本文探讨了一个看似错误却蕴含着有趣数学原理的命题：如果某件事发生的概率是1/n，尝试n次后，成功的概率是多少？文章通过抛硬币和抽奖等例子，推导出一个近似的概率公式，即当n足够大时，成功的概率约为63%。文章解释了这个概率背后的数学原理，以及如何利用该公式进行估算，并强调了该公式在实际应用中的局限性，例如n较小的情况以及成功次数的分布。

🤔 **概率公式推导：**文章通过分析抛硬币和抽奖等例子，推导出一个近似的概率公式：(1-1/n)^n，其中n代表尝试次数，该公式计算的是失败的概率，成功的概率则为1减去失败的概率。

🍀 **当n足够大时，成功的概率约为63%：**当n趋近于无穷大时，(1-1/n)^n的极限值为1/e，约等于0.37，这意味着失败的概率约为37%，成功的概率则约为63%。

💡 **公式的实际应用及局限性：**该公式适用于n较大时的概率估算，但当n较小时，估算结果可能不够准确。文章也指出，即使成功的概率约为63%，平均成功的次数仍然是1，并非100%保证成功一次。

📊 **不同n值下的概率：**文章举例说明了当n分别为5、10和20时，成功的概率分别约为67%、65%和64%，说明随着n的增大，成功的概率逐渐逼近63%。

Published on November 18, 2024 1:14 AM GMT

Many of you readers may instinctively know that this is wrong. If you flip a coin (50% chance) twice, you are not guaranteed to get heads. The odds of getting a heads are 75%. However you may be surprised to learn that there is some truth to this statement; modifying the statement just slightly will yield not just a true statement, but a useful and interesting one.

It's a spoiler, though. If you want to figure this out as you read this article yourself, you should skip this and then come back. Ok, ready? Here it is:

 

The math:

Suppose you're flipping a coin and you want to find the odds of NOT flipping a single heads in a dozen flips. The math for this is fairly simple: The odds of not flipping a single heads is the same as the odds of flipping 12 tails. which is 

$$(1/2{)}^{12}\approx 0.000977$$

The same can be done with this problem: you have something with a 1/10 chance and you want to do it 10 times. The odds of not getting it to happen even once is the same as the odds of it not happening, 10 times in a row. So 

$$(9/10{)}^{10}\approx 0.35$$

If you learned some fairly basic probability, I doubt this is that interesting to you. The interesting part comes when you look at the general formula: The probability of not getting what you want (I'll call this $1-p$, because $p$ would be the probability of the outcome you want) is

$$1-p=(1-\frac{1}{n}{)}^n$$

Where $n$ in our case is 10, but in general is whatever number you hear when you hear the phrase "It's a one-in-$n$ chance, and I did it $n$ times, so it should be $100\%$"

 

Hold on a sec, that formula looks familiar...

"$(1-\frac{1}{n}{)}^n$ ..." I thought to myself... "That looks familiar..." This is by no means obvious, but to people who have dealt with the number $e$ recently, this looks quite similar to the limit that actually defines that number. This sort of pattern recognition led me to google what this limit is, and it turns out my intuition was close:

$$\underset{n\to \infty }{lim}(1-\frac{1}{n}{)}^n=\frac{1}{e}$$$$\frac{1}{e}\approx 0.37$$

So it turns out: for any n that's large enough, if you do something with a $1/n$ chance of success $n$ times, your odds of failure are always going to be roughly $37\%$, which means your odds of success will always be roughly $63\%$.

So, if something is a $1/n$ chance, and I did it $n$ times, the odds should be... $63\%$.

Isn't that cool? I think that's cool.

 

What I'm NOT saying:

There are a couple ways to easily misinterpret this, so here are some caveats:

The average number of successes, if you try a $1/10$ chance $10$ times is $1$. This is always true for any value of $n$. I believe this is the cause of why some folks to think the odds are $100\%$ if you try enough times. Your odds of succeeding at least once may be $63\%$, but you also have a chance of succeeding twice, thrice, etc. This means 1 success on average, not a 100% chance of success once.This is post explores how probability behaves as you try a less-likely chance, more often. Obviously for any given $1/n$ chance, trying it more often will increase your odds of succeeding at least once.This relies on a limit as $n\to \infty$. How useful is this in the real world where $n$ is probably a rather small number like 5, 10, or 20? Well you can try out the formula to find out. Personally I think the formula is good enough for $n\ge 4$ (within 5%). You could use 65% as a nice round number instead of 63% if you want to round up, too.

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