Vsauce! Kevin here, and I’m about to stretch my winkey until it snaps. Actually, I’ve
got 3 of these anti-stress emoji people called winkeys, only one will survive and you will
witness an incredibly simple problem that almost everyone gets wrong. And at the end,
get to the bottom of one of the most famous probability problems ever.
It’s time to meet the winkeys: orange, purple, and blue. They’ve been sentenced to death,
but just one will be pardoned. The other two are walking the green mile, but they don’t
know which two it will be. All they know is that I, Warden Kevin, get to decide.
Right now, each winkey’s chance of survival is 1 out of 3 -- so, ⅓ + ⅓ + ⅓ = 1,
with every winkey having an equal shot at living. There’s no paradox here… yet.
The fairest, least-biased way to choose which winkey lives is to put three slips of paper
into a hat and randomly draw a color. I’ll mix around the slips, and… here we go. But…
what if they know I drew a color, yet I don’t tell them which one it is? These winkeys would
be left to wonder who will live, who will die, and what their own personal odds of survival
really are --
Orange: Kevin… Kevin! You can tell me. I can keep a secret. Tell me who will live.”
Kevin: My winkey is talking to me. And I think I recognize that voice. A keeper of internet
history. How’d you end up on death row!? Anyway, I can’t tell you who survives. It’s
just not fair -- and I DEFINITELY can’t tell you whether YOU live or die.
Orange: “I understand, you bear a heavy burden. Just tell me one of the winkeys that
will be executed. If purple lives, tell me blue. If blue lives, tell me purple. And if
I’m going to live, just choose which of the other two you’d like to name. Do this
for me, Kevin -- pwease.”
Kevin: I… I don’t know, orange winkey. It really doesn’t seem fair.
Orange: “Please, Kevin. Pretty please. You are so handsome. And those glasses look so
good on you.”
Kevin: Okay -- what’s the harm? I can’t possibly give Orange Winkey any useful information
here. Okay... Blue will be executed. Right now!
I didn’t mean to put his body back
that way, upside down. But you know what? Warden Kevin is a cruel warden.
Orange: “Kevin, you fool. Now I know that my odds are 50/50 -- I will live, or purple
will live. There is no alternative. You stupid, stupid man. He’s so dumb. And also fat.”
Kevin: Oh no. He’s right! I got tricked by a winkey. Now, there are only two choices.
One will live and one will not -- so his odds improved from 1 out of 3 to 50/50, right?
WRONG, ORANGE WINKEY. You still only have a 1 out of 3 chance to live, and purple winkey
is actually twice as likely to win the pardon. This may remind you of the Monty Hall Problem
and we will get to why that’s important later, but for now let’s break down the
only possible scenarios that orange winkey faced. Remember that I can’t tell Orange
whether he lives or dies:
Purple lives, I name blue (⅓) Blue lives, I name purple (⅓)
Orange lives, I name blue (⅙) Orange lives, I name purple (⅙)
When I named blue, we can compare the two possible outcomes after naming blue: scenario
#1, where I name blue and purple lives, has a 1 out of 3 probability of occurring. Scenario
#3, where I name blue and orange lives, has a 1 out of 6 chance of happening. Since ⅓
is twice as large as ⅙ -- and since they’re the only two possible outcomes after naming
blue, meaning their probabilities have to add up to 1 -- purple actually has a 2 out
of 3 chance of survival, and orange only has a 1 out of 3 chance of avoiding execution.
The probabilities of two events that appear to be equally likely just… aren’t. And
we can’t handle that.
Alright, it’s time to reveal which winkey will live and which winkey will die. This
slip of paper determines our survivor… and.. purple lives! Sorry, orange! -- “Kevin,
please!” -- Time for you -- “No, have mercy!” -- to BECOME history. “My body
is stretching… do not stretch me, Kevin. Noooooo! Was it ‘cause I called you fat?
I am dead.”
Welcome to the Three Prisoners Problem... which is actually the Bertrand’s Box Paradox.
And kind of the Monty Hall Problem. But mostly the Bertrand’s Box Paradox -- which came
out the same year the Eiffel Tower opened, Van Gogh painted The Starry Night, and Nintendo
Koppai first started producing playing cards. They eventually made video games.
In 1889, 70 years before recreational math columnist Martin Gardner started playing mind
games with prisoners on their execution date, French mathematician Joseph Bertrand published
Calcul des probabilités, in which he posed a paradox that became the basis, the genesis,
for Monty Hall and will require me to stick my winkeys in pudding.
Let me to explain.
After making me a plate of tendies and my favorite dipping sauce, my mom secretly hid
winkeys in these three large candy dishes full of pudding. One of them contains two
winkeys of the same color. Another one contains two winkeys of a different color. And then
one contains one winkey each of those two colors. I’m pretty sure mom said one winkey
color is Yellow, but I’m not sure what the other one is…
This is so much grosser than I thought it would be. Ughhh! Let’s reveal this winkey’s
color.
“Thank you for licking me clean, Kevin. Your tongue is so smooth.”
How are you alive?!?
“You can’t kill me, Kevin. We’re friends now.”
Ok, well, clearly we have an Orange winkey… somehow back from the dead. And we know the
second winkey color is yellow. But what are the chances that the other winkey in this
pudding is also orange?
We know that the three dishes are orange-orange, orange-yellow, and yellow-yellow. Each combination
is equally likely to be chosen at the start. We also know that this can’t be the yellow-yellow
dish because I just pulled out the creepiest orange winkey on the planet. That means it’s
coming from orange-orange, or from orange-yellow, and both had the same chance of being selected.
Once we pull the orange winkey, we know that the odds the other winkey in this dish is
orange is 50/50, because I’ve clearly chosen
either the orange-orange pudding or the orange-yellow
pudding. Right?
Wrong. Again. For the same mathematical reason as our prisoner problem.
This is the mathematical foundation of all these scenarios: our inability to recognize
that what appears to be 50/50 really isn’t.
Hold on, before I explain further, let’s just find out real quick where the winkeys
are because I actually don’t know. Okay
this was orange. Orange… ah! So it was orange-orange!
Jar number 2… ugh, I can’t even tell. Ok, that’s yellow. Now is this one orange,
or is this the yellow-yellow jar? Time to find out… looks orange, looks orange, let’s
see. Yep, there’s the other orange, so that was our yellow-orange… which means that
this has to be the yellow-yellow. Yellow… ughgh… yellow. I really have to clean up
this pudding or else we can’t continue the video. So, I’ll be right back.
Alright, now that the Vsauce mukbang portion of the video is complete, let’s get to the
root of this problem.
Bertrand conceived his original paradox with gold and silver coins in boxes. He had the
same basic setup we used: gold-gold, gold-silver, and silver-silver, with the player reaching
into a random box, drawing a gold coin, and wondering about the odds of that box containing
another gold coin.
Once we pull a gold coin -- or an orange winkey, or one of the prisoners to be executed -- it
feels like our odds of whatever comes next are an equal 50/50. We’re just choosing
between two leftover options that appear to be equally likely: we know the other coin
is either gold or silver. But look, if you pull a gold coin you either pulled gold coin
1, 2 or 3. And at this point you conceptually know that you’re not dealing with the box
with two silver coins. You don't know exactly what it is, but you do know what it's not.
If you choose Gold 1, the other coin is Gold 2. If you choose Gold 2, the other coin is
Gold 1. If you choose Gold 3, you’ll get Silver 1. So 2 out of 3 times you draw gold,
it’ll be paired with another gold, and only 1 time out of 3 will you get stuck with silver.
But it’s almost impossible to shake that notion of two equivalent options even when
the math shows us that it’s wrong. It’s the same reason why the Monty Hall Problem
is a Problem.
The Prisoner Problem and Monty Hall add a story to the math and an additional third
party. It seems like that would make it more complex, but it actually helps you understand
the probability more easily. Adding a second player’s perspective lets us see that The
Warden will never reveal who lives and Monty Hall will never reveal the door with the money.
Despite that additional information, we still get it wrong.
Why is it so hard when the solution is right in front of our face?
Probably because it is right in front of our face -- and it’s why very, very smart people
often get it more wrong than right.
I mentioned this in my “What Is A Paradox?” video but when Marilyn vos Savant wrote in
Parade magazine that switching doors in The Monty Hall Problem resulted in a ⅔ chance
of success instead of being an obvious 50/50 trade, Parade received over 10,000 letters
telling her she was wrong. About 1,000 of them were signed by PhDs, and she claims many
were sent on official letterheads from college math and science departments. And I still
get comments on my video from people all over the world telling me that I’m wrong.
Massimo Piattelli-Palmarini, author of “Inevitable Illusions,” said that, “No other statistical
puzzle comes so close to fooling all the people all the time,” and research in 1995 by Granberg
and Brown found that only 13% of people correctly switched Monty Hall’s doors.
In a 2003 paper in the Journal of Experimental Psychology, Stefan Krauss and X.T. Wang described
these particular problems -- from Bertrand’s Box to Monty Hall -- as “cognitive illusions,”
stating that they “demonstrate people’s resistant deficiency in dealing with uncertainty.”
They found that only 3% in the control group solved the problem with mathematically correct
reasoning, and that a significant problem was perspective. When participants were able
to think of the problem from the perspective of Monty Hall instead of the game show contestant
-- or thinking as me, Warden Kevin, instead of as the prisoner Orange Winkey -- “Hello”
-- their chances of choosing the right pathway improved, vaulting them from simple non-Bayesian
thinking into more complex Bayesian analysis.
In the 1990s, it appeared that there was no way to break the mathematical illusion of
Bertrand’s Box, The Three Prisoners, or the Monty Hall Problem. Explaining the math
just didn’t seem to work. It turns out the math isn’t the problem.
We are.
Uncertainty is hard for us. Changing perspectives is harder. You’re watching this video because
you’re a smart, curious person -- and by altering your perspective, you don’t just
dismiss a mathematical illusion that has perplexed us for over a century because you’ve identified
its flaw… you don’t just SOLVE it… you embrace it, you absorb it, you understand
it, and you ultimately use it to your advantage.
You KNOW it.
And I know that I have SO much pudding to eat.
And as always -- OHH! -- thanks for watching.
“Pudding! I mean… surprise.”