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知晓答案却仍难以置信的谜题

The Riddle That Seems Impossible Even If You Know The Answer

There is a riddle that is so counterintuitive,
有个非常反直觉的谜题
it still seems wrong even if you know the answer.
即使告诉你谜底 仍让你难以置信
– You’d think it’s an almost impossible number.
你会认为这几乎是个不可能的数字
– I feel like you probably hit me with some truth bomb.
你好像在朝我扔“真相炸弹”一样
– I mean, if you’re trying to create controversy
如果你想让大家争论起来
and you’re trying to confuse people,
或者让大家陷入困惑
you’re gonna succeed.
你会成功的
(both laughs)
[同时发笑]
– There are a bunch of YouTube videos about it,
有许多讲解这个谜题的油管视频
but I find all of them either incorrect or incomplete.
不过我发现它们都是错误或不完整的
So in this video, I’m going to dive deeper
所以本视频 我将更进一步
and explain it fully.
全面地解释该谜题
Here is the setup.
谜题的设定是……

监狱
(suspenseful music)
[悬疑音乐]
Say there are 100 prisoners numbered 1 to 100.
假定有100个囚犯 编号1-100
Slips of paper containing each of their numbers
在纸张上分别写上他们的号码
are randomly placed in 100 boxes in a sealed room.
随机放置在一个密室的100个箱子中
One at a time, each prisoner is allowed to enter the room
每次让一个囚犯进入该房间
and open any 50 of the 100 boxes,
打开100个箱子中的50个
searching for their number.
寻找自己的号码
And afterwards, they must leave the room exactly as they found it,
一找到自己的号码就必须立即离开房间
and they can’t communicate in any way
并且不许与其他囚犯
with the other prisoners.
进行任何形式的交流
If all 100 prisoners find their own number
如果100个囚犯都在房间中
during their turn in the room,
找到了自己的号码
they will all be freed.
他们将被全部释放
But if even one of them fails to find their number,
可只要有一个囚犯没找到
they will all be executed.
全部囚犯都将被处决
The prisoners are allowed to strategize
囚犯进入房间前
before any of them goes into the room.
是可以一起制定策略的
So what is their best strategy?
那么最好的策略是什么呢?
If they each search for their own number randomly,
如果每名囚犯随机打开箱子
then each prisoner has a 50% chance of finding it.
则每人有50%的几率找到自己的号码
So the probability that all 100 prisoners find their numbers
100名囚犯都找到自己号码的概率就是
is 1/2 times 1/2 times 1/2 a hundred times
1/2×1/2×1/2… 这样乘100次
or 1/2 to the power of 100.
即1/2的100次方
This is equal to 0.00000000 30 zeros,
等于0.00000000… 30个0
and then an eight.
最后是个8
To put this probability into perspective,
把这个概率解释得形象点就是
two people have a better chance of picking out
两人从地球上所有海洋与沙漠中
the same grain of sand from all
挑选出同一粒沙子的概率
the beaches and deserts on earth
都比囚犯用随机开箱的方法
than by escaping this way.
逃出监狱的概率大
But what if I told you that with the right strategy,
可如果我告诉你 只要用正确的策略
there’s a way to raise their chances to nearly one in three.
他们就能将逃出几率提升至近1/3呢?
It improves their odds of a random chance
比随机开箱的逃出概率
by nearly 30 orders of magnitude.
提升了30个数量级
That’s like taking a millimeter and scaling it up
这就像把一毫米扩大到了
to the diameter of the observable universe.
可观测宇宙的直径那么长
– But they can only coordinate this strategy beforehand.
不过他们只能在进房间前商量?
– Correct.
没错
– Is this true?
真能有1/3?
– Yes.
真的
– Teach me.
教我
– This is not a trick question.
这可不是道陷阱题
The solution just involves an incredible feature of math.
解决方案其实包含了对数学的绝妙运用
So what is this mathematical strategy?
这个数学策略究竟是什么呢?
Well, if you don’t already know the answer,
如果你还不知道答案
feel free to pause the video here and try it for yourself.
可以暂停视频自己尝试一下
And if you don’t come up with it, don’t worry,
如果你还是想不出来 别担心
you’re in good company.
你不是一个人
Even the person who came up with this riddle,
就连想出该谜题的
computer scientist Peter Bro Miltersen,
计算机科学家Peter Bro Miltersen
he didn’t even think of this strategy
也是在他同事提示他后
until a colleague pointed it out.
才想出了这个策略
Miltersen, ultimately published this problem in a paper
Miltersen最终在论文中发表了该谜题
where he generously left the solution
且大方地将解决方案作为练习题
as an exercise for the reader.
留给读者自行解决
So here is the solution.
那么以下就是解决方案
Pretend you are one of the prisoners,
假如你是囚犯之一
when you go into the room,
当你进去房间时
open the box with your number on it,
打开写着你号码的箱子
the number on the slip inside probably won’t be yours,
里面纸张上的号码不一定是你的
but that’s okay.
不过问题不大
Go to the box with that number on it.
继续打开与纸张上数字相同的箱子
look at the number inside,
看看里面的号码
then go to the box with that number on it, and so on.
再继续找对应的箱子 以此类推
Keep doing this until you find the slip with your number.
直到你找到写有你号码的纸张
If you find your number,
如果你找到了自己的号码
that essentially tells you
也就意味着
to go back to the box where you started.
你会回到你最初打开的箱子
It closes the loop of numbers you’ve been following.
于是 你经过的所有数字形成了闭环
But if you’ve found your number, then you’re done.
不过 你找到了自己的号码 已经齐活了
You can stop and leave the room.
你不用回到起点 可以直接离开房间
This simple strategy gives over a 30% chance
这个简单的策略使全部囚犯
that all the prisoners will find their number.
都找到号码的概率上升到了30%以上
– The entire pool has a 30…
总概率是百分之三十…
– Everyone can find their number 31% of the time.
所有人都找到自己号码的概率是31%
– What?
啥?
– But how does it work?
那么原理是什么呢?
The first thing to notice is that all boxes
首先要清楚 所有箱子
become part of a closed loop.
都是某个闭环的一部分
The simplest loop would be
最简单的闭环是
a box that contains its own number.
内外号码一致的箱子
If you’re prisoner number one and you go to box one,
如果你是囚犯1号 你去打开1号箱
it contains slip one, then you’re done.
里面纸张也是1号 那就成功了
Your number was part of a loop of one,
你的号码是单个数字闭环中的一部分
but you could also have a loop of two.
你也可能会遇到两个数字的闭环
Say box one points to box seven
比如1号箱中有数字7
and box seven points back to box one.
7号箱中有数字1
Or you could have a loop of three, or four, or five,
也可以是三个数字闭环 或四个 五个
or any length all the way up to 100.
至多是由100个数字形成的闭环
The longest loop you could have would connect
最长的闭环是把所有数字
all the numbers in a single loop.
用一个闭环串联
But more generally, any random arrangement
不过通常情况下 若随机排列
of the slips in these boxes
箱子中的纸张
will result in a mixture of some shorter
结果会是有的闭环短
and some longer loops.
有的闭环长
When you start with a box labeled with your number,
当你从标有你号码的箱子开始搜
you are guaranteed to be on the loop
也就确保自己进入了
that includes your slip.
你号码所处的闭环中
So the thing that determines whether or not you find your slip
所以决定你是否能找到自己号码的
is the length of the loop.
是闭环的长度
If your number is part of a loop that is shorter than 50,
如果你号码所处的闭环有少于50个数字
then you will definitely find your slip.
那你肯定能找到自己的号码
But if your number is part of a loop that is 51 or longer,
可如果你号码所处闭环有51个数字或更多
you are in trouble.
就麻烦大了
You won’t find it before you’ve exhausted
你按规则费力翻找50个箱子后
the 50 boxes you’re allowed to search.
会一无所获
When you open the box labeled with your number,
当你打开标有自己号码的箱子时
you are in fact starting
其实是从闭环中
at the farthest point on the loop from your slip.
离你的号码纸最远的地方开始的
You wanna know where is the slip that points to this box,
你想知道和这个箱子号码一致的纸张在哪
but to find it, you have to follow the loop of numbers
可为了找到它 你必须跟随闭环中的数字
all the way around to the end.
一步步到达终点
That means if the prisoners follow this strategy
这就意味着 这些囚犯用这个策略时
and the longest loop is 51,
若最长闭环有51个数字
not just one or two prisoners
那么不止一两个囚犯
will fail to find their number,
会找不到自己的号码
but all 51 on this loop won’t make it.
而是该闭环的51个囚犯都会失败
They make it to the box just before the box with their slip,
他们离装有自己纸张的箱子只有一步之遥
but they have to stop searching there.
却只能止步于此
(both laughs)
[同时发笑]
So the probability that all of the prisoners succeed
所以所有囚犯都成功的概率
is just the probability that a random arrangement
就等于100个号码随机排列后
of a hundred numbers contains no loops longer than 50.
没有闭环超过50个数字的概率
Now I promised that this probability would come out
我能肯定这概率是
to around one in three,
大约1/3
but how do we calculate it?
可具体怎么算呢?
Well, imagine writing down all the different ways
假如你要写下所有能将
that you could connect 100 boxes to form a loop
100个数字连接成一个闭环
of length 100.
的不同方法
So you could have box one points to box two,
可以是 1号箱中数字指向2号箱
box two points to box three to box four, and so on,
2号箱指向3号 再指向4号 以此类推
all the way to 100,
直到100号箱
and then box 100 would point back to box one,
然后100号箱内的号码让你回到1号箱
or you could have something random.
或随机排列
Box five points to box 99 to box 17 and so on,
5号箱指向99号 再指向17号 以此类推
and let’s pick the last one,
最后的箱子是
is 63.
63号
And box 63 points back to box five.
63号箱指回5号箱
So how many different arrangements of these a hundred boxes
所以这100个箱子到底有多少种
or permutations could you have?
不同的排列组合方式呢?
Well, for the first box,
先看第一个箱子
I have 100 different boxes that I could choose from.
我有100个不同的箱子可以选
The second box, because I’ve already used one,
第二个箱子 因为已经选掉一个
I can only pick from 99 boxes,
只能从剩的99个箱子中选
and the next one, I can pick from 98 boxes, and so on,
下一个箱子从剩的98个中选 以此类推
down to the very last box.
直到选完最后一个
I don’t really have a choice.
没得选了
There’s only one box left I could put in the last position.
能放最后的箱子只剩一个了
So the total number of different permutations
那么看下有多少种不同的排列
would be 100 times 99, times 98, times 97,
100×99×98×97
all the way down to one.
一直乘到1
That is just 100 factorial.
也就是100的阶乘
There are 100 factorial different ways
有100!种不同的方法
that you could create a loop of a hundred boxes.
让你创造出由100个箱子连成的闭环
But what we can’t forget
可不要忘记
is that these are not just lines of numbers.
这些不只是一串数字
They are loops.
而是闭环
So some of these lines that look different
所以有些看起来不一样的排列
are actually the same loop.
其实是同一个闭环
For example, two, three, four, five, and so on
例如 2 3 4 5…
up to 100 and then 1
到100 最后回到1
is the same thing as 1, 2, 3, 4, 5 to 100.
和1 2 3 4 5… 到100是一个意思
You can rearrange the way you write these numbers
你可以用100种不同的方法
a hundred different ways,
重新排列这些数字
but they all represent the same loop.
可它们全代表同一个闭环
So the total number of unique loops of length 100
所以 由100个数字组成的不同闭环
is 100 factorial divided by 100.
有100!除以100个
So what is the probability that any
那么100个箱子随机与纸张组合后
random arrangement of 100 boxes
会出现由100个数字组成的闭环
will contain a loop of length 100?
的概率有多大?
Well, it’s just equal to the number of
答案是我们刚计算的
unique such loops that we just calculated,
不同闭环的总数
100 factorial over 100,
100!/100
divided by the total number of ways
除以你能将100个号码放入
that you could put a hundred slips in 100 boxes,
100个箱子可以使用的方法总数
which is 100 factorial.
也就是100!
So the answer is 1 over 100.
所以结果是1/100
So there is a 1% chance that a random arrangement of slips
有1%的几率 随机安排纸张后
results in a loop of length 100,
会出现由100个号码组成的闭环
and this is a general result.
这是个普遍结果
The probability that you get a loop of length 99
你得到由99个数字组成的闭环的概率
is 1 over 99.
是1/99
The probability that you get a loop of length 98
得到由98个数字组成的闭环的概率
is 1 over 98.
是1/98
So the probability that there is a loop longer than 50
那么出现比50个数字多的闭环的概率
is 1 over 51 plus 1 over 52
是1/51+1/52
plus one over 53, et cetera.
+1/53 以此类推
Add all these up, and it equals .69.
全加起来 是0.69
There is a 69% chance of failure for the prisoners,
有69%的几率囚犯们会失败
meaning a 31% chance of success
意味着有31%的概率会成功
where the longest loop is 50 or shorter.
即最长闭环等于或少于50个数字
– I still find it difficult to believe.
我还是觉得难以置信
– [Derek] This feels a bit like magic.
这方法听起来有点像魔法
Using the loop strategy,
只要使用闭环策略
all the prisoners are more likely to find their numbers
所有囚犯都找到自己号码的概率
than even just two prisoners choosing at random.
就会大于两个囚犯随机开箱后都成功的概率
So using the loop strategy,
那么使用闭环策略时
what is the probability that each prisoner alone
单个囚犯找到自己号码的概率
finds their number?
是多大呢?
It is still 50%.
仍然是50%
Each prisoner can still only open half the boxes,
每个囚犯都只能打开一半箱子
so their individual chance is still 1/2,
所以单人概率仍是1/2
but these probabilities
可单人概率
are no longer independent of each other.
不再独立于其他囚犯
Imagine running this experiment a thousand times over.
想象一下 将这个实验进行一千次
If everyone is guessing randomly,
如果每个人都随机开箱
you’d expect that on most runs
可推测 大部分情况下
around 50 prisoners would find their number.
约50个囚犯会找到自己的号码
On lucky runs, the number would be a bit higher,
幸运的话 较多囚犯能找到自己的号码
on unlucky runs, a bit lower.
不幸的话 找到的人少点
But using the loop strategy,
但如果使用了闭环策略
all of the prisoners would
31%的情况下
find their numbers 31% of the time.
所有囚犯都能找到自己的号码
And 69% of the time, fewer than 50 find their number.
还有69%的情况下 找到号码的人不超过50个
The prisoners all win together
要么就是全部囚犯一起成功
or the majority loses together.
要么就是大部分囚犯都找不到
That’s how this strategy works.
这个策略就是这么运行的
– Why are you assuming that your number will always be
为什么你假设你的号码会永远
on the loop that you’re on?
在你选择的闭环内?
– I feel like- – I don’t understand that.
– 我觉得 – 我不理解
– This is a key question, right?
这个问题很关键 对吧?
– ‘Cause I feel like it’s possible to
因为我觉得你有可能
start and go on a complete loop
从一个闭环开始到结束
and not come back to your own number
都没有找到自己的号码
because you got on the wrong loop
因为你选错闭环了
and then you’d have to get on another loop,
那么你就不得不换个闭环重新找
so I don’t know that I’d buy this.
所以我对这个策略深表怀疑
– Right, right, right, right, right, right.
对对对对对对!
Now, the big question everyone asks is
大家最常问的问题就是
how do you know that
怎样才能确保
if you start with a box with your number on it
从有你号码的箱子开始
you are guaranteed to be on the loop
你就一定处在写有你号码的纸张
that contains your slip?
所在的闭环?
Well, if you think about it,
你想想看
the slip that says 73, if anyone sees that,
若有人看到了写着73的纸张
they will definitely go to the box with the number 73.
意味着一定也有个箱子上写着73
So the slip and box with the same number
所以有同样号码的纸张和箱子
essentially form a unit.
成为了一组
They’re like a little Lego brick.
它们像乐高积木一样连接在一起
And then every slip is hidden inside another box.
每张纸藏在另一个箱子里
So as I start laying out slips and boxes randomly,
我现在把纸张和箱子随机摆放
you can see that there’s no way
如你所见 它们一定是
that we can end up with a dead end.
有始有终的
It’s not like you can just get to a box and then stop
你不可能选了个箱子 就没下文了
because every box contains a slip
因为每个箱子里都有一张纸
and that points at another box.
用以指向下一个箱子
So the only way for you to see only boxes
所以 唯一让你走进房间
when you walk into the room
只看到箱子的方式
is for every slip to be contained within a box,
就是把所有纸张都分别放进箱子里
and that necessarily will mean
这也绝对意味着
that we are forming loops.
闭环的形成
So when I start with box 73,
所以若我从73号箱开始
I must eventually find slip 73,
最终一定会找到73号纸
because then and only then
因为这样 也只有这样
will I be directed to go back to box 73
我才能被引回73号箱
which closes the loop.
使闭环形成
– Who is the warden to this prison?
这监狱的监狱长是谁?
And what kind of sadistic mathematical warden are you dealing with here?
这得是个多么变态的数学狂监狱长?
This is awful.
太糟糕了
– Now, what if there is a sympathetic prison guard
现在 如果一个具有同情心的监狱守卫
who sneaks into the room before any of the prisoners go in?
在所有囚犯进房间之前悄悄溜进去
Well, then they can guarantee success for the prisoners
守卫只需交换两个箱子里的纸
by swapping the contents of just two boxes.
就能确保囚犯们一定成功
That’s because there can be at most one loop
因为最多只会有一个闭环
that is longer than 50,
有多于50个数字
and you can break it in half just by swapping
你交换该闭环任意两个箱子的纸张
the contents of two boxes.
就可以将其一分为二
And now I have two separate loops
现在我就得到了两个
that are each shorter than 50.
小于50个数字的闭环
But what if there was a malicious guard
可如果有个恶毒的守卫
who figured out that the prisoners
发现了囚犯们
were going to use this loop strategy?
打算使用闭环策略呢?
Well, then they could put the numbers in boxes
那么守卫就要通过摆放箱子内的纸张
to ensure they formed a loop longer than 50.
来确保有个闭环大于50个数字
In this case, are the prisoners doomed?
这种情况下 囚犯们必死无疑了?
Surprisingly, no.
令人意外的是 不会
They can counter by arbitrarily renumbering the boxes.
他们可以用随意给箱子编号的方式予以反击
They could, for example, add five to each box number.
比如 他们可以把箱子上的号码都加5
The loops are set both by the locations of the slips
闭环会被纸张的放置影响
and by the box numbers.
也会被箱子上的数字影响
Renumbering the boxes is essentially
重新给箱子编号本质上
the same as redistributing the slips.
和重新摆放纸张是一个意思
So the problem is back to a random arrangement of loops,
所以问题又变回了随机安排的闭环
meaning the prisoners are back to their
这意味着囚犯们还是
31% chance of survival.
有31%的几率存活
Now, what happens if you increase the number of prisoners?
那么 增加囚犯数量会发生什么呢?
– Fun fact, nobody knows if
有趣的是 没人知道
as you have more and more prisoners
当囚犯数量越来越多时
it’s going towards a limit,
概率会趋于某个极限值
or if it’ll eventually go down to zero, or what?
还是最终下降到零 还是其他
– That is my friend Matt Parker,
这是我的朋友Matt Parker
and I think what he meant to say
我觉着他该说的是
is we know exactly what happens
即使增加囚犯数量
as you increase the number of prisoners.
我们也可以明确地知道结果
With a thousand prisoners each allowed to check 500 boxes,
1000个囚犯 每人能开500个箱子
you might expect their chance of success
你可能以为成功的几率
to drop dramatically,
会大幅下降
but you can calculate it like we did before,
可若像我们之前那样计算
and it comes out to 30.74%,
结果是30.74%
only half a percentage point lower than for 100 prisoners.
与100个囚犯相比 几率只降了0.5%
For 1 million prisoners,
100万个囚犯的话
the probability of success is 30.685%,
成功几率是30.685%
which is only a little higher than for 1 billion.
比十亿个囚犯的成功几率高一点点
Of course, their bigger problem would be
当然 他们面临的更大问题是
the time it takes to open all the boxes.
打开箱子需要耗费的时间
So your probability of winning this game
所以你赢下这场游戏的概率
does indeed approach a limit.
确实接近一个固定数值
So what is that limit?
什么数值呢?
The formula we’ve been using
我们所使用的公式
is one minus the chance of failure,
是1减去失败的几率
which is the series 1 over 51 plus 1 over 52,
也就是 1-(1/51+1/52…)
and so on, up to 1 over 100.
以此类推 直到1/100
We can depict this series as the sum of areas of rectangles,
我们可以把括号里的数列用柱状图的面积总和表示出来
and there is a curve that follows
因每个柱高度不同
the heights of these blocks.
形成了一条曲线
That curve is one over X.
这个曲线是1/x
The area under that curve from 50 to 100
横坐标50-100区域曲线下的面积
approximates the area of all the rectangles.
接近于该区域所有柱的面积总和
And as the number of prisoners goes to infinity,
当囚犯数量趋于无穷大时
it becomes a better and better approximation.
这个面积越来越趋近于一个值
So to find the probability of failure,
所以要得出失败的概率
we can just take the integral of one over X
我们可以直接取从n到2n
from n to 2n.
1/x的积分
And we find that it’s equal to the natural logarithm of two.
可以发现 结果等于2的自然对数
This gives a probability of success
进而推出成功的概率
of one minus the natural log of two,
是1-In2
which is about 30.7%.
大约为30.7%
The bottom line is
结论就是
that no matter how many prisoners you have,
无论有多少囚犯参与
you’ll always end up with above a 30% chance
使用闭环策略逃脱成功的几率
of escaping using this strategy.
一直大于30%
And that feels really wrong.
这听起来很不合理
I mean, at first it seemed essentially impossible for all 100 prisoners
最开始 100个囚犯都找到自己的号码
to find their numbers.
似乎是不可能的
But now we’re seeing that you could have
可现在 我们知道可以有
a hundred, a million,
一百个 一百万个
or any arbitrarily large number of prisoners
或者任一巨大数量的囚犯参与
with at least a 30% chance that they all find their numbers.
都有至少30%的几率全部找到自己的号码
The beauty of the loop strategy
闭环策略的绝妙之处在于
is linking everyone’s outcomes together,
把所有囚犯的结果绑定在一起
instead of each prisoner walking in
取代了每个囚犯走进房间
with their own 50-50 shot.
各自面对五五分的概率
Following the same loops means that
使用闭环策略意味着
they have the exact same chance of finding their number
他们和处在同一闭环的其他囚犯
as everyone else in their loop.
成功找到号码的概率一样
And once the boxes and slips are arranged,
并且一旦刻意摆放箱子与纸张
that chance is set at either 100% or 0%.
成功几率要么是100% 要么是0%
With this strategy,
用这个策略
you can’t ever get close to winning with only a few
不会因为只有一小部分人没找到号码
people missing their numbers.
而功败垂成
You can only fail hard or succeed completely.
囚犯要么彻底失败 要么集体成功
《Ve元素》
Now, if you like solving puzzles even outside of
除了威胁生命的监狱谜题外
life-threatening prison situations,
只要你喜欢解谜
well, you’ll love Brilliant, these sponsor of this video.
你就会喜欢本视频的赞助商Brilliant
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Brilliant有网站版和软件版
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其中有解决问题的技能
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in math and science.
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他们在许多领域都有超赞课程
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从统计学到天体物理学 再到逻辑学
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如果你喜欢监狱谜题
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甚至另一种数学狂监狱长
They’ve also got a joy of problem solving course
他们也有一堆有趣的解题课程
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最有趣的数学谜题
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每节课都基于你之前学习的东西设置
to give you an in-depth understanding
任何课程都会让你
with any course you take.
理解得很深入
Your knowledge is constantly developed
你的知识储备会不断增加
and tested through interactive experiments and quizzes.
并通过互动课和试题来得到检测
And if you get stuck, there’s always a helpful hint.
如果你卡住了 还有提示帮助你
Head over to brilliant.org/veritasium
快去brilliant.org/veritasium
to check out all these courses
查看所有课程
and test your instincts after learning about
并看看学习100个囚犯谜题后
the 100 prisoners riddle.
你的直觉力是否有所改变
If you click through right now,
如果你现在点击链接
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减免20%的订阅年费
So I wanna thank brilliant for supporting Veritasium,
我想感谢Brilliant赞助《Ve元素》
and I want to thank you for watching.
也谢谢你们的收看

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视频概述

监狱中的100个囚犯将进行一个挑战,胜则全员释放,败则集体处决,他们该怎么做呢?

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视频来源

https://www.youtube.com/watch?v=iSNsgj1OCLA

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