  # Would You Switch Envelopes? (No Goats)

Somebody is holding two envelopes, one with some amount of money, the other with twice as much.

He hands you one envelope, you open it and see there is 10 dollars. He asks you if you want to switch envelopes.

One envelope has x dollars and the other has 2x dollars. If you get the x dollar envelope, you stand to gain x dollars if you switch. If you get the 2x dollar envelope, you stand to lose x dollars if you switch.

But why then, if you do this experiment over and over again, will you make more money in the long-run by switching?

Not as tricky as picking goats, but a little weird nonetheless. It seems logical that you should switch, but if you try to explain why… it gets a little tricky.

So do you switch or not?

The mindfuck, it burns.

It’s essentially just probability, assuming there is no body language or visual cues you can pick up on (like one envelope being heavier, fatter etc) then there is no way to beat the system. Assuming this trial is repeated multiple times, the best bet would probably be to remain consistent, AKA always switch, or always stick with the first envelope.

Kick him in the balls and take both.

But really I a gambler. Just as long as I got their aren’t any goats or whammies, I 'm going to press my luck and take the other envelope.

And you make more money because you never had the money to begin with.

OK so you have 10 bucks. You only have 2 envelopes so you can’t repeat over and over…
Keep the 10, worst you lost 10 from the other envelope. Best you gained 5. Enough for a free breakfast.

Stick with one strategy. Always switch or never switch. Regardless, you’ll make money pending there are no side bets.

If there are only 2 choices, wouldn’t it be 50/50 either way? Therefore it’s not logical to always switch or keep the original, it wouldn’t matter in terms of probability.

[quote]LankyMofo wrote:
If there are only 2 choices, wouldn’t it be 50/50 either way? Therefore it’s not logical to always switch or keep the original, it wouldn’t matter in terms of probability.[/quote]

My reasoning for staying consistent in the case of multiple choices is that over a large sampling, the 50/50 will average out, by always keeping the one you’re given OR always switching, you keep the 50/50 probability, when you start changing back and forth randomly on a whim you fuck with the 50/50 probability because if you stay consistent, it’s always 50/50, but if you’re constantly jumping back and forth you have to factor int he probability that on that specific instance you made the right decision. Think of it this way, if there are 8 envelopes , 7 of which contain 1 dollar and 1 of which contains 100 dollars, what’s the best strategy? Pick a different envelope each time (say theyre organized side by side, left to right in positions 0-7), or always pick the same envelope eg. always take the envelope at position 5? You know that sooner or later the 100 dollar envelope will pop up at position 5, but if you randomly take position 1, 2, 6, 4, 7, 5 and so on you’ll just be “dancing” with it and your chances of you choosing the same position that the envelope with the payout popped up on THAT particular instance is far less likely.

[quote]LankyMofo wrote:
If there are only 2 choices, wouldn’t it be 50/50 either way? Therefore it’s not logical to always switch or keep the original, it wouldn’t matter in terms of probability.[/quote]

No. See the example given above where you open an envelope with \$10. If you switch and the other envelope has \$20, you gain \$20. If you switch, and you end up with \$5, you only lose out on \$5.

The probability for both outcomes is 50/50, but the expected returns are different.

DB

Maths geek to the rescue:

Let X denote your monetary return in the game.

X = 10 if you don’t switch
X = 5 or 20 if you do switch
i.e. either the envelopes are {5,10} or {10,20}.

Now if we don’t switch we always get 10.
If we switch our expected returns (i.e. average returns in repeated experiments) are:
E[X] = 5Prob({5,10} occurs)+20Prob({10,20} occurs)

Now without better information we will assume that {5,10} and {10,20} are equally likely for the envelopes, i.e.
Prob({5,10} occurs)=Prob({10,20} occurs)=1/2

So,
E[X] = 51/2+201/2 = 12.5 > 10

So as a strategy for maximising the expected returns over repeat occurences one should always switch.

Intuition: Half the time when you switch you’ll get 5 dollars the other half the time you get 20 dollars so on average you get:
5/2+20/2 = 12.5

[quote]dollarbill44 wrote:
LankyMofo wrote:
If there are only 2 choices, wouldn’t it be 50/50 either way? Therefore it’s not logical to always switch or keep the original, it wouldn’t matter in terms of probability.

No. See the example given above where you open an envelope with \$10. If you switch and the other envelope has \$20, you gain \$20. If you switch, and you end up with \$5, you only lose out on \$5.

The probability for both outcomes is 50/50, but the expected returns are different.

DB[/quote]

Well yeah, the returns are different but that doesn’t change the probability of picking the higher paying envelope being 50/50 each time you pick.

[quote]JLu wrote:
LankyMofo wrote:
If there are only 2 choices, wouldn’t it be 50/50 either way? Therefore it’s not logical to always switch or keep the original, it wouldn’t matter in terms of probability.

My reasoning for staying consistent in the case of multiple choices is that over a large sampling, the 50/50 will average out, by always keeping the one you’re given OR always switching, you keep the 50/50 probability, when you start changing back and forth randomly on a whim you fuck with the 50/50 probability because if you stay consistent, it’s always 50/50, but if you’re constantly jumping back and forth you have to factor int he probability that on that specific instance you made the right decision. Think of it this way, if there are 8 envelopes , 7 of which contain 1 dollar and 1 of which contains 100 dollars, what’s the best strategy? Pick a different envelope each time (say theyre organized side by side, left to right in positions 0-7), or always pick the same envelope eg. always take the envelope at position 5? You know that sooner or later the 100 dollar envelope will pop up at position 5, but if you randomly take position 1, 2, 6, 4, 7, 5 and so on you’ll just be “dancing” with it and your chances of you choosing the same position that the envelope with the payout popped up on THAT particular instance is far less likely.[/quote]

You could also go on a roll by switching and keep picking the higher paying envelope. I’m just saying the probabilities are the same whether you consistently switch, consistently don’t switch, or do a little mix and match.

[quote]wushu_1984 wrote:
Maths geek to the rescue:

Let X denote your monetary return in the game.

X = 10 if you don’t switch
X = 5 or 20 if you do switch
i.e. either the envelopes are {5,10} or {10,20}.

Now if we don’t switch we always get 10.
If we switch our expected returns (i.e. average returns in repeated experiments) are:
E[X] = 5Prob({5,10} occurs)+20Prob({10,20} occurs)

Now without better information we will assume that {5,10} and {10,20} are equally likely for the envelopes, i.e.
Prob({5,10} occurs)=Prob({10,20} occurs)=1/2

So,
E[X] = 51/2+201/2 = 12.5 > 10

So as a strategy for maximising the expected returns over repeat occurences one should always switch.

Intuition: Half the time when you switch you’ll get 5 dollars the other half the time you get 20 dollars so on average you get:
5/2+20/2 = 12.5[/quote]

I don’t understand all the math mumbo jumbo but this statement:

“Now if we don’t switch we always get 10.”

is inherently wrong. If we don’t switch it could be 5, 10, or 20, depending on the values of the envelopes.

I’m not saying I’m definitely right and that it definitely doesn’t matter, because I don’t definitely know. I’m just saying I’ve yet to see any explanation that convinces me otherwise.

Nevermind, I get it.

[quote]wushu_1984 wrote:
Maths geek to the rescue:

Let X denote your monetary return in the game.

X = 10 if you don’t switch
X = 5 or 20 if you do switch
i.e. either the envelopes are {5,10} or {10,20}.

Now if we don’t switch we always get 10.
If we switch our expected returns (i.e. average returns in repeated experiments) are:
E[X] = 5Prob({5,10} occurs)+20Prob({10,20} occurs)

Now without better information we will assume that {5,10} and {10,20} are equally likely for the envelopes, i.e.
Prob({5,10} occurs)=Prob({10,20} occurs)=1/2

So,
E[X] = 51/2+201/2 = 12.5 > 10

So as a strategy for maximising the expected returns over repeat occurences one should always switch.

Intuition: Half the time when you switch you’ll get 5 dollars the other half the time you get 20 dollars so on average you get:
5/2+20/2 = 12.5[/quote]

your system has three variables
\$5 would be x
\$10 would be 2x
\$20 would be 4x.

the original scenario only has x and 2x.
with only x and 2x would have a 50/50 chance each time of gaining 1x or losing 1x. assuming x is held constant, you come out even

edit: even meaning in the long run you gain 1.5x on average every time

[quote]LankyMofo wrote:
I don’t understand all the math mumbo jumbo but this statement:

“Now if we don’t switch we always get 10.”

is inherently wrong. If we don’t switch it could be 5, 10, or 20, depending on the values of the envelopes.

I’m not saying I’m definitely right and that it definitely doesn’t matter, because I don’t definitely know. I’m just saying I’ve yet to see any explanation that convinces me otherwise.[/quote]

In the question he said “He hands you one envelope, you open it and see there is 10 dollars.” So if you don’t switch you get the 10 dollars. In the sense of repeated experiments we only count the experiments when you open it and see 10 dollars. This is the “given information”.

So:
Half the time when you switch you’ll get 5 dollars the other half the time you get 20 dollars so on average you get:
5/2+20/2 = 12.5
This is conditional on the face that everytime the envelope given has 10 dollar in it.

[quote]_lb wrote:
your system has three variables
\$5 would be x
\$10 would be 2x
\$20 would be 4x.

the original scenario only has x and 2x.
with only x and 2x would have a 50/50 chance each time of gaining 1x or losing 1x. assuming x is held constant, you come out even
[/quote]

The given information is one of the envelopes has 10 dollars i.e. the envelope in your hand.

So given the setup of {x,2x} then we have that:
10 = x or 10 = 2x
1st case implies {10,20} as the envelope values
2nd case implies {5,10} as the envelope values

You don’t consider 4x.

The setup implies two possible envelope values {10,20} or {5,10}.

[quote]wushu_1984 wrote:
_lb wrote:
your system has three variables
\$5 would be x
\$10 would be 2x
\$20 would be 4x.

the original scenario only has x and 2x.
with only x and 2x would have a 50/50 chance each time of gaining 1x or losing 1x. assuming x is held constant, you come out even

The given information is one of the envelopes has 10 dollars i.e. the envelope in your hand.

So given the setup of {x,2x} then we have that:
10 = x or 10 = 2x
1st case implies {10,20} as the envelope values
2nd case implies {5,10} as the envelope values

You don’t consider 4x.

The setup implies two possible envelope values {10,20} or {5,10}.[/quote]

This makes sense, I think I didn’t read the op carefully enough.

[quote]LankyMofo wrote:
Nevermind, I get it.[/quote]

It’s good to know you can admit when you’re wrong. Now head over to Funniest Guy on T-Nation and mop up the mess you created there.

Now say you aren’t allowed to see what amount you get with the first envelope.

Now it doesn’t matter if you switch or not, your expected value is the same whether you switch or not.

Once you open the envelope, you raise your profit 25%.

Interesting…

What if, when you open the envelope, instead of money, there is a baby rattlesnake?

DB