The Game You Win By Losing (Parrondo’s Paradox)

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
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over here. Other than that, have a great… day…

100 thoughts on “The Game You Win By Losing (Parrondo’s Paradox)”

  1. 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. 😉


  2. So the presumption is that you, whilst even in finance spend on even on the higher loss game in an attempt to recoup the losses on the first game and use the loss in the first game to get you even to win on the second game.

  3. At 3:27, he tells us to add the numbers of the roulette wheel. I know the answer, but I'm not gonna spoil it. I will give a hint, however: Roulette used to be referred to as "The Devil's Game."

  4. If you have a series of numbers and you wana know the sum of them, this is how you can do it quickly: take half of the highest number and multiply it with the highest number plus one. So: (36:2)x(36+1)=666

  5. May I ask what your camera setup is for your videos? Have seen a couple of them and it feels like you are using two cameras at the same time.
    Are they connected to a PC while recording with an HDMI Grabber or are they saving it on their own Memory cards/SSDs first?

  6. Sooo… it's basically like the strategy in Mario Kart were you slow down on purpose to get last place and, with that, a better chance of rolling good items which in return help you make up for more than enough of the time lost.

  7. So… In other words, don't bet M/3 on game B? Also, woudn't you have to bet 20 times with 1 dollar at a time to get to 99… 98.. if you bet $100 and lost.. you would be at $0. And if you can choose how much your betting, why not just always bet 11$ on game B… If you always have to bet MAX, then your going to end up with 0 after 1 loss. The only way this works is, you walk in with $100, forced to bet MAX, choose game B2 with 74% winrate, and win, then leave.

  8. So in the example at 5:30 you do combined bets on even plus a colour, right? Even if the zeros count it's not like every even will be a win. If you hit the even but miss the colour (or hit the colour and miss the even) you don't win, you get your bet back (assuming both bets are the same).

  9. My guess is that both games rules work like this you get to pick a random 3 numbers that equal a even number but you can only pick or get 3 odd numbers both games work the same so you have to play both games to win

  10. totally usless video esp considering you lose on green unless you bet on one of the green u dont win if u bet even or odd and it comes up zero so wtf is this even? do the math with LOSING on a green

  11. Penguins don't live in the south pole, they live all around the southern hemisphere, from the edges of Antarctica, to the beaches of South America and Australia, and Africa.

  12. That's the point. You created an interdependence in the rules. You could have also postulated that you get a reward when you lose both types of games consecutively. And it's not just about winning/losing but the expected value.

  13. So, since the zeroes aren't going to add to the total, count them out.

    1 + 36 = 37
    2 + 35 = 37

    You can simplify this to 37 * 18 (Half the numbers on the board)

    37*20 = 740
    740 – 74 (37*2) = 666

    And this is how I do math in my head.

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