## Mathematical Conundrum or Not? Number Six

• 1.1k
Not sure I can find another conundrum that will generate as much discussion as our last one; however, it is time to move on.

So this one is called the Two Envelopes Paradox and this is how it goes:

You are playing a game for money. There are two envelopes on a table.
You know that one contains $X and the other$2X, [but you do not
know which envelope is which or what the number X is]. Initially you
are allowed to pick one of the envelopes, to open it, and see that it
contains $Y . You then have a choice: walk away with the$Y or return
the envelope to the table and walk away with whatever is in the other
envelope. What should you do?

What should you do?
• 6.8k
By £2X do you mean twice what’s in the other envelope?
• 1.1k

Yes.
• 6.8k
OK, so the initial answer is that it doesn’t matter as there’s a 50% chance of having picked the more valuable envelope and switching doesn’t change the odds.

The paradox supposedly arises when you consider that switching into the more valuable envelope doubles your winnings, whereas switching into the less valuable envelope halves your winnings, so there’s more to gain than there is to lose, and as each is equally likely, switching is the better choice.
• 6.8k
As a bet it seems like a 2:1 payout? You have £10. You bet £5 on heads. If you win you get £15 back. And you're betting on a coin toss. Is that right?
• 1.1k

I am not sure that is quite right, as you have added information that was not in the OP. In your example you have a starting amount that is known, but in our case you don't know your starting amount.

You could have X or 2X. If you have X and you switch then you get 2X but lose X so you gain X; so you get a +1 X. However, if you have 2X and switch then you gain X and lose 2X; so you get a -1 X.
• 6.8k
I am not sure that is quite right, as you have added information that was not in the OP. In your example you have a starting amount that is known, but in our case you don't know your starting amount.

You bet $\frac{x}{2}$ on heads. If you win you get $\frac{3x}{2}$ back. The odds of winning are $\frac{1}{2}$.
• 6.8k
You could have X or 2X. If you have X and you switch then you get 2X but lose X so you gain X; so you get a +1 X. However, if you have 2X and switch then you gain X and lose 2X; so you get a -1 X.

The amount you have is $x$. The other envelope contains either $2x$ or $\frac{x}{2}$. If it's $2x$ then you gain $x$ by switching. If it's $\frac{x}{2}$ then you lose $\frac{x}{2}$ by switching.
• 1.1k
You just said the same exact thing I did, as it can only be half X if you start with 2X. Which still results with a -1 X.
• 1.1k
• 6.8k
You just said the same exact thing i did

It's not. Mine is closer to how one would understand it were we to look in the envelope before deciding.

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.

Am I right with this reasoning? If so, why would not looking first make a difference? The odds are 50% that the other envelope contains twice as much whether we look or don't.
• 1.1k
It's not. Mine is closer to how one would understand it were we to look in the envelope before deciding.

It is the same thing, the math comes out exactly the same.
• 6.8k
It is the same thing, the math comes out exactly the same.

It's not the same thing.

I'm saying that if I have £10 then I either lose £5 by switching or gain £10.

You're saying that if I have £10 and the other envelope contains £5 then I lose £5 by switching and that if I have £5 and the other envelope contains £10 then I gain £5 by switching.

Notice that in my example it's better to switch whereas in your example it isn't.
• 6.8k
So there are two different ways to describe the situation, each leading to a different conclusion:

1. I have $y$ and the other envelope contains either $2y$ or $\frac{y}{2}$.

2. Either I have $x$ and the other envelope contains $2x$ or I have $2x$ and the other envelope contains $x$.
• 1.1k
I'm saying that if I have £10 then I either lose £5 by switching or gain £10.

There are two possibilities in this case either X = 5 or X = 10

Case one X = 5

If you have 2X and switch then you are left with 5 bucks. You had 10 and now you have 5. You got a -X.

Case two X = 10

If you have X and switch then you are left with 20 bucks. You had 10 and now you have 20. You got a +X.

Exactly what I said before.
• 6.8k

I have £10. If X is 5 then I lose £X by switching. If X is 10 then I gain £X by switching.

So it's either -£X or +£X. This is symmetrical.

My reasoning is:

I have £10. If X is 5 then I lose £5 by switching. If X is 10 then I gain £10 by switching.

So it's either -£5 or +£10. This is not symmetrical.
• 1.1k
My reasoning is:

I have £10. If X is 5 then I lose £5 by switching. If X is 10 then I gain £10 by switching.

So it's either -5 or +10. This is not the same.

Look here:

Case one X = 5

If you have 2X and switch then you are left with 5 bucks. You had 10 and now you have 5. You got a -X.

Case two X = 10

If you have X and switch then you are left with 20 bucks. You had 10 and now you have 20. You got a +X.

Exactly what I said before.

Case one X = 5. You lose 5 bucks.

You had 10 and now you have 5. Means you lost 5 and in that case X = 5. Therefore you got a -X.

Case two X = 10. You gain 10 bucks.

You had 10 and now you have 20. Means you gained 10 and in that case X=10. Therefore you got a +X

The same outcome as you.

You have to understand that X is a variable.
• 1.1k
The fact that X is a variable is what makes this different, as even if you know the amount in one envelope you still don't know if that amount it is X or 2X. So I guess the question is: Does the new information of an amount shift the odds at all, given that you still don't know if it is X or 2X?
• 6.8k
Case one x = 5. You lose 5 bucks.

You had 10 and now you have 5. Means you lost 5 and in that case X = 5. Therefore you got a -X.

Case two x = 10. You gain 10 bucks.

You had 10 and now you have 20. Means you gained 10 and in that case X=10. There you got a +X

The same outcome as you.

Yes, and the X that you lose is £5 and the X that you gain is £10. Therefore, there's more to gain than there is to lose.
• 1.1k
Only if you have X.
• 6.8k
Only if you have X.

Obviously you'll only win if you have X. My point is that by switching there's a 50% chance of gaining an extra £10 and a 50% chance of losing £5. Those odds favour a switch.
bold
italic
underline
strike
code
quote
ulist
image
url
mention
reveal
tweet

#### Welcome to The Philosophy Forum!

Get involved in philosophical discussions about knowledge, truth, language, consciousness, science, politics, religion, logic and mathematics, art, history, and lots more. No ads, no clutter, and very little agreement — just fascinating conversations.