The coin flip conundrum - Po-Shen Loh

翻译人员: Zibo Huang 校对人员: Sylvia He

“喔，唷！世界必须围绕轴心循环，

人人跟着转，无论顺转或逆转......” -拜伦，《唐·璜》

当莱特兄弟必须对

When the Wright brothers had to decide

who would be the first to fly their new airplane

谁先驾驶新型飞机飞离沙丘作出选择时，

off a sand dune, they flipped a coin.

他们抛出了一枚硬币

That was fair:

这很公平：

we all know there’s an equal chance of getting heads and tails.

我们知道得到正面和 反面的几率是相同的，

But what if they had a more complicated contest?

但如果这是一个更为复杂的竞赛呢？

What if they flipped coins repeatedly,

像将一枚硬币不停地抛出，

so that Orville would win as soon as two heads showed up in a row on his coin,

呈现两次正面时奥维尔胜，

and Wilbur would win as soon as heads was immediately followed by tails on his?

先有正面然后背面时威尔伯胜。

Would each brother still have had an equal chance to be the first in flight?

这样每人仍然拥有相同的胜率吗？

At first, it may seem they’d still have the same chance of winning.

貌似他们仍然有相同的胜率。

There are four combinations for two consecutive flips.

抛硬币两次会有四种不同的组合。

And if you do flip a coin just twice,

如果只将硬币抛出两次 （非连续抛），

there’s an equal chance of each one -- 25%.

那么每一种组合 有相同的概率 -- 25%。

So your intuition might tell you that in any string of coin flips,

所以你的直觉告诉你，无论硬币顺序，

each combination would have the same shot at appearing first.

每一种组合的最先出现的概率都相同。

Unfortunately, you’d be wrong.

不幸的是，这样想你就错了。

Wilbur actually has a big advantage in this contest.

事实上，连续抛时威尔伯 拥有更大的优势。

Imagine our sequence of coin flips as a sort of board game,

想象一下抛硬币正反的排列 是一个图版游戏。

where every flip determines which path we take.

每一次硬币的抛出 决定了以后路径的方向。

The goal is to get from start to finish.

目标是从出发点走到终点，

The heads/tails board looks like this.

正/反 的图版就像这样，

And this is the head/head board.

而这是 正/正 的图版。

There’s one critical difference.

这里有一处至关重要的区别：

Heads/heads has a move that sends you all the way back to the start

正/正 会把你送回最初点，

that heads/tails doesn’t have.

而 正/反 则没有。

That’s why heads/heads takes longer on average.

这就是为什么得到 正/正 一般需要更多时间。

So we can demonstrate that this is true using probability and algebra

我们可以通过概率和代数来计算，

to calculate the average number of flips it would take to get each combination.

连续抛时得到每一种组合 需要的平均次数来证明这个结论。

Let’s start with the heads/tails board,

让我们从 正/反 图版开始。

and define x to be the average number of flips to advance one step.

将 x 定义为前进一步的 硬币平均抛出数。

Focus only on the arrows.

只看着箭头，

It has two identical steps,

有两个关键的步骤：

each with a 50/50 chance of staying in place or moving forward.

停留或前进都拥有 50/50 的概率。

Option 1: If we stay in place by getting tails, we waste one flip.

选择1 ：如果我们得到反面， 我们就浪费了一次抛的机会，

Since we’re back in the same place,

然后我们回到了相同的位置。

on average we must flip x more times to advance one step.

我们平均需要再抛出 x 次来前往下一步。

Together with that first flip,

在第一次抛出的基础上，

this gives an average of x + 1 total flips to advance.

前往下一步的总抛出数 是平均数 x+1。

Option 2: If we get heads and move forward,

选择2：如果我们得到正面并向前移动，

then we have taken exactly one total flip to advance one step.

那么我们只抛出 一次硬币就前往了下一步。

We can now combine option 1 and option 2 with their probabilities

现在我们可以将选择1选择2 和他们的概率结合起来，

to get this expression.

得到这个表达式。

Solving that for x gives us an average of two moves to advance one step.

解表达式得到平均需要 抛两次硬币前往下一步。

Since each step is identical,

因为两个步骤相同，

we can multiply by two and arrive at four flips to advance two steps.

所以我们将平均数 乘以二，平均四次完成图版。

For heads/heads, the picture isn’t as simple.

对于 正/正 来说， 事情并没有那么简单。

This time, let y be the average number of flips to move from start to finish.

这一次，我们将 y 定义 为从开始到结束所需抛出平均数。

There are two options for the first move, each with 50/50 odds.

第一步拥有两个选择， 每一个都有 50/50 的几率。

Option 1 is the same as before,

选择 1 和上文相同，

getting tails sends us back to the start,

得到反面则重新开始。

giving an average of y+1 total flips to finish.

从而完成整个图版的 次数是平均数 x+1。

In Option 2, there are two equally likely cases for the next flip.

选择2：两个相同 概率的情况前往下一步，

With heads we’d be done after two flips.

抛出正面则游戏结束， 两次抛出完成版图。

But tails would return us to the start.

但是抛出反面则游戏重新开始。

Since we’d return after two flips,

所以在2次抛出后重新开始的情况下，

we’d then need an average of y+2 flips in total to finish.

我们一共需要平均数 y+2 次抛出才能完成版图。

So our full expression will be this.

所以我们完整的表达式会是这样：

And solving this equation gives us six flips.

解出等式得到平均数为 6，

So the math calculates that it takes an average of six flips to get heads/heads,

所以数学告诉我们， 得到 正/正 平均需要 6 次抛出，

and an average of four to get heads/tails.

而得到 正/反则需要 4 次抛出。

And, in fact, that’s what you’d see if you tested it for yourself enough times.

事实上，长时间的实验 也会得到这个结论。

Of course, the Wright brothers didn’t need to work all this out;

当然，莱特兄弟并不需要解决这个问题。

they only flipped the coin once, and Wilbur won.

他们只需要抛出一次硬币， 然后威尔伯胜利。

But it didn’t matter: Wilbur’s flight failed,

但这并不重要， 因为威尔伯的飞行失败了，

and Orville made aviation history, instead.

而奥维尔创造了航空历史。

Tough luck, Wilbur.

真不走运，威尔伯。