Recall that event L is when you start with 2X and event K is when you start with X

Since we don't know which it is upon seeing the money we will consider this an uninformative prior and give each a fair 50% likelihood. Then our sample space is [K,L]

In the event of L our expected gain loss sample space is [-X, 0]

In the event of K our expected gain loss sample space is [X,0]

That is the same even if you go the 1/2 route.

Let's try running a simulation on that structure.


The Result are:

x: 2528
-x: 2510
0: 4962
```

In the event of L our expected gain loss sample space is [-X, 0]

In the event of K our expected gain loss sample space is [X,0] — Jeremiah

Which is:

In the event of L our expected gain loss sample space is [-5, 0]

In the event of K our expected gain loss sample space is [10,0]

So there's more to gain by switching than there is to lose.

You are forgetting that X is a variable.

You are forgetting that X is a variable. — Jeremiah

No I'm not, hence why I'm saying it's 5 in some situations and 10 in others.

X is an unknown, so you can't treat it as a known.

At any rate I think the probability of me going to get something to eat is very high at this time.

X is an unknown, so you can't treat it as a known. — Jeremiah

I'm not treating it as known. I'm saying that there's more to gain than there is to lose. I don't need to know what X is to know that if I have £10 then by switching I am equally likely to gain £10 as to lose £5.

Let's try running a simulation on that structure.

K <- c("X",0)
L <- c("-X",0)
q <- c(K,L)
w <- sample(q, 10000, replace = TRUE)
sum(w == "X")
sum(w == "-X")
sum(w == 0)

The Result are:

x: 2528
-x: 2510
0: 4962
``` — Jeremiah

I don't know R, but from what I can gather that's just selecting an outcome at random and showing that the number of times you win is equal to the number of times you lose? That's not in question. What I'm saying is that the amount you win is greater than the amount you lose.

We've established that where X is gained, X = 10 and where X is lost, X = 5.

So the 2,528 gains amounts too £25,280 (the initial £10) + £25,280 (the gain) = £50,560 and the 2,510 losses amounts to £25,100 (the initial £10) - £12,550 (the loss) = £12,550.

So that's a total of £63,110 for switching compared to £49,620 for not switching.

And if we even it out to x: 2,500, -x: 2,500, 0: 5,000 we get £50,000 for not switching and £62,500 for switching. This is the same x1.25 figure that my result gave for switching.

If we open the envelope and see £10 then we know that the other envelope contains either £5 or £20. By switching there's a 50% chance of losing £5 and a 50% chance of gaining £10. Switching seems like the better option. — Michael

Might as well just not even bother looking in the first envelope then, and pick the second one from the start. Looking at the contents seems not to affect this decision. This tells me it doesn't matter if you switch.

My solution is to switch only if the amount in the envelope is an odd number.

My solution is to switch only if the amount in the envelope is an odd number. — noAxioms

Genius. :lol:

I don't know R, but from what I can gather that's just selecting an outcome at random and showing that the number of times you win is equal to the number of times you lose?. — Michael

That is because mine is set up correctly while yours is set up incorrectly. You are adding up sums of known numbers. Where mine is adding up an unknown, because X is an unknown.

We've established that where X is gained, X = 10 and where X is lost, X = 5. — Michael

You are letting your examples confuse you once more. What we have establish is that -X and X are equally likely to occur. If X = 5 then that is -5 or 5. If X = 10 then that is -10 or 10. If X = 15456783134.1346854654 then that is -15456783134.1346854654 or 15456783134.1346854654. That is what it means to be an unknown variable. What you are suggesting is that for each X there are two possible outcomes, and well that does not fit the definition of a function. You are not considering that L and K are independent events and that -X and X are conditional events.

|-X| = |X|, which means -X cannot equal -5 while at the same time X is equal to 10, as |-5| does not equal |10|. If X = 10 then -X=-10 and if X= 5 then -X=-5.

I really don't know many more different ways I can say this.

You are letting your examples confuse you once more. What we have establish is that -X and X are equally likely to occur. If X = 5 then that is -5 or 5. If X = 10 then that is -10 or 10. If X = 15456783134.1346854654 then that is -15456783134.1346854654 or 15456783134.1346854654. That is what it means to be an unknown variable. What you are suggesting is that for each X there are two possible outcomes, and well that does not fit the definition of a function. You are not considering that L and K are independent events and that -X and X are conditional events. — Jeremiah

You explained the scenario yourself:

Let's call case one L and case two K. Where L is the event you start with 2X and K is the event you start with X. These are two different events.

You open it up and see you have 10 bucks:

In the event of L if you switch then you gain -X, which in this case is a loss of 5 bucks. If you don't switch then you gain 0.

In the event of K if you switch then you gain +X, which is gain of 10. If you don't switch then you gain 0. — Jeremiah

We play 5,000 games. 2,500 of those games are case L and 2,500 of those games are case K. Assuming we switch, the only possible outcome for case L is a loss of £5 and the only possible outcome for case K is a gain of £10. Our total earnings for these 5,000 games is £62,500. Whereas if we don't switch then our total earnings is £50,000. Switching is more profitable.

Whether you benefit by switching or not is dependent on what envelope you just chose, which must be Envelope X or Envelope 2X where X is a given sum of money. So, switching can only ever take you from Envelope X to Envelope 2X (+X) or from Envelope 2X to X (-X).

Let X=10.

If you choose Envelope X then you get 10 and switching gains you ten. If you choose Envelope 2X then you get 20 and switching loses you 10.

The same applies for all X.

Let X=10.

If you choose Envelope X then you get 10 and switching gains you ten. If you choose Envelope 2X then you get 20 and switching loses you 10. — Baden

Let the amount in my envelope = £10.

If I chose Envelope X then I get £10 and switching gains me £10. If I chose Envelope 2X then I get £10 and switching loses me £5.

No. 20 and 5 are incompatible possibilities given you've already chosen an envelope. 20=4(5). There is no 4X.

No 20 and 5 are incompatible possibilities given you've already chosen an envelope. 20=4(5). There is no 4X. — Baden

I still don't know what you mean by this.

I'll put it another way:

Let the amount in my envelope = £10.

If X = 10 then I have £10 and switching gains me £10. If X = 5 then I have £10 and switching loses me £5.

I don't understand why you don't understand. There is no point in this scenario after an envelope is chosen where X( £5) and 4X(£20) are compatible possibilities. The second choice (to switch) is dependent on the first, which limits you to Envelope X or Envelope 2X.

There is no point in this scenario after an envelope is chosen where X( £5) and 4X(£20) are compatible possibilities. — Baden

I don't understand why you're saying that X is 5. We don't know what X is. All we know is that there's £10 in my envelope. The other envelope either has £5 or has £20, depending on whether my £10 envelope is Envelope X or Envelope 2X.

There can never be a possible spread from £5 to £20 given that you've chosen an envelope already. Given that you've already chosen and you've chosen only either X or 2X, switching can only take you from X to 2X(+X) or from 2X to X(-X). So no matter what X is you can only gain or lose that X by switching.

@Michael I'll leave it for now as we're coming at it from different angles. I think you are treating the first and second event as independent and I am treating them as dependent (the second dependent on the first). Maybe someone else has a more formal way of putting that.

If my £10 envelope is Envelope X then switching to Envelope 2X gains me £10.
If my £10 envelope is Envelope 2X then switching to Envelope X loses me £5.

Yes, it's true that in the first case it's a gain of X and in the second case it's a loss of X, but in the first case X is 10 and in the second case X is 5, so it's wrong to say that the gain and loss are equivalent.

There's more to gain than there is to lose by switching.

Take any given pair of envelopes X+2X
Let X=10 (for example)
If you choose Envelope X you see £10 and switching to Envelope 2X gains you £10
If you choose Envelope 2X you see £20 and switching to Envelope X loses you £10.

Should you switch? As there is a 50% chance you have already chosen Envelope X and a 50% chance you have already chosen Envelope 2X, it doesn't matter.

We work with what we know, and what we know is that our chosen envelope has £10. Therefore we can't consider the possibility that X is 10 and that our chosen envelope is 2X. The only thing we can consider is:

If my £10 envelope is Envelope X then switching to Envelope 2X gains me £10 and if my £10 envelope is Envelope 2X then switching to Envelope X loses me £5.

We know the chosen envelope has £10 in your scenario and we also know that one of the envelopes contained an amount X and one of the envelopes contained an amount 2X, and therefore we know we can only move from X to 2X or from 2X to X by switching. Therefore we know we have a 50% chance of gaining X by switching and a 50% chance of losing X by switching and that therefore there is no point in switching.

You're using X to refer to both the X in the scenario where switching gets you less and in the scenario where switching gets you more. But given any given amount you see in front of you after already choosing an envelope, those are not the same X.

So, you have 10. But only one scenario applies to your choice, either the one where you lose by switching in which case X is five and 2X is 10, or you gain by switching in which X is 10 and 2X is 20. You can't be in both scenarios at the same time and X has a different value in both, so your calculation fails.

What should you do? — Jeremiah

It depends on my utility function U. Utility functions map one's wealth to levels of 'utility', ie satisfaction. They are generally assumed to be monotonically increasing and concave down.

The expected dollar gain from switching after observing the envelope contains $X is positive, as it is 0.5 (2X - X) + 0.5 (X/2 - X) = 0.25 X

However the expected gain in utility from switching is

(0.5 U(W + 2X) + 0.5 U(W + X/2)) - U(W + X)

where W is my wealth prior to receiving the envelope.

This may be positive or negative, depending on the slope and curvature of the utility function U in the interval between S+X/2 and W+2X. If the slope is high and curvature is strong, expected utility will be maximised by not switching, otherwise by switching.

I'd draw a diagram, only I'm no good at doing that on a web page. But if you draw a few diagrams yourself, hopefully you'll see what I mean.

What should you do? — Jeremiah

The expected value of each envelope is $(X + 2X)/2 = $3X/2 = $1.5X. You should be indifferent between switching or not.

However consider this variation. If you choose an envelope, and then the other envelope is emptied and randomly filled with either half or double the amount of the envelope you chose, then the expected value of the other envelope is $(X/2 + X*2)/2 = $5X/4 = $1.25X. Since $1.25X > $X, you should switch.

So, you have 10. But only one scenario applies to your choice, either the one where you lose by switching in which case X is five and 2X is 10, or you gain by switching in which X is 10 and 2X is 20. You can't be in both scenarios at the same time and X has a different value in both, so your calculation fails. — Baden

So if X is 5 then I lose £5 by switching and if X is 10 then I gain £10 by switching. Nowhere am I saying that it's both 5 and 10 at the same time, so I don't understand your objection.

I'm just saying that there's a 50% chance that I'm in situation 1 (X is 5) and a 50% chance that I'm in situation 2 (X is 10). Switching is effectively betting on a coin toss, but with a 2:1 payout (I lose £5 if I lose and win £10 if I win).

We know the chosen envelope has £10 in your scenario and we also know that one of the envelopes contained an amount X and one of the envelopes contained an amount 2X, and therefore we know we can only move from X to 2X or from 2X to X by switching. Therefore we know we have a 50% chance of gaining X by switching and a 50% chance of losing X by switching and that therefore there is no point in switching. — Baden

The issue is that in using this variable X you're hiding the inequality. This is your argument above:

A1. I have £10.
A2. If I have 2X then I lose X by switching.
A3. If I have X then I win X by switching.
A4. There's a 50% chance of winning X and a 50% chance of losing X by switching.
A5. Therefore there's no benefit to switching.

Now replace the variable with the actual value:

B1. I have £10.
B2. If I have 2X then I lose £5 by switching.
B3. If I have X then I win £10 by switching.
B4. There's a 50% chance of winning £10 and a 50% chance of losing £5 by switching.
B5. Therefore there's a benefit to switching.

You're conflating different values of X when comparing the expected gain and loss. A5 is an invalid inference because it treats a win of X and a loss of X as symmetrical which is wrong because the X you win is greater than the X you lose.

The expected value of each envelope is $(X + 2X)/2 = $3X/2 = $1.5X. You should be indifferent between switching or not. — Andrew M

It looks like you're assuming the player's utility function is the identity function, which would be unusual and unrealistic. Even if we assume that, the quoted calculation doesn't take account of all the available information. There is new information, which is the known dollar value in the opened envelope. That changes the utility calculation.