response.

Our models only work with what we know. In your case, you only know that a fair coin has a 50% chance of landing heads. In my case, I only know that there's a 50% chance of my envelope being envelope X and that my envelope contains £10.

What we know are the constants. What we don't know are the variables.

Not going to get dragged into some other scenario with you.

You are in effect demanding that I figure out what's wrong with your analysis and explain it to you. I'd love to. I'd love to have your help doing just that, once you admit that my analysis shows yours is wrong, even if we're neither of us quite sure why.

It is a simple fact that what's in the other envelope does not change, and that each envelope has one of two values.

I'm trying to think of a new way to explain the fallacy being committed, but I think I've hit a roadblock (not in the matter itself – I understand it, but am not sure how to explain it, given that the above hasn't worked).

Here is as blunt as I can put it.

In deciding that switching is a "good bet," you must average the amount gained (lost) from switching across two situations. This requires you to take the value of what you picked first (say 10) as constant across those two situations, so that you can determine how much you would get as a portion of this (double, or half). The problem is that if you do this, then you are changing the value of X across the two situations, and treating things as if there is one epistemic possibility, say that X = 5, and another, say that X = 10. While this is on one sense so (and is reflected already in saying it the way I've put it: that you may have either drawn X or 2X), the problem comes when you try to use this changed value to average the two situations.

You can't do this, since it presumes there are two epistemic possibilities, each of which has a distinct value for X. This is not so: we know a priori that whatever X is, it is fixed. Hence the epistemic possibilities appropriately described are that, given some value X, with we have drawn that, or double that. They are not that given we have drawn Y, the true value of X is either double that or equal to that. This is fallacious, since it sneaks into our epistemic space two distinct values for X, across which we compare the value of Y as if it were an independent variable. Hence the illusion that we are 'getting' 1.25Y on average, as I explained in my first post.

––––

This conceptual explanation seems to be getting us nowhere, so I wonder if another tack might work. First is the empirical: when the game is modeled, this demonstrates that you are wrong. Multiple trials show the two strategies converging. Hence, you should be able to appreciate that there is an error in your reasoning in fact, though you can't pinpoint it.

Another way to go might be to devise other ways of framing the problem that demonstrate how thinking that a switching strategy is beneficial is absurd. For instance, if we always switched, then which envelope we picked first to look at would always determine which we chose (viz. the other one). But then, looking at 1 would result in picking 2, and looking at 2 would result in picking 1. In either case, what we picked would be advantageous given what we looked at. But then, this leads to the absurd conclusion that beginning with either envelope to look at (and hence de facto picking whichever envelope) is equally advantageous. All switching does is change which envelope is picked; the switcher is thus committed both to thinking that switching is advantageous and indifferent.

All switching does is change which envelope is picked; the switcher is thus committed both to thinking that switching is advantageous and indifferent. — Snakes Alive

That's clever.

Here's a new version that also calculates the average payouts as ratios to X.

That is not a Classical statistics interpretation. Classical statistics is based on using estimated variance.

However, while we are on this topic, there are five steps in the Baysian method. The last step in a Baysian method is to check your model against the data. It requires empirical verification.

If you simulated a data set for the two envelopes, then did a Classical analysis and a Baysian analysis they would both come out to the same solution.

Even, and this is important, if your prior was a loss of 5 and a gain of 20. The MCMC chain would drag the misleading prior back on track to reflect the data. A Baysian method is not an excuse to skip empirical verification.

The problem is that if you do this, then you are changing the value of X across the two situations, and treating things as if there is one epistemic possibility, say that X = 5, and another, say that X = 10. While this is on one sense so (and is reflected already in saying it the way I've put it: that you may have either drawn X or 2X), the problem comes when you try to use this changed value to average the two situations.

You can't do this, since it presumes there are two epistemic possibilities, each of which has a distinct value for X. This is not so: we know a priori that whatever X is, it is fixed. — Snakes Alive

When you average it you change the value of the first envelope chosen, despite the fact that the question is "what should I do, given that I have £10 in my envelope?".

These are incompatible premises:

1. X = 10 (or X = 5)
2. My envelope contains £10
3. There's a 50% chance of winning X and a 50% chance of losing X.

You have to pick from 1 and 3 or 2 and 3, and given that the situation I'm modelling is my current situation where my envelope contains £10, I opt for 2 and 3.

Although you could pick 1 and 2 and abandon 3, but then that leaves us with:

4. There's a 100% chance of winning X (or a 100% chance of losing X).

So we conclude that switching is advantageous (or sticking is advantageous).

Multiple trials show the two strategies converging.

That's because although it's a 2:1 payout with an even chance of winning the bet when you lose is twice as big as the bet when you win. Obviously you're going to break even. It would be like flipping a coin and hiding the result (heads) and having one person bet £10 on heads and another £20 on tails and offering them both a 2:1 payout. It's an advantageous bet for both people (given that they don't know the result or about each other's offer), despite the fact that the average payout is £0.

So with this in mind I have a new strategy for repeated games. You switch on your first game and then in subsequent games only if your chosen envelope is less than or equal to half the maximum seen.

http://sandbox.onlinephpfunctions.com/code/7c601c1099f15ea278eb6ccb4922e6c96b7ca644


Surprise surprise, there's a gain of .25.

That's because although it's a 2:1 payout with an even chance of winning the bet when you lose is twice as big as the bet when you win. Obviously you're going to break even. It would be like flipping a coin and hiding the result (heads) and having one person bet £10 on heads and another £20 on tails and offering them both a 2:1 payout. It's an advantageous bet for both people (given that they don't know the result or about each other's offer), despite the fact that the average payout is £0. — Michael

There is some misunderstanding here. The number of trials is irrelevant to the effectiveness of the strategies. If multiple trials cause you to break even on average, so will one.

Also, a 2:1 payout would be getting $30 on a win and nothing on a loss, after betting $10.

Also, a 2:1 payout would be getting $30 on a win and nothing on a loss, after betting $10. — Snakes Alive

Yes, so we bet £5 of our £10 and if we win we get £15 back (+£5 left over = £20) and if we lose we get £0 back (+£5 left over = £5).

There is some misunderstanding here. The number of trials is irrelevant to the effectiveness of the strategies. If multiple trials cause you to break even on average, so will one. — Snakes Alive

Then why does this work?

It works because you limited the selection of numbers to below 400, and so funneled the choice to switching only when lower numbers relative to that range appear. This is not a feature of the original example, and if there were no a priori range from which the values were chosen, the strategy would have no benefit.

It works because you limited the selection of numbers to below 400, and so funneled the choice to switching only when lower numbers relative to that range appear. This is not a feature of the original example, and if there were no a priori range from which the values were chosen, the strategy would have no benefit. — Snakes Alive

There's always going to be a maximum because there's a finite number of games. One (or more) of them is going to have the highest randomly selected X.

And my strategy doesn't require knowing that 400 (or whatever) is the maximum. The choice is determined by prior envelopes, not by some limit as to what can possibly be selected.

I would prefer not to get into this topic, because it's off-subject and the errors made here are strictly orthogonal to the errors driving the original misconception.

But it doesn't matter if there's a finite number of games, or how many games there are. What matters is you baked a hard limit into the selection of X. Wherever X can be at most, say 400, then one knows a priori never to switch when what one sees in the envelope >400 (since these can only be the 2X).

It doesn't matter that the strategy doesn't require knowing. Wherever there is some limit, albeit unknown, the strategy will funnel you towards figuring out what that limit is.

The strategy relies on something not present in the original example, and so is irrelevant.

So you're saying that if there is no maximum then my switching strategy wouldn't make a difference, whereas if there is a maximum then my switching strategy as modeled above does offer a benefit?

I'd have thought that this is true only if there are diminishing returns as the number of possible X envelopes increases (as the limit approaches infinity?), but as far as I can tell there isn't. There's always a gain of .25. Although the most I can test up to is 9,223,372,036,854,775,807, so what do I know?