A real poser this time

This one will probably (I hope) start a discussion, aka "pissing match", so here it is:

You are presented with 2 envelopes, each containing a non-zero amount of money. You are told that one envelope contains twice as much as the other one, and then you are asked to choose an envelope. Before opening your envelope, you are given the choice to switch with the other one.

The questions is: which is more advantageous, to switch or to keep your original envelope?

it is 50/50

Originally posted by PBE6
This one will probably (I hope) start a discussion, aka "pissing match", so here it is:

You are presented with 2 envelopes, each containing a non-zero amount of money. You are told that one envelope contains twice as much as the other one, and then you are asked to choose an envelope. Before opening your envelope, you are given the choice to switch with ...[text shortened]... e.

The questions is: which is more advantageous, to switch or to keep your original envelope?

Are you always given the option to switch, regardless of your initial choice of envelope?

Originally posted by DoctorScribbles
Are you always given the option to switch, regardless of your initial choice of envelope?

Yes, no matter which envelope you choose you are always given the choice to switch (before you open the envelope, of course).

Originally posted by Chessteen16
it is 50/50

That's what I originally thought, and I have seen reasoning which makes that answer seem obvious, but I tried running a test on Excel and my results indicate that switching is consistently the better option.

Originally posted by PBE6
I tried running a test on Excel and my results indicate that switching is consistently the better option.

LMAO! Surely you're kidding. Would you humor us by posting your code, the results, and the statistical analysis you performed on the results to conclude that switching is better?

The explanation holds for 3 or more envelopes, and you are shown one of the wrong ones.
For 2 it's still 50/50. Unless of course the wrong one is opened anyway, then it's 100/0 😵

Originally posted by DoctorScribbles
LMAO! Surely you're kidding. Would you humor us by posting your code, the results, and the statistical analysis you performed on the results to conclude that switching is better?

Sure thing, loud mouth.

Here are my column headings:

n; envelope1; random; envelope2; stay; switch; result;

And here's what they mean:

n = trial number (1,2,3...)
envelope1 = 1/rand() where rand() is a random number between 0 and 1
random = rand()
envelope2 = IF(rand()<0.5, 2*envelope1, 0.5*envelope1) - this randomly decides if envelope2 is twice as big or half as big as envelope 1
stay = sum of the contents of envelope1 over all the trials
switch = sum of the contents of envelope2 over all the trials
result = 1 if envelope1 is bigger, and 0 is envelope2 is bigger (for each individual trial)

In my simulation, it is assumed that the player stays with the first envelope.

As you would expect, and as DoctorScribbles so flatulently alluded to, over 5000 trials the sum in the "results" column remained between 2450-2550 almost every time (45 times out of 50). However, the "stay" column, that is the total money kept by staying, was less than the "switch" column almost every time (44 times out of 50).

So DoctorScribbles, rather than LYAO, I encourage you expound your own theory, do your own simulation, and share it with the group.

Originally posted by PBE6
However, the "stay" column, that is the total money kept by staying, was less than the "switch" column almost every time (44 times out of 50).

That is because your simulation is incorrect. Every time a number smaller than 0.5 is chosen envelope 2 gets TWICE the money env. 1 gets. When the number is greater, is gets HALF the money. So, assuming the random number generator is correct, when the total of env1=x, the total of env2=(2+0.5)/2*x = 1.25*x. That's why you always get greater results for envelope 2.

The correct way to do it would be:

r = rand()
IF(r<0.5, envelope1 = rand(), envelope2 = rand())
IF(r<0.5, envelope2 = 2*envelope1, envelope1 = 2*envelope2)

or something like that. If you do it that way you will see that both envelopes have about equal chance of being bigger and thus it doesn't matter which you choose and thus whether you change,

Come to think of it, you could also balance it by using:

IF(rand()<(1/3), 2*envelope1, 0.5*envelope1)

and it will give you for result about 1666, one third of the number 5000 you used.

Originally posted by PBE6
So DoctorScribbles, rather than LYAO, I encourage you expound your own theory, do your own simulation, and share it with the group.

piderman has given a valid critique of your implementation, and he has given an alternative that I fully endorse as being valid, so I will continue to laugh my ass off if you continue to think that it is better to switch envelopes.

Originally posted by PBE6
This one will probably (I hope) start a discussion, aka "pissing match", so here it is:

You are presented with 2 envelopes, each containing a non-zero amount of money. You are told that one envelope contains twice as much as the other on ...[text shortened]... is more advantageous, to switch or to keep your original envelope?

Here's a similar problem (I think piderman beat me to it, though):

You are presented with a £20 note and an envelope. You are told that one is worth twice as much as the other, with an equal probability that the envelope or the note is worth more. You can either take the envelope or the note. At first you pick the note, but then you're given an opportunity to change your mind.

Assuming you want to maximise your expected gain, should you stick or switch?

Originally posted by DoctorScribbles
piderman has given a valid critique of your implementation, and he has given an alternative that I fully endorse as being valid, so I will continue to laugh my ass off if you continue to think that it is better to switch envelopes.

Actually, I retract my full endorsement of
IF(rand()<(1/3), 2*envelope1, 0.5*envelope1)

It should be this:
IF(rand()<(1/2), 2*envelope1, 0.5*envelope1)

But piderman's critque of PBE6's code remains valid.

Originally posted by piderman
That is because your simulation is incorrect. Every time a number smaller than 0.5 is chosen envelope 2 gets TWICE the money env. 1 gets. When the number is greater, is gets HALF the money. So, assuming the random number generator is correct, when the total of env1=x, the total of env2=(2+0.5)/2*x = 1.25*x. That's why you always get greater results for env ...[text shortened]... envelope1)

and it will give you for result about 1666, one third of the number 5000 you used.

In the code you provided, IF(r<0.5, envelope1=rand(), envelope2=rand()) puts random amounts in both envelopes. This is not what happens in the problem. The amount in one envelope will always be twice as big as the amount in the other. Now, picking the big one in the first place is a whole different kettle of fish.

On one hand, the chance of picking the bigger envelope at the beginning is still 50/50, right? But one interesting thing to note is that there is a lower bound on the amount of money in the other envelope, but no upper bound. Is the risk really symmetrically distributed? It doesn't seem like it.

Now, as I've said before the answer is supposed to be 50%, and I've seen cogent arguments to this effect. But a test is always necessary to confirm a theory. So let's get someone who can prove it's 50% to do so, with a sound theory and convincing simultation.

Originally posted by PBE6
In the code you provided, [b]IF(r<0.5, envelope1=rand(), envelope2=rand()) puts random amounts in both envelopes. This is not what happens in the problem. The amount in one envelope will always be twice as big as the amount in the other. Now, picking the big one in the first place is a whole different kettle of fish.

On one hand, the chance of picking ...[text shortened]... et someone who can prove it's 50% to do so, with a sound theory and convincing simultation.

[/b]

I can't tell if you're serious or joking. If you're really serious that you're not convinced that switching gives you no benefit, I'll provide both a theoretical proof and a reproducible Excel simulation to back it up. But I do hope that you're joking - the 50/50 result should be completely obvious.

Originally posted by DoctorScribbles
piderman has given a valid critique of your implementation, and he has given an alternative that I fully endorse as being valid, so I will continue to laugh my ass off if you continue to think that it is better to switch envelopes.

Scribbles:

piderman has given a valid critique of your implementation

He sure did give me his critique, where's yours?

and he has given an alternative that I fully endorse as being valid

It's not. It doesn't match the question. Did you read it?

so I will continue to laugh my ass off if you continue to think that it is better to switch envelopes.

I actually stated in my original post that I believed the answer to be that it doesn't matter if you stay or switch, but that my simulation made me want a more rigourous solution.

But at least if you keep laughing your ass off you'll lose some of that much needed weight. You obviously aren't breaking a sweat writing solutions.