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# The two envelopes problem

factor-fiction | 15:15 Thu 10th Nov 2011 | Riddles
The Deal or No Deal thread reminded me of two tricky probability/choice problems. One is the Monty Hall goat and cars problem which has been covered on here several times before so I'll leave that one.The other is the Two Envelopes problem:

Suppose you are in a Game Show. The host has 2 identical looking envelopes The host then tells you that one envelope contains twice as much as the other but neither of you knows which is which.

He asks you to choose one envelope and says you can open it and he will then offer you the chance to 'swap' or 'stick'.

You pick an envelope and open it to find it contains £2400. You think
"Mmm. The other envelope contains either £4800 or £1200. Should I swap or stick?"

What would you do?

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I've just read this all the way through, now I need to go and lie down.....
http://en.wikipedia.o...Two_envelopes_problem
15:24 Thu 10th Nov 2011
Question Author
High O_G. Maybe I should not have mentioned an amount in the original example. I just thought it might be easier for people to visualise numbers rather than talk about £X or £Y.

But I think I did say that the rules were specified at the outset and the amount in the chosen envelope wasn't revealed until after the rules had been set out and the envelope had been selected. The original post said:

"Suppose you are in a Game Show. The host has 2 identical looking envelopes The host then tells you that one envelope contains twice as much as the other but neither of you knows which is which. He asks you to choose one envelope and says you can open it and he will then offer you the chance to 'swap' or 'stick'"
If we are going to assume that:

1) people do not care about risk (counter to reality)
2) there is only one swap allowed (implied, as you get to open first the envelope after you choose it)
3) this is a one shot deal (no multiple rounds of the game, as some examples have assumed)

then there is no issue here. You one the first envelope and it contains X. You know the second envelope contains 2X or X/2 with equal probability. So the expected value is .5(2X) + .5(X/2) = 1.25X. Since this is greater than X you always swap.

The complicated problem (as in the wiki article) is when you can't open the envelope and can swap back and forth repeatedly.

If we allow the realistic aspect of risk aversion (hardly a "human frailty" but a well established fact of life) we get back to the fact that the choice to swap or not is an individual one.
"you OPEN the first envelope" gah!!
PS for an example of how risk aversion affects the choice see the first example in this wiki article; very similar to the example here:

Question Author
Ah, but in that case why even bother opening the envelope, drb? Your theory works whatever the value of X. If it's thought to be better to swap regardless of the value of X what is the point of opening the first enevelope
Correct. As long as you are risk neutral there is no need to know the value of X.

Again, the interesting parts of the problem arise when you can swap back and forth repeatedly (only interesting if you CAN'T open the first envelope), and if you allow risk aversion (a contestant will react differently to X = 100 vs. X = 100,000 and therefore will want to open envelope 1.)
Question Author
... well if there's no need to open the first envelope and you'd always swap, then that gives me a problem. That means that on average the second envelope is more attractive than the first. But how can that be when you are picking them at random?
"Attractive" only to a risk-neutral contestant, of which there are none (no, I'm not going to get off that point!!)

But, "first" and "second" can only be defined after you've picked, and therefore aren't random. They are like Schroedinger's envelopes!
"Human frailty" is a well established fact of life. However I don't feel taking it into consideration is helpful in deducing the correct answer here. IMO it just adds noise.

I'd need some convincing that swopping was not beneficial in the long term (i.e. multiple offers) since the spreadsheet shows it is. And it is the long term that allows you to uncover any bias. Sure the original question implies the rules are set out in advance but that doesn't seem to bee the issue. The rules are about one amount being double the other, they say nothing about the total amount of cash involved in this particular round.

I agree it is an interesting one though. For a 'one off' round one would think the same benefit would hold as for the loner term, but that is far less clear. I don't understand yet why knowing the value of the first envelope appears to make a difference as common sense is telling me it ought not.
Geezer. The procedure you model in your spreadsheet does not reflect the way the game show host acts.

The conclusion you draw from it would be valid if the organisers put £ 2,400 in an envelope then spun a coin to determine whether £ 1,200 or £ 4,800 goes into the second.

But they don't. They decide (we don't how) two amounts, one twice the other, to go into two envelopes. They then pass one of these chosen at random to the contestant. The difference between the two procedures does make a difference!

I realise that you always gave the contestant a £ 2,400 envelope because that figure was given in the OP, but Factor30 has explained that this figure wass just an example. The game show would provide poor entertainment, if successive contestants received the same amount each time!

So can you agree that the real problem is : Two envelopes are prepared, one containing twice the amount of the other. One of these, chosen at random, is given to a contestant who sees the amount and may then either keep it or exchange it for the other?
To test the benefits of swapping, your spreadsheet might be something like this
Column 1 - a random amount, chosen between limits
Column 2 - an amount twice Column 1
Column 3 - a random choice of Col 1 or Col 2
Column 4 - the amount not put into Col3
The total of Col 3 is the amount won with no swaps
The total of Col 4 is the amount won with swapping
Question Author
I agree J-J!

I spent many days going down the route of working out X then 2X and X/2 options and using spreadsheets and simulated outcomes even though I felt intuitively it seemed wrong that exchanging one mystery envelope for another one would be more beneficial than sticking. After wrestling with it for a week I realised that I'd been going off track and (ignoring human frailties/attitude to risk issues) there was no mathematical case for swapping.

Yes, J-J, there are only 2 envelopes so simulating outcomes based on 3 values (in this case 1200, 2400 and 4800) is not valid. If we know one envelope contains 2400 then the other contains only one of the other values- ie 1200 or 4800.

Suppose (unknown to the participant) the 2 envelopes were 2400 and 4800. If he picked 2400 he'd gain 2400 by swapping. If he picked 4800 he'd lose 2400 by swapping. So it's just a 50/50 gamble as to whether he'll lose £2400 or gain £2400 by swapping.

Similarly, consider what happens if (unknown to the participant) the 2 envelopes were 2400 and 1200. If he picked 2400 he'd lose 1200 by swapping. If he picked 1200 first he'd gain 1200 by swapping. So it's just a 50/50 gamble as to whether he'll lose £1200 or gain £1200 by swapping.

Another way to think of it is, rather than one player playing lots of times, have 1000 people all playing with the same pair of envelopes- one containing £X and the other £2X. Roughly half would pick the £X envelope and half would pick £2X. Those that pick the £X envelope would gain £X by swapping; those that pick the £2X envelope would lose £x by swapping.. So on average the gains and losses from swapping cancel out.

Anyone for the Monty Hall goats and cars problem now? Not me!
Factor30 - "there are only 2 envelopes so simulating outcomes based on 3 values (in this case 1200, 2400 and 4800) is not valid" That's a brilliant, succinct argument - why didn't I think of it? I've remembered another paradox and adapted it to this problem. I'll work on it and post later.
I don't think we disagree with the logic, but I disagree the specifics given as clarification should have been ignored That is new information, and it has an effect on the conclusion one draws. And I defend the conclusion I came to.

Certainly if the question is posed without the additional limitations the benefit of the swop disappears, as I agreed earlier.

"simulating outcomes based on 3 values (in this case 1200, 2400 and 4800) is not valid", is true for either a single go, or when the amount if the first envelope is not constant. Otherwise each round can vary between 2 scenarios, and the 3 values are valid.

I personally like considering the fixed first envelope value variant. It seems to highlight a difference between a single round, and multiple rounds, which is intriguing.

It's obvious if you think about it - If there was an automatic advantage inherent in the act of swapping, you'd be able to pick the first envelope - pause for a moment without even looking at the contents, and then swap it for the other one - which of course makes no sense whatsoever.

What's amazing is that something so simple can be so complicated.
Question Author
I agree, ludwig. Even though I know the answer, niggling little doubts emerge occasionally as I find myself puzzling over it every now and again.

There is a similar problem I must dig out. Something to do with a bag contains 3 balls and you are told there's at least one white and at least one black, and you spot a black ball in the bag. I can't remember the rest but I know I saw conflicting solutions.

There was also one on Answerbank a year or so ago where I recall there is a family with two children (I think) and you know at least one is a girl. You knock on the door and girl answers. What is the probability the other child is a boy. I can't remember it exactly but I know it caused a lot of debate and I don't think I agreed with the answer given. Again I'll see if I can remember it.

The best of all is the Monty Hall 'Goats and Cars' problem . Look it up if you don't already know it
Question Author
Here's the 3 bags of balls problem, Ludwig, if you fancy a go.

You are shown 3 identically looking bags.
You are told that in the first bag there are 2 white balls, in the second bag there is 1 white ball and 1 black ball, and in the third bag there is 2 black balls.
The bags are mixed up.
Without looking, you reach into one of the bags at random and pull out a ball.

If this ball is black, what is the probability now that the remaining ball in that bag is also black?

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