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.

Real quick, I did a video with Donut Media

where they brought me out to rip around a

racetrack and talk about carbon fiber, aerodynamics,

and carbon ceramic brakes. Y’know. Awesome

things. Go watch the video on their channel

— link down in the description below — and

tell them Vsauce2 sent you in the comments!

Okay so 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.

Woah! Ahhh! Hello there, my little penguins.

I have some suggestions. Suggestion #1, please

hit the like button if you want to like the

video. That’s helpful to me and it takes

a split second of your time. Also, subscribe

if you aren’t yet subscribed so you can

watch more Vsauce2 videos. And if you want

to take our relationship to the next level,

hit that notification bell. Because then you’ll

know exactly when I upload a new video. So,

ring-a-ding-ding that bell, as they say. No

one says that. But now I do. If you want to

watch more videos that I’ve made, just click

over here. Other than that, have a great…

day…

Quick point — those added green spaces throw off the math a little bit and make game B2 very slightly biased toward the player. With a 38.36% chance of a player in the B1/B2 Markov Chain playing B1, we can calculate (4/38)*(.3836) to find B1's share of the overall win: .04037. We can do the same for B2, which he plays 61.64% of the time: (29/38)*(.6164) for a share of .47041. When adding those two together, .04037 + .47041, we get a winning probability of 0.51078, or 51%.

How do we tweak this to get a truly losing Game B? We take away those additional green spaces. By giving a player just 27 spaces to win on B2, the value becomes (27/38)*(.6164) = 0.43796, putting the combination of B1 and B2's payoffs well below the 50% threshold and turning it into a proper loser.

tldr; Make the green spaces losers in our B2, and my roulette version of Parrondo's biased coin flips align properly.

Also, check the footnote in the video about the North Pole penguins. 😉

HAVE A GREAT DAY.

Jesus!, I had one of those penguin toy's as a kid…

X + X = ✔️

No no no, two wrongs don’t make a right, three rights make a left!

10:46 102 = 3*34 so you can't play B1

If odd was wrong and even was right, odd + odd (2 wrongs) = even (1 win).

powerful knowledge right there…thanks

Why is 00 even?

IF YOU LOSE YOU WIN BUT IF YOU WIN YOU LOSE AHHH

At school we learned that – *

~~=+ & +*+=+ but – *+=~~(*could be /). I haven't seen the whole video but that what I first thought ofAll i hear is blah blah blah.

7:13 game a 1 once then game b 2 times

Because -1-1=1

TL;DR Lost compesation ?

This video is explaination of this https://en.wikipedia.org/wiki/Parrondo%27s_paradox

Penguins live in the south pole

You just lost the game.

2:06 What are those penguins at the bottom doing

I mean or you could just get lucky on game A

Same principal as how a pen clicks up and down. Alternating notches. Also, said pen mechanism rotates and you do not notice it.

Watching this as an old crupie makes me laugh so hard, my job was to make people loose . 😀

I'm ready to take our relationship to the next level. Come pick me up and let's blow each other.

I am not to smart for this

Always bet on black

haram af

What? Can you repeat that?

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ok, lemme just save everyone some time, the reason loosing twice makes you win is cuse double negative = positive 🤷♂️

"Two wrongs doesn't make a right"

Vsauce:allow me to introduce my self

I don’t like you because you’re not Vsause

How the penguins stopped at 2:07

i'm off to the fruity at spoons yee

there are no penguins at the northpole

“*There are no penguins in the North Pole” 💀

I just noticed how recently this was uploaded and I'm happy

666

Given that some, if not most, of the assets in the stock/crypto market are somewhat connected, would this work when applied to trading?

= 666

No you're not !!!

30 seconds in its politics

A game with a 0 % of winning is a game impossible to win so you will always lose no matter what you do; that would be my definition. The games he proposed never had a 0% of that…

38

But u. Can make. It 37 playing. 00/0

Penguins aren’t in the North Pole

Because two negatives make a positive

Doesn't this guy sound like the protagonist from How to Train Your Dragon?

Turning my back two times turns me to face it 180°+180° is 360 right

Or maybe you could just keep adding a dollar to your balance (or removing a dollar from your balance) when it is an odd number as you play game B and skip game A altogether.

"game A" is just a way to destroy a dollar to make your balance even again. But if you're allowed to walk away from game B and come back with a different balance the game isn't a losing game in the first place.

In summary, game B

given the assumption that you can leave and come back with a different balance (which you almost certainly wouldn't be allowed to do if this was a real casino game). Game A is just a pointless waste of money and a distraction.isn't a losing gameTwo losing games don't make a winning game. Game A is just a sneaky way to hide the fact that you are either cheating or that game B is broken.

"So instead, I made a series of games using 3 roulette wheels with a total of 114 spaces and 1 Markov chain.

Because … PENGUINS." — Jake, 2019

"And this is getting dreadful, so let's bring back our penguins and lighten the mood." — Jake, 2019

2 negatives can also equal a positive in words ex: when you are not not allowed to do something

Games Vsauce is going to tawk about:

The game you win a waqf by winning a waqf.

The game you lose by winning.

The game you win by buning.

The game you win by eating the other player.

The game you lose a waqf by losing a waqf.

The game you win by fhqwhgads.

The game you you.

The game where people eat accumulates.

The the the game game game.

when you said that playing 2 losing games alternating results in winning, i just thought of -2^2=4

Man.. we need to go to the casino together

72

No, It is -n × -n = n (negative × negative = positive)

tHiS gAmE cAn'T eXiSt

I may not know what any of this meant, but I do know that penguins are not in the North Pole.

0:22

HOWBut that's like two wrongs make a right, and we are teached that two wrongs

DONTmake a right!but do they make a

left2 L’s make a W

Chandler would like that

0:56 Whats that song?

so that's where n-(-n)=+n

the odds of winning are actually 26.32% odds of loosing are 26.32% and odds of not loosing nor winning are 47.36% so after a long game of playing it is same as 50/50 will you go with profit or not

Hey Dongers, Michael's Math Magic here

But isn't there still that possibility that you have a huge chain of losses and are out of cash

Ezy question double negative = positive

I really don’t know what I just watched.

Except counting cards in blackjack lol.

I'm so sorry because my brain is too little

Where did you get that penguin thing

This doesn't make sense because game B is not guaranteed to win $3.

47.5%- loss automatically

10%- just very likely

@5:00 you can’t win on the green zeroes unless you bet on them… right?

Nah, you're wrong it's 50/50. You win or you don't it's not that hard.

as someone who has studied mathematics, I am wondering whether you have been studying mathematics or something similar (cs, physics)? Your understanding about probability theory and logic is far beyond the average joe

1000000th view

9:14 penguins dont live on the northpole they live on the southpole

My friend be u have 76% chance of not losing.. Not of winning

Casino rules

Don't let him him

im at 0:35 and im sure its just a cheap game hack

Anyone here after casino update tryna find life hacks to win roulette

2:06 oh shoot look at the bottom of the stairs

Lol the sum is 666

2 loses makes 1 win.

Blackjack defies your claim about independent randomness in casinos.

Wow i saw this through much faster than normally, am i getting smarter every time? Nah not really…

616 is the number of the beast not 666 -nice try vsauce

9:13

SOUTH pole. Penguins don't live on the north pole.

10:48 the 102 -> 105, you would have to play game A again because 102 is a multiple of 3

Literally every video

Kevin:*

puts a question*Also Kevin:

N O10:18 you can play game b cause 100 is even and then get 103 lose a to 102 then 105 etc it gets higher faster cause you start higher

Imagine any of the Vsauce crew being your teachers. I don’t have words for how awesome it would be I’d want to go to school so bad

DONUT!?

2 wrong situations can make a right? Example! Situation 1: Your Boyfriend/Girlfriend leaves you. You feel bad so the situation is wrong. Situation 2: Your Boyfriend/Girlfriend abuse you. Once again. You are in a wrong situation. But by interacting, once both situations happen, you are in a better situation than before (completely unrelated to my personal life. That is just where my brain went for an example)

Its not really 2 losing games. Its 2 losing games and 1 winning one, or lets just say 1 losing one and 1 winning one because we don't care about the second losing one

Health is by my side..,

for (1+2+3…+35+36) we do (1+36)/(n/2) where n is the total number of terms which all equals

566

0:25 ´cuz – * – = + v:

I dont know why… but it's exactly like when you multiply two negative numbers and you get a positive result.