Vsauce! Kevin here with two games. If you play Game A -- you are guaranteed to lose.

If you play Game B -- you are guaranteed to lose.

But.

If you alternate back and forth between playing Game A and playing Game B -- you are guaranteed

to win.

How? How can two losing games combine into one winning one? Is it just a cheap game hack

or something more? And does this mean that grandma was wrong and it’s actually possible

for two wrongs to make a right? Welcome to Parrondo’s Paradox.

Physicist Juan Parrondo who created this paradox demonstrated with flipping a

biased coin. A coin that, instead of having 50/50 odds of landing on heads or tails has

49.5/50.5 odds. So slightly unfair, biased odds. And… that’s not gonna work for us.

Because precision-biased coins that look like normal coins are likely impossible, or as

statisticians Andrew Gelman and Deborah Nolan wrote in a 2002 paper, “You can load a die

but you can’t bias a coin.”

So, instead, I made a series of games using 3 roulette wheels with a total of 114 spaces

and 1 Markov Chain. Because…PENGUINS.

Seriously, this paradox can be visualized using a penguin slide toy. The same mechanism

that moves these penguins to the top allows 

us to turn two losing games  into our own money-printing

machine. This is… this is really loud. I’ll explain how later but for now let’s actually

talk about our games.

Okay, Game A works like this: it’s a game of chance in which your odds of winning are

slightly lower than your odds of losing. The American roulette wheel has 38 spaces -- 18

red and 18 black, and these two 2 green zeroes. The rules of Game A are that you can only

bet on red or black. Either way, our odds of winning are 18/38, or 47.36% -- so a little

less than 50/50. Our 53.6% chance of losing means the house edge is 5.2%, which we’ll

round down to 5%. Because of that 5% losing edge, theoretically, every time we bet $1

on this game, we’ll lose about 5 cents. If we start with $100 and keep playing this

game, we’ll be totally broke after about 2,000 spins. In the long run, when we play

Game A, we’re guaranteed to lose.

Come onnnn, red!

Since definitely losing sounds terrible, let’s give Game B a shot.

Before we do that you may want to add up all the numbers on a roulette wheel and just tell

me the sum in the comments.

Okay. B is composed of two games of chance, each with different odds. You’ll still bet

$1, but the wheel that you play depends on how much money you have left. We’ll call

these B1 and B2. You’ll play B1 only if your total money is a multiple of M and…

let’s say that M = 3. So if your leftover money is a multiple of 3, like 93 or 81 or

66, then you have to play wheel B1. If your bankroll balance is not a multiple of 3, then

you’ll play wheel B2.

Here’s the catch.

With B1, you’re only allowed to make what’s called a Corner bet -- choosing an intersection

of 4 numbers, so you’ll win $1 if the ball lands on any of those four. So like, a corner

bet of 26, 27, 29, or 30. A corner bet means your odds of winning are just 4 out of 38,

or… about 10%. So when you play B1, you’re going to lose 90% of the time. Losing isn’t

guaranteed, but you’re going to lose a lot more often than not. Game B1 is a pretty bad

game for the player.

Don’t worry, though, B2 is much better. You get to choose a combination of winning

spaces -- red or black AND odd or even. So, if you choose red and evens, you’d win every

time the ball lands on a red space or an even space, even if it’s a black even space.

The two green spaces are also winners for you on B2. This allows B2’s odds to shift

significantly in your favor, with 18 reds and 9 black evens plus the two 2 greens giving

you a total of 29 chances to win out of 38 possible spaces -- and that, my friends, is

76%.

The good news is that since there are more possible money counts that aren’t multiples

of 3 than there are, you’ll be playing the 

much-friendlier B2 a lot more  than the nearly-impossible

B1. So that means most of the time, with Game B, you’ll be winning, right? No.

Here’s the thing.

B1 is a loss 90% of the time. And since you’ll also lose roughly every 1 out of 4 times playing

Game B2, overall you’re guaranteed to lose if you just play the B Games. Game B is actually

a Markov Chain, a stochastic process that takes a situation like ours -- different states

with varying probabilities -- and generates a loss by B1’s disproportionate effect on

our fortunes.

Let me explain.

The key is that even though there are only three possible states of our money balance

-- a multiple of 3, like 90 or 93, or two possible states in between multiples of 3,

like 91 and 92 -- our probabilities of playing the bad B1 game vs. the better B2 game aren’t

a simple ⅓ and ⅔. What’s surprising is that a Markov Chain analysis, and if you

want to learn more about Markov chains I’ll link you to a course on those in the description,

shows that our probability of playing the bad B1 game is actually closer to 40%. And

the winning edge that the good B2 game gives us isn’t enough to make up for the terrible

B1.

All of that is to say that Game A is a simple guaranteed loser. And Game B is a kinda complex

guaranteed loser. And… this is getting dreadful so let’s bring back our penguins and lighten

the mood. Our flightless bird friends will help us visualize how we leverage two negatives

to create a positive.

In a 2000 paper titled, “Parrondo’s paradoxical games and the discrete Brownian ratchet,”

Derek Abbott, Parrondo and others, described a process known as the “flashing brownian

ratchet” as an analogy for how the paradox works. Basically, directed motion is achieved

by alternating between a sawtooth and a flat potential. I’ll show you. These penguins

clearly reach the top of the slide, even though they themselves aren’t moving, because they’re

being carried back and forth between  

two downward-sloped-sawtooth shapes and a flat shape. These shapes slope

downward. And this shape here -- sorry, penguin! -- is flat. Yet the penguins climb to the

top by alternating between them. Ok, that’s loud, sorry penguins. Wheeee! The end. Excuse

me one second, Happy Feet.

Think of Game A as our flat shape because the odds of winning Game A are close to 50/50

with a slight bias toward losing. Think of Game B as our sawtooth shape because the odds

are much steeper in Game B1 and much better in Game B2, creating a distinct asymmetry.

Look at that! But instead of moving plastic penguins uphill, by alternating between playing

Game A and playing Game B, we, the player, are carried upward by our combination of losing

games into winning.

To put it another way, we can get really simple and math-y about this. Back to the North Pole

with you! Say you start with $100, and each time you play Game A, you lose $1. That’s

it -- done. That’s just how Game A works in this simplified scenario. You’re a loser

with no hope of winning. And if you played Game A 100 times, your financial trajectory

goes downward until you’re broke.

There are two strains of Game B: if the amount of money you have left is an even number,

like $82, then you win $3. If it’s an odd number -- say, you’re down to $71 -- then

you lose $5. If you only played Game B, you’d be broke even faster than if you’d just

auto-lost Game A. At least with Game A you’d get to play that game 100 times.

Both Game A and Game B are clearly 100% losing games, but when you switch back and forth,

you can make them profitable. Check it out. Play Game A and you’re down to $99. That’s

an odd number, so you don’t want to play Game B or else you’d lose another $5…

instead you can play Game A again and lose another dollar. Now you’re at $98 -- when

you switch to Game B, you win $3, and suddenly you’re up to $101. Lose Game A to get to

$100, win Game B to get to $103. You can repeat a cycle of A/B switching to amass infinite

wealth despite playing two games that on their own are guaranteed losers.

But is that really a paradox? When we examine the best way to flip coins or spin roulette

wheels, it seems like just a cheap trick to manually hack a couple of games. It’s not.

The paradoxical situation is that you can alternate playing Parrondo’s two losing

games randomly and it will still produce a winning one.

We can use Stan Wagon’s Parrondo Paradox wolfram simulator to prove it. This simulator

uses Parrondo's biased coin flip odds and allows us to set the number of flips and how

many times we want to repeat that experiment. Parrondo's numbers for Games A, B1 and B2

are almost identical to our roulette example. So, we set the number of flips in a game and

how many times we'll repeat that game, and each time we click New Run, the simulator

crunches those numbers... which is a lot faster than flipping an impossible biased coin 2

million times.

Let’s set this to run 1 repetition of 1,000 flips. When we play in a specific pattern

-- BBABA -- we clearly win over time.

But what happens if we choose Game A or Game B randomly? Most of the time we win over 1,000

flips, occasionally we don’t -- that's just variance. But when we run 2,000 simulations

of 1,000 flips, a pattern emerges: we see a clear upward slope. We don't win as much

as when we carefully orchestrate the best approach to the two games, but... we win.

Despite having no plan and no specific hack, alternating randomly between Games A and B

yields a long-term win.

So even if it takes a while, even it’s done randomly, by alternating between these two

losing games, we actually get a winning result. We get to the top of the slide -- like our

penguins.

But here’s the question.

Can you use this strategy to guarantee that you win at a casino? No. Parrondo's Paradox

depends on being able to interact within two games, and you just can't do that in a casino.

Real casino games like slot machine spins or the way roulette is actually played are

based on entirely separate events -- one outcome will never influence the outcome of the next

round or a different game, like a roulette result will never steer you toward an advantageous

round of Pai Gow.

By guaranteeing that games never intersect, casinos avoid the possibility of an exploit

or a weird Parrondo situation. The outcome of every 'bettable' event in a casino depends

on nothing before it or after it. It exists in its own impenetrable bubble that pops as

soon as that round is over and then is blown up again -- and that keeps the games fair

and keeps the casino’s math predictable.

The independence of each game is actually one of the reasons you so rarely see new games

at casinos. It’s just hard to come up with games that are totally independent, give the

house enough of an edge that they win over time but also give the player enough of a

chance that they want to play and have fun playing.

What’s exciting about Parrondo’s Paradox is not how to win money playing two games

that no casino would never allow. It’s figuring out how to apply its surprising property to

other fields of study. Researchers are working on real world applications for it in disciplines

ranging from quantum mechanics to biogenetics.

And for the rest of us, it can be helpful to realize that two losing games can become

one winning one. That, in a way, as weird as it sounds, two wrongs can make a right.

And even what seems, by all accounts, like a totally hopeless situation can, undeniably,

mathematically, be turned into a winning one.

And as always, thanks for watching.