Check your intuition The birthday problem David Knuffke

Imagine a group of people.

How big do you think the group
would have to be

before there’s more than a 50% chance
that two people in the group

have the same birthday?

Assume for the sake of argument
that there are no twins,

that every birthday is equally likely,

and ignore leap years.

Take a moment to think about it.

The answer may seem surprisingly low.

In a group of 23 people,

there’s a 50.73% chance that
two people will share the same birthday.

But with 365 days in a year,

how’s it possible that you need such
a small group

to get even odds of a shared birthday?

Why is our intuition so wrong?

To figure out the answer,

let’s look at one way a mathematician

might calculate
the odds of a birthday match.

We can use a field of mathematics
known as combinatorics,

which deals with the likelihoods
of different combinations.

The first step is to flip the problem.

Trying to calculate the odds
of a match directly is challenging

because there are many ways you
could get a birthday match in a group.

Instead, it’s easier to calculate the odds
that everyone’s birthday is different.

How does that help?

Either there’s a birthday match
in the group, or there isn’t,

so the odds of a match
and the odds of no match

must add up to 100%.

That means we can find
the probability of a match

by subtracting the probability
of no match from 100.

To calculate the odds of no match,
start small.

Calculate the odds that just one pair
of people have different birthdays.

One day of the year will be
Person A’s birthday,

which leaves only 364 possible birthdays
for Person B.

The probability of different birthdays
for A and B, or any pair of people,

is 364 out of 365,

about 0.997, or 99.7%, pretty high.

Bring in Person C.

The probability that she has
a unique birthday in this small group

is 363 out of 365

because there are two birthdates
already accounted for by A and B.

D’s odds will be 362 out of 365,
and so on,

all the way down to W’s odds
of 343 out of 365.

Multiply all of those terms together,

and you’ll get the probability
that no one shares a birthday.

This works out to 0.4927,

so there’s a 49.27% chance that no one in
the group of 23 people shares a birthday.

When we subtract that from 100,
we get a 50.73% chance

of at least one birthday match,

better than even odds.

The key to such a high probability
of a match in a relatively small group

is the surprisingly large number
of possible pairs.

As a group grows, the number of possible
combinations gets bigger much faster.

A group of five people
has ten possible pairs.

Each of the five people can be paired
with any of the other four.

Half of those combinations are redundant

because pairing Person A with Person B
is the same as pairing B with A,

so we divide by two.

By the same reasoning,

a group of ten people has 45 pairs,

and a group of 23 has 253.

The number of pairs grows quadratically,

meaning it’s proportional to the square
of the number of people in the group.

Unfortunately, our brains
are notoriously bad

at intuitively grasping
non-linear functions.

So it seems improbable at first that 23
people could produce 253 possible pairs.

Once our brains accept that,
the birthday problem makes more sense.

Every one of those 253 pairs is a chance
for a birthday match.

For the same reason,
in a group of 70 people,

there are 2,415 possible pairs,

and the probability that two people
have the same birthday is more than 99.9%.

The birthday problem is just one example
where math can show

that things that seem impossible,

like the same person winning
the lottery twice,

actually aren’t unlikely at all.

Sometimes coincidences aren’t
as coincidental as they seem.

想象一群人。

你认为这个群体
必须有多大,

才能有超过 50% 的机会
让群体中的两个人

有相同的生日?

为了论证的目的
,假设没有双胞胎

,每个生日的可能性相同

,忽略闰年。

花点时间考虑一下。

答案似乎出奇的低。

在 23 人的小组

中,两个人生日相同的概率为 50.73%

但是一年有 365 天

,你怎么可能需要
这么小的一群人

来获得共同生日的机会呢?

为什么我们的直觉如此错误?

为了找出答案,

让我们看看数学家

计算生日比赛几率的一种方法。

我们可以使用称为组合学的数学领域

它处理
不同组合的可能性。

第一步是翻转问题。

尝试直接计算比赛的几率
是具有挑战性的,

因为有很多方法
可以让您在小组中获得生日比赛。

相反,计算
每个人的生日不同的几率更容易。

这有什么帮助?

组中要么有生日
比赛,要么没有,

所以
比赛的几率和没有比赛的几率

加起来必须是 100%。

这意味着我们可以

通过
从 100 中减去不匹配的概率来找到匹配的概率。

要计算不匹配的几率,请
从小处着手。

计算只有一
对人生日不同的几率。

一年中的某一天是
甲的生日

,乙只剩下 364 个可能的生日

。甲和乙或任何一对人的生日不同的概率

是 365 中的 364,

大约 0.997,或 99.7%, 相当高。

把人 C 带进来。


在这个小组中有一个唯一生日的概率

是 365 中的 363,

因为
A 和 B 已经占了两个生日

。D 的几率是 365 中的 362,
依此类推,

所有的 低至 W
的 365 中的 343 的几率。

将所有这些项相乘

,您将得到
没有人共享生日的概率。

计算结果为 0.4927,

因此
23 人组中没有人共享生日的概率为 49.27%。

当我们从 100 中减去它时,
我们得到至少一场生日比赛的 50.73% 的

机会,

比偶数赔率要好。

在相对较小的组中如此高的匹配概率的关键

是可能的配对数量惊人地多。

随着组的增长,可能组合的数量
会更快地增加。


人一组有十对可能。

这五个人中的每个人都可以
与其他四个人中的任何一个配对。

这些组合中有一半是多余的,

因为将人 A 与人
B 配对与将 B 与 A 配对相同,

因此我们除以 2。

同样道理

,10 人一组有 45 对

,23 人一组有

253 对。对数呈二次方增长,

这意味着它
与组内人数的平方成正比。

不幸的是,众所周知,我们的大脑

不善于直观地掌握
非线性函数。

因此,起初 23
个人可能产生 253 对可能的配对似乎是不可能的。

一旦我们的大脑接受了这一点
,生日问题就更有意义了。

这 253 对中的每一对都有机会
进行生日比赛。

同理,
在 70 人的群体中,

有 2415 对可能的配对

,两个
人生日相同的概率超过 99.9%。

生日问题只是一个例子
,数学可以

证明看似不可能的事情,

比如同一个人中
了两次彩票,

实际上根本不可能。

有时巧合并不
像看起来那么巧合。