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.
