The coin flip conundrum - Po-Shen Loh

譯者: Ting-Yen Tsai 審譯者: Helen Chang

「好吧，世界必須繞著地軸轉， 人人跟著轉，頭或尾」──拜倫勳爵

When the Wright brothers had to decide

當萊特兄弟必須決定

who would be the first to fly their new airplane

誰要第一個駕駛新飛機 從沙丘上升空

off a sand dune, they flipped a coin.

他們擲硬幣，很公平

That was fair:

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

因為我們都知道硬幣 出現正反面的機率相同

But what if they had a more complicated contest?

但假設他們採行更複雜的比試呢？

What if they flipped coins repeatedly,

如果他們重複拋擲硬幣

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

並規定：只要「連續出現 兩次正面」由奧維爾獲勝

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

而「擲出正面後擲出反面」 由韋伯獲勝呢？

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

兄弟倆還會有相同的首飛機會嗎？

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

首先，他們贏的機率似乎相同

There are four combinations for two consecutive flips.

連擲兩次共有四種可能發生的情況

And if you do flip a coin just twice,

而且如果只擲兩次

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

就會有相同的機率（各 25%）

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

所以你的直覺或許會告訴你 在任何連續的硬幣投擲下

each combination would have the same shot at appearing first.

兩種組合（正正和正反） 先出現的機率相同

Unfortunately, you’d be wrong.

但很不幸，你是錯的

Wilbur actually has a big advantage in this contest.

韋伯其實在這個對賭中有很大的優勢

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

想像硬幣投擲結果的序列是某種桌遊

where every flip determines which path we take.

也就是說每一次拋擲 都是由我們進行的結果來決定

The goal is to get from start to finish.

目標是從開始到結束（有人勝出）

The heads/tails board looks like this.

那麼「擲出正面然後反面」的 路徑就會長成這樣

And this is the head/head board.

而這個是「連續兩次擲出正面」的路徑

There’s one critical difference.

這裡有一個關鍵差別

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

「連續出現兩次正面」是一個 讓你回歸到初始狀態的動作

that heads/tails doesn’t have.

而「正面後擲出反面」並不是

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

這就是為什麼平均來說 連續擲兩次正面需要花較久時間

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

我們可以運用機率和代數 來證明這為什麼是正確的

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

計算成功達成兩種情況 平均需要的拋擲次數

Let’s start with the heads/tails board,

我們從「正面後擲出反面」的 路徑來開始吧

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

我們定義 x 是進到下一步 所需要的平均拋擲次數

Focus only on the arrows.

注意看箭頭的部分

It has two identical steps,

這裡有兩個一樣的步驟

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

分別有一半的機率 會待在原地或往前一步

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

第一種可能，如果我們得到反面 將會待在原地，浪費一次拋擲

Since we’re back in the same place,

因為我們會回到原來的位置

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

平均來說我們必須 再擲 x 次來前進到下一步

Together with that first flip,

加上原本的第一次拋擲

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

平均需要 x+1 次的拋擲來前進

Option 2: If we get heads and move forward,

第二種可能，如果我們 得到正面並向前進

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

那我們就只需要擲一次 就可以前進到下一步

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

結合這兩個可能的機率

我們可以得到以下的解釋

to get this expression.

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

解出 x 的值為二，告訴我們 要進到下一步平均需要擲兩次

Since each step is identical,

因為每一步都是等價的

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

我們可以將二乘以二 得出前進兩步需要擲四次

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

至於「連續擲出兩次正面」 這個就不是那麼簡單了

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

這次，將從開始到結束 需要的平均拋擲次數設為y

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

第一次拋擲會有兩種結果 各有一半的機率

Option 1 is the same as before,

第一種可能和先前一樣

getting tails sends us back to the start,

得到反面讓我們回到原點

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

得出平均需要 y+1 次的拋擲來勝出

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

第二種可能，下一次投擲 會有兩種出現機率相同的可能

With heads we’d be done after two flips.

得到正面奧維爾就會勝出

But tails would return us to the start.

但得到反面會讓我們回到原點

Since we’d return after two flips,

因為我們擲完兩次又回到原點

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

接下來我們平均就需要 y+2 次來產生勝利者

So our full expression will be this.

所以我們完整的解釋如下

And solving this equation gives us six flips.

解出這個等式會得到六次拋擲

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

所以數學計算告訴我們 平均需要六次才能連續擲出兩次正面

and an average of four to get heads/tails.

而擲出正面後擲出反面 平均只需要四次拋擲

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

而且事實上，這也將會是 你進行足夠多次試驗後的結果

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

當然，萊特兄弟不需要擲這麼多次

they only flipped the coin once, and Wilbur won.

他們只擲了一次硬幣，由韋伯獲勝

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

但這不要緊，因為他的飛行失敗了

and Orville made aviation history, instead.

而奧維爾寫下了飛行史的扉頁

Tough luck, Wilbur.

倒楣的韋伯

如果你喜歡這部影片 看看這個清單裡的影片吧