Vsauce! Kevin here, with two envelopes. One of them contains twice as much money as the
other one, and you get to choose which one you want to take.
But before you open your envelope and find out whether you’ve won the smaller prize
or the larger prize, you have the opportunity to switch to the other envelope... so, do
you? Do you switch?
The answer is… No. Irrefutably, 100% no.
In our scenario, it makes absolutely no statistical
difference which envelope you choose. Each one could contain the lower amount of money
or the higher amount of money with equal probability,
and you have zero information to help you
choose one over the other. Right?
Wrong. The answer is irrefutably, 100% yes -- switch the envelopes. And there’s
a mathematical
explanation for why. Right?
Wrong.
WHAT IS HAPPENING HERE!?
Let’s...let’s... let's start at the beginning.
The origin of the Two Envelopes Paradox is just as confusing as the problem itself. In
1953, Belgian mathematician Maurice Kraitchik posed a scenario in his book about recreational
mathematics in which two men compared the values of neckties their wives bought them,
with neither man knowing the prices. Also in 1953, a math book credits a similar puzzle
using playing cards to physicist Erwin Schrodinger -- ever hear of Schrodinger’s cat? Where
a cat in a box is both alive and dead until you open the box and find out? Yeah, same
guy -- and in our problem, the same forces at play are the same whether we're using playing
cards, neckties, or envelopes.
Oh also, I’m wearing my brand new Birthday Paradox shirt. I have a floating space baby
bracketed by the chart detailing the probability of at least two people sharing a birthday
vs. the number of people. It’s in my new Vsauce2 store -- link below. Wear it, for
all the space children.
So look. The logic for not switching envelopes is pretty straightforward. Given that you
don’t know anything about either one -- or even what the higher and lower values are
inside them -- one is as good as the other. But the math in favor of switching is… kinda
compelling.
Let’s say that the envelope you choose contains X dollars. We don’t know how much X is,
just that there’s something in the envelope we can call X. The probability of that envelope
containing the smaller value is ½, and the probability of it containing the larger amount
is ½. It's 50%.
The other envelope either contains twice that amount -- 2X -- or half that amount, 1/2X.
If X turns out to be the higher value, then the other envelope contains 1/2X. If your
envelope's X is the lower value, then the other envelope contains 2X. Make sense?
I’ve talked about expected value in a few other Vsauce2 videos. It’s the average of
the possible outcomes of a series of probabilistic events, which we can use to identify the most
advantageous course of action. So, to find the expected value of the envelope we didn’t
choose, we add the value of the only two possibilities
weighted by their probability. Like this:
½ (2X) + ½ (1/2X) = 5/4 X equals five over four X. Half of the time the other envelope
will contain 2X, and half of the time the other envelope will contain one-half X…
which equals an expected value of… five-fourths X.
5/4 X is 25% more than just X, so our best possible choice is to switch to the other
envelope.
Cool. We did it. BUT THAT MAKES NO SENSE.
You have a simple, clean 50/50 chance of choosing the envelope with more money inside. Switching
can’t change that... but the math just told us that it can.
Don’t worry, though. It gets weirder. What if I said you could switch envelopes again
if you wanted to? Well, then you’d just go through the same process over and over
again, always switching, never opening, forever and ever and ever and erver and nerver and
blerver -- those aren't words but you get the point. Because switching makes the most
mathematical sense, with switching gaining you a 25% advantage. And I just proved it.
Sooooooo....
Let’s come back to math later. It’s time for logic.
Logician Raymond Smullyan poses two scenarios for the paradox: First, the risk of gaining
twice the amount in your envelope is worthwhile when the alternative is only losing half of
it. You’re gaining X or you’re only losing half of X. The potential payoff is twice as
high as your risk of loss. Makes perfect sense.
But one envelope contains X, and the other contains 2X… so you’re either gaining
X or losing X. Also makes sense.
But both can’t be true. So we’re going to do this problem again using economics professor
Barry Nalebuff’s variation.
Let’s say that we give an envelope to Player 1, and then flip a coin. If it’s heads,
we put double that amount in the second envelope and give it to Player 2. If it’s tails,
we put half that amount in the envelope and give it to Player 2. Neither player knows
how much is in their envelope or the outcome of the coin toss. Once they open their envelopes
in secret, they can switch with each other if they both agree to switching.
Player 1 opens their envelope and finds $10. With equal probability, they think that switching
to Player 2’s envelope will get them $20 (for a $10 gain) or take them down to $5 (for
a $5 loss). Player 1 decides to switch because switching gives them a greater possible reward
relative to what they might lose.
Player 2, who’s in another room, opens their envelope and finds $5. They think there’s
a 50% chance that switching to Player 1’s envelope will get them $10 (for a $5 gain)
or get them only $2.50 (for a $2.50 loss). So… Player 2 also thinks that what they
stand to gain is greater than what they might lose. They are both absolutely sure that their
upside is better than their downside.
But it’s obvious that BOTH players can’t have an advantage in this game. That’s impossible!
If you’re confused at this point… good. Martin Gardner, one of history’s greatest
puzzle-solvers and paradox-explainers, knew the answer but admitted that there just wasn’t
a simple way to communicate
the flaw in mathematical reasoning.
And sadly, that’s where our journey ends… NO IT’S NOT. Because we CAN figure this
out.
Think of it this way: the total sum of money in the envelopes is three units -- a small
unit, and then a larger value that’s made up of two smalls -- that combine to form a
total of X. If you’ve got the smaller envelope, its value is x/3, and if you have the larger
envelope, its value is 2X/3. When you switch from small to large, you gain x/3, and when
you switch from large to small you lose x/3… So, the expected value of switching is: ½
* [(2X - X)/3] + ½ [(X - 2X)/3], which equals… 0.
There’s just no value in switching. It doesn’t help us and it doesn’t hurt us -- which
is probably what you knew until we got math involved. The Two Envelopes problem is a lot
more than just X and 2X -- which is what makes it a falsidical paradox.
Because there appears to be a gain from switching when we fail to consider that the states of
the two envelopes don’t actually reflect simple values of X and 2X. A 2X envelope is
only 2X when it’s the larger amount, and the ½ X envelope is only ½ X when it’s
the smaller amount. Without recognizing that difference, basic algebra playing off our
common sense fools us into thinking that each switch gives you a gain of 5/4, or 25%.
It’s really a matter of perspective. Depending on which envelope you’re holding, which
one you think you have, which one you think you don’t have, and which one you want to
wind up with, your course of action might be incredibly clear… or not. And then clear
again. And then not again.
Until you realize that no matter if you diagnose it mathematically, no matter if you deconstruct
it logically, no matter what you do, when it comes to the two envelope paradox, you
were right all along.
And as always,
thanks for watching.