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输即是赢(帕隆多悖论)

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

Vsauce! Kevin here with two games.
《维瑟科普》! Kevin这里有两款游戏
If you play Game A —
如果你玩游戏A
you are guaranteed to lose.
你保证会输
If you play Game B —
如果你玩游戏B
you are guaranteed to lose.
你依然保证会输
But, if you alternate back and forth
但是 如果你来回交替地
between playing Game A and playing Game B —
玩游戏A和游戏B的话
you are guaranteed to win
那你保证能赢
How? How can two losing games
怎么回事?怎么两个必输之局
combine into one winning one?
联合起来就成了胜局呢?
Is it just some cheap game hack or something more?
这是游戏里的小伎俩吗 还是别的什么东西?
And does this mean that grandma was wrong
它是否意味着我们的经验出现了问题
and it’s actually possible for two wrongs to make a right?
还是真的有可能出现负负得正这种事?
Welcome to Parrondo’s Paradox.
欢迎一起来探讨 帕隆多悖论
Real quick, I did a video with donut media where they brought me out
插播一下 我和donut媒体一起做了个视频
to rip around a racetrack
视频里 他们带我绕着赛道疾驰
and talk about carbon fiber, aerodynamics, and carbon ceramic brakes.
谈论碳纤维 气体力学 陶瓷制动器
you know awesome things.
你懂的 一些神奇的东西
Go watch the video on their channel —
大家可以去他们的频道 观看这个视频
link down in the description below
链接就在视频简介里
and tell them that Vsauce 2 sent you in the comments
然后在评论里告诉他们 是《维瑟科普2》推荐的
Okay, physicist Juan perón Doe
好了 物理学家胡安·帕隆多
who created this paradox
创造了这个悖论
demonstrated it by flipping a biased coin.
他通过抛一枚有偏硬币来演示这个悖论
A coin that, instead of having 50/50 odds
这枚硬币的正反面朝上的概率
of landing on heads or tails has
不是50/50
49.5/50.5 odds.
而是49.5/50.5
So slightly unfair, biased odds.
几近均等而稍微有偏的概率
And… that’s not gonna work for us.
然后……这对我们来说是不可能的
Because precision-biased coins
因为这种精密设计成有偏
that look like normal coins are likely impossible,
看起来却很正常的硬币 几乎不可能存在
statisticians Andrew Gelman and Deborah Nolan
统计学家安德鲁.格尔曼和德博拉·诺兰
wrote in a 2002 paper,
在2002年的论文中写道
“ You can load a die
你可以给骰子注铅
but you can’t bias a coin.”
但你没办法使硬币有偏
So, instead, I made a series of games
因此 为了替代有偏硬币 我做了一套游戏
using 3 roulette wheels with a total of 114 spaces
用到了总共有114个小室的三个轮赌盘
and 1 Markov Chain.
和一个马尔科夫链
Because…PENGUINS.
因为……企鹅
Seriously, this paradox can be visualized
真的 我们可以借助一个企鹅滑梯玩具
using a penguin slide toy.
来想象这个悖论
The same mechanism that moves these penguins to the top
同样的原理 将这些企鹅移动到顶部
allows us to turn two losing games
让我们把两个必输的游戏
into our own money-printing machine.
变成赢钱的机器
This is… this is really loud.
这……这个装置太吵了
I’ll explain how later
我待会儿再解释是怎么做到的
but for now let’s actually talk about our games.
现在我们先来谈谈这些游戏
Okay, Game A works like this:
好了 游戏A是这样运作的
it’s a game of chance
这是个看运气的游戏
in which your odds of winning are slightly lower
因为你赢的几率
than your odds of losing.
比输的几率略低一点
The American roulette wheel has 38 spaces —
美式轮赌盘有38个小室
18 red and 18 black, and these two 2 green zeroes.
18个红的和18个黑的 还有这2个绿色的零
The rules of Game A are that
游戏A的规则是
you can only bet on red or black.
你只能押红色 或者押黑色
Either way, our odds of winning are 18/38,
无论选哪个 获胜的概率都是18/38
or 47.36 %
也就是47.36%
so a little less than 50/50.
所以是略低于50/50的
That means the house edge is 5.2 %,
这意味着庄家优势是5.2%
which we’ll round down to 5%.
我们四舍五入成5%
Because of that 5 % losing edge, theoretically,
从理论上讲 因为这5%的庄家优势
every time we bet $ 1 on this game,
在这个游戏里 我们每赌1美元
we’ll lose about 5 cents.
就将输掉大约5美分
If we start with $100 and keep playing this game,
如果我们从100美元开始玩 并且一直玩这个游戏
we’ll be totally broke after about 2,000 spins.
那么我们将会在玩第2000次时彻底破产
In the long run, when we play Game A,
从长远考虑 当我们玩游戏A时
we’re guaranteed to lose.
我们就输定了
Come onnnn, red!
来吧 红色!
Since definitely losing sounds terrible,
既然必输听起来这么可怕
let’s give Game B a shot.
那我们就来试试游戏B吧
Before we do that
在开始之前
you may want to add up all the numbers on a roulette wheel
你不妨把轮赌盘上的所有的数字加起来
and just tell me the sum in the comments.
然后在评论区中告诉我总和
Okay.B is composed of two games of chance,
好的 B是由两个博弈游戏组成的
each with different odds.
两者都有不同的胜率
You’ll still bet $1,
你还赌一美元
but the wheel that you play
但是你玩哪个轮赌盘
depends on how much money you have left.
取决于你还剩下多少钱
We’ll call these B1 and B2.
我们把这两个轮赌盘叫做B1和B2
You’ll play B1 only if your total money is a multiple of M
只有当你的总钱数 是M的倍数时 你才会玩B1
and let’s say that M = 3.
假设M等于3
So if your leftover money is a multiple of 3,
所以 如果你剩下的钱是3的倍数时
like 93 or 81 or 66,
比如93或者81 或者66
then you have to play wheel B1.
你就得玩轮赌盘B1
If your bankroll balance is not a multiple of 3,
如果你的资金余额不是3的倍数
then you’ll play wheel B2.
那么 你将会玩轮赌盘B2
Here’s the catch.
这里请注意
With B1, you’re only allowed to make what’s called a Corner bet —
玩B1时 你只能去下所谓的角注——
choosing an intersection of 4 numbers,
选择四个数字的交叉点
so you’ll win $1 if the ball lands on any of those four.
如果球落在这4个数字中的任何一个上 你将赢得1美元
So, like a corner bet of 26 27 29 and 30.
比如 这是一个26 27 29和30的角注
A corner bet means your odds of winning
角注 就代表你赢的几率
are just 4 out of 38,
仅为4/38
or… about 10%.
也就是…… 大约10%
So when you play B1,
所以当你玩游戏B1时
you’re going to lose 90 % of the time.
你将会有90%的次数会输
Losing isn’t guaranteed,
输不是必然的
but you’re going to lose a lot more often than not.
但是你会经常输
Game B1 is a pretty bad game for the player.
对玩家而言 游戏B1非常糟糕
Don’t worry, though, B2 is much better.
不过不要担心 B2要好的多
You get to choose a combination of winning spaces-
你可以选择一个你认为有胜算的小室组合
red or black and odd or even.
红色或黑色 奇数或偶数
So, if you choose red and evens,
所以 如果你选择红色和偶数
you’d win every time the ball lands on a red space
那么每次小球落在红色或偶数小室上
or an even space, even if it’s a black even space.
即使是落在黑色的偶数小室上 你都会赢
The two green spaces are also winners for you on B2.
B2中的两个绿色小室也是你的获胜小室
This allows B2’s odds
这就让B2的胜率
to shift significantly in your favor
变得对你十分有利
With 18 reds and 9 black evens
18个红色小室 9个黑色偶数小室
plus the two greens
再加上两个绿色小室
giving you a total of 29 chances
一共给了你29次机会
to win out of 38 possible spaces —
在38个可能的小室中获胜——
and that, my friends, is 76%.
朋友 也就是说 获胜几率是76%
The good news is that since there are more possible money counts
好消息是 因为你的资金计数 不是3的倍数的可能性
that aren’t multiples of 3 than there are,
比是3的倍数更大
you’ll be playing the much-friendlier B2 a lot more
所以 你更多时候是在玩更友好的B2
than the nearly-impossible B1.
而不是玩几乎不可能赢的B1
So that means most of the time, with Game B,
所以这就意味着 你玩游戏B的大部分次数
you’ll be winning, right? No!
你都会赢 对吗?不!
Here’s the thing.
事情是这样的
B1 is a loss 90% of the time.
B1有90%的次数会输
And since you’ll also lose roughly
又因为你玩游戏B2时
every 1 out of 4 times playing Game B2,
大概每四次会输一次
overall you’re guaranteed to lose
总的来说 如果你只玩游戏B的话
if you just play the B Games.
你肯定会输
Game B is actually a Markov chain,
游戏B实际上是一个马尔科夫链
a stochastic process that takes a situation like ours
是一个随机过程 以我们的情况为例
— different states with varying probabilities —
——不同的情况有不同的概率——
and generates a loss by B1’s disproportionate effect
这个随机过程通过BI对我们财产的不等比例的影响
on our fortunes.
造成了资金上的损失
Let me explain. The key is that
让我来解释一下 关键在于
even though there are only three possible states of our money balance
即使我们的资金余额 只有3种可能的情况
— a multiple of 3, like 90 or 93,
——一种是3的倍数 比如90或93
or two possible states in between multiples of 3,
另两种可能的情况介于3的倍数之间
like 91 and 92 —
比如91和92——
our probabilities of playing the bad B1 game vs. the better B2 game
但是 我们玩糟糕的游戏B1和友好的游戏B2的概率
aren’t a simple ⅓ and ⅔.
并不是简单的1/3和2/3
What’s surprising is that a Markov Chain analysis,
令人惊讶的是马尔科夫链分析
and if you want to learn more about Markov chains,
当然如果你想学习更多关于马尔科夫链的知识
I’ll link you to a course on those in the description,
可以在视频简介里找到相关课程的链接
shows that our probability of playing the bad B1 game
分析表明 我们玩糟糕的游戏B1的概率
is actually closer to 40 %.
实际上接近40%
And the winning edge that the good B2 game gives us
而友好的游戏B2带给我们的优势
just isn’t enough to make up for the terrible B1.
并不足以弥补糟糕的游戏B1
All of that is to say that
所有的这些都说明
Game A is a simple guaranteed loser.
游戏A是个简单的绝对败局
And Game B is a kind of complex
而游戏B是个有点复杂的
guaranteed loser. And…
绝对败局
this is getting dreadful
这太可怕了
so let’s bring back our penguins
所以让我们回到企鹅的话题
and lighten the mood.
然后放松一下心情
Our flightless bird friends will help us visualize
我们这位不会飞的朋友 将会帮助我们想象
how we leverage two negatives
如何利用两个不利的局面
to create a positive.
来创造有利的局面
In a 2000 paper titled,
在2000年的一篇论文
“ Parrondo’s paradoxical games and the discrete Brownian ratchet, ”
《帕隆多悖论与离散布朗棘轮》中
Derek Abbott, Parrondo and others,
德雷克·阿伯特和帕隆多 还有其他一些人
described a process known as the “ flashing Brownian ratchet”
描述了一个“闪烁的布朗棘轮”的过程
as an analogy for how the paradox works.
用来类比帕隆多悖论如何运作
Basically, directed motion is achieved
基本上 定向运动是通过
by alternating between a sawtooth and a flat potential.
在锯齿形和扁平形的力之间交替来实现的
I’ll show you.
我来向你演示
These penguins clearly reach the top of the slide,
这些企鹅很显然能到达滑梯的顶端
even though they themselves aren’t moving,
尽管它们自己并没有移动
because they’re being carried back and forth between
因为它们被带着
two downward-sloped-sawtooth shapes and a flat shape.
在两个向下倾斜的锯齿形和一个平面形状之间来回移动
These shapes slope downward.
这些形状都向下倾斜
And this shape here — sorry, penguin!
然后这个形状——抱歉 企鹅
— is flat.
——是平的
Yet the penguins climb to the top
然而企鹅是通过在这两种形状中来回交替
by alternating between them.
从而爬上顶端的
Ok, that’s loud, sorry penguins.
好的 这太吵了 抱歉了企鹅们
Wheeee! The end.
喔——结束
Excuse me for one second, Happy Feet.
打扰一下 快乐的大脚
Think of Game A as our flat shape
我们将游戏A想象成平面形状
because the odds of winning Game A are close to 50/50
因为玩游戏A赢的几率接近50/50
with a slight bias toward losing.
但输的几率略大于50%
Think of Game B as our sawtooth shape
把游戏B想象成锯齿状
because the odds are much steeper in Game B1
因为在游戏B1中的获胜的几率非常小
and much better in Game B2,
在游戏B2中获胜几率则大得多
creating a distinct asymmetry.
这形成了明显的不对称
Look at that! But instead of moving plastic penguins uphill,
看那儿!但是 我们并不是要通过交替地玩游戏A和游戏B
by alternating between Game A and Game B,
来将企鹅向上移动
we, the player, are carried upward
我们作为玩家 是通过把必输的游戏联合起来
by our combination of losing games into winning.
变成胜局 从而爬上胜利的顶端
To put it another way,
换言之
we can get really simple and mathy about this.
我们可以对此进行非常简单和数学化的处理
Back to the North Pole with you!
回到一开始
Say you start with $100,
假设你一开始有100美元
and each time you play Game A,
并且每次你玩游戏A
you lose $ 1.
都会输掉一美元
That’s it — done.
就是这样 结束了
That’s just how Game A works in this simplified scenario.
这就是游戏A在这个简化场景中的运作方式
You’re a loser with no hope of winning.
你是一个没有胜利希望的输家
And if you played Game A 100 times,
如果你玩了100次游戏A
your financial trajectory goes downward
你的经济曲线会一直下滑
until you’re broke.
直到你破产
There are two strains of Game B:
游戏B有两种情况:
if the amount of money you have left is an even number,
如果你剩下的钱是偶数
like $82, then you win $3.
比如82美元 那么你将赢得3美元
If it’s an odd number — say, you’re down to $ 71–
如果是个奇数 比如你只剩71美元了
then you lose $5.
那么你将输掉5美元
If you only played Game B,
如果你只玩游戏B的话
you’d be broke even faster than
那么比起只玩会自动输掉的游戏A
if you’d just auto-lost Game A.
你甚至破产得更快
At least with Game A
至少是游戏A的话
you’d get to play it 100 times.
你还能玩个100次
Both Game A and Game B
游戏A和游戏B
are clearly 100 % losing games,
显然都是100%会输的游戏
but when you switch back and forth,
但是 当你来回切换着玩A和B
you can make them profitable.
你就可以赢钱
Check it out. Play Game A and you’re down to $99.
来看一下 玩游戏A 你的钱降到99美元
That’s an odd number,
这是个奇数
so you don’t want to play Game B
所以你不想玩游戏B
or else you’d lose another $5…
否则你还会再另外输掉5美元
instead you play Game A again
相反 你再玩一次游戏A
and just lose another dollar.
然后又输掉1美元
Now you’re at $98 —
现在你有98美元
when you switch to Game B,
当你转到玩游戏B时
you win $ 3,
你赢了3美元
and suddenly you’re up to $ 101.
你的钱突然就涨到了101美元
Lose Game A to get to $100,
玩A输掉 变成100美元
win Game B to get to $103.
玩B赢了 变成103美元
You can repeat a cycle of A/B switching
你可以重复一个A/B转换的周期
to amass infinite wealth
来积累到无穷的财富
despite playing two games that on their own
尽管单独玩这两个游戏时
are guaranteed losers.
你肯定是个输家
But is that really a paradox?
但这是一个真正的悖论吗?
When we examine the best way to flip biased coins
当我们研究抛有偏硬币
or spin biased roulette wheels,
或旋转有偏的轮赌盘的最佳方式时
it just seems like just a cheap trick
这看起来似乎只是一个低级伎俩
to manually hack a couple of games.
可以用来手动破解几个游戏
It’s not.
这不是真正的悖论
The paradoxical situation is that
悖论的情况是指
you can alternate playing Parrondo’s two losing games randomly
你可以随机轮流地玩帕隆多的两个必输游戏
and it will still produce a winning one.
并且它最终仍然会变成胜局
We can use Stan Wagon’s Parrondo Paradox wolfram simulator to prove it.
我们可以用Stan Wagon的帕隆多悖论元素模拟器来证明
This simulator uses Parrondo’s biased coin flip odds
这个模拟器使用帕隆多的有偏硬币抛掷概率
and allows us to set the number of flips
并且允许我们设置抛掷次数
and how many times we want to repeat that experiment.
以及我们想要重复实验的次数
Parrondo’s numbers for Games A, B1 and B2
游戏A B1和B2的帕隆多概率数字
are almost identical to our roulette example.
和我们的轮赌盘例子几乎相同
So, we set the number of flips in a game
因此 我们设置一次游戏的抛掷次数
and how many times we’ll repeat that game,
以及游戏重复的次数
and each time we click New Run,
每次我们单击New Run
the simulator crunches those numbers…
模拟器都会处理这些数字
Let’s set this to run 1 repetition of 1,000 flips
我们将其设置为点击一次运行 就重复抛1000次
when we play on the specific pattern
当我们按一种特定模式来玩时
— BBABA — we clearly win over time.
比如 BBABA式 很明显最后我们赢了
But what happens if we choose Game A or Game B randomly?
但如果我们随机选择游戏A或B呢 会发生什么?
Most of the time we win over 1,000 flips,
在抛掷的1000次里 大多数情况下我们都会赢
occasionally we don’t — that’s just variance.
偶尔会输 这只是方差
But when we run 2,000 simulations of 1,000 flips,
但当我们运行模拟重复2000次 每次抛掷1000次时
a pattern emerges: we see a clear upward slope.
一个模式出现了:我们看到一个明显上升的斜率
Despite having no plan and no specific hack,
尽管没有任何计划 没有特定的破解方法
alternating randomly between Games A and B
但是在游戏A和B之间随机切换
yields a long-term win.
也会获得一个长期的胜利
So even if it takes a while,
所以 即使它要花费不少时间
even if it’s done randomly,
即使它是随机进行的
by alternating between these two losing games,
但是通过交替地玩这两个必输的游戏
we actually get a winning result.
我们实际上能得到一个胜利的结果
We get to the top of the slide–
我们到达了滑梯的顶端——
like our penguins.
就像这些企鹅一样
But here’s the question.
但问题来了
Can you use this strategy to guarantee
你能用这个策略来保证
that you win at a casino? No.
你在赌场大获全胜吗?不能
Parrondo’s Paradox depends on
帕隆多悖论取决于
being able to interact within two games,
两个游戏之间 能够相互影响
and you just can’t do that in a casino.
而这在赌场中是做不到的
Real casino games like slot machine spins or the way roulette
真正的赌场游戏 比如老虎机旋转或者轮赌盘
is actually played are based on entirely separate events
实际上是基于完全独立的事件
–one outcome will never influence
一个结果永远都不会影响
the outcome of the next round or a different game,
下一轮或者另一种游戏的结果
like a roulette result will
就像轮赌盘的结果
never steer you toward
永远不会引导你进入
an advantageous round of Pai Gow.
有利的牌九回合
By guaranteeing that games never intersect,
通过保证游戏之间不相互影响
casinos avoid the possibility of an exploit
赌场避免了被利用
or a weird Parrondo situation.
或者出现怪异的帕隆多情形的可能性
The outcome of every ‘bettable’ event in a casino
赌场上 每一场赌局的结果
depends on nothing before it or after it.
都不取决于任何发生在它前面或后面的结果
It exists in its own impenetrable bubble
它存在于自己无法穿透的气泡中
that pops as soon as that round is over
这轮游戏一结束 气泡就会破裂
and then is blown up again —
然后 这个气泡又会被吹大
and that keeps the games fair
这保证了游戏的公平性
and keeps the casino’s math predictable.
也让赌场的数学计算保持可预测性
The independence of each game
每个游戏的独立性
is actually one of the reasons
实际上是你很少在赌场
you so rarely see new games at casinos.
看到新游戏的原因之一
It’s just…It’s just hard to come up with games
因为……因为我们很难创造出这样一款
that are totally independent,
完全独立的游戏
give the house enough of an edge
既要给足庄家优势
that they win over time
让他们最终赢得游戏
but also give the player enough of a chance
又要让玩家有足够的机会
that they want to play and have fun playing.
让他们愿意玩并且玩得开心
What’s exciting about Parrondo’s Paradox
帕隆多悖论最让人兴奋的地方
is not how to win money playing two games
不是怎样通过玩两个游戏赢钱
that no casino would ever allow.
因为没有赌场会允许这种情况发生
It’s figuring out
而在于要研究清楚
how to apply its surprising property
如何将这种神奇的特性
to other fields of study.
应用到其他研究领域
Researchers are working on
研究人员正致力于
real world applications for it in disciplines
研究它在现实世界中的应用
ranging from quantum mechanics to biogenetics.
范围从量子力学到生物遗传学规律
And for the rest of us,
对于我们其他人来说
it can be helpful to realize that
意识到两个必输的游戏
two losing games can become one winning one.
可以变成一个胜利游戏 是很有帮助的
That, in a way,
这听起来很奇怪
as weird as it sounds,
也就是说 在某种程度上
two wrongs can make a right.
负负可得正
And even what seems, by all accounts,
即使是在一个所有人都说
like a totally hopeless situation
看起来完全没有希望的情况下
can, undeniably, mathematically,
也可以毫无疑问地 数学上地
be turned into a winning one.
转化为一场胜局
And as always, thanks for watching.
一如既往 谢谢大家的观看
Hello, Ahhh, Hello there, my little penguins.
嗨 嗷 好了我的小企鹅们
I have some suggestions.
我有一些建议
Suggestion #1,
建议1
please hit the like button
如果你喜欢这个视频
if you want to like the video.
请点赞
That’s helpful to me
这对我来说是极大的鼓励
and it takes a split second of your time.
而这只需要你花一点点时间
Also, subscribe if you aren’t yet subscribed
如果你还没有订阅节目 欢迎订阅
so you can watch more Vsauce2 videos.
这样你就能看到更多《维瑟科普2》的视频
And if you want to take our relationship to the next level,
如果你想进一步增进我们的关系
hit that notification bell.
点击通知铃声
Because then you’ll know exactly
因为这样你就能确切地知道
when I upload a new video.
我什么时候上传了新视频
So ring-a-ding-ding that bell, as they say.
就像他们说的 会响起“叮叮”的铃声
No one says that. But now I do.
好吧 没人说过 但是现在我说了
If you want to watch more videos that I’ve made,
如果你想观看更多我做的视频
just click over here.
就点击这里
And the… Other than that,
另外 祝大家
have a great day…
生活愉快
(laugh)

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译制信息
视频概述

带你走进负负得正的神奇世界,看看赌场奥妙躲过骗局,想想未来科学之光……

听录译者

收集自网络

翻译译者

千思与

审核员

审核员BA

视频来源

https://www.youtube.com/watch?v=PpvboBJEozM

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