"It's a 10% chance which I did 10 times, so it should be 100%"

post by egor.timatkov · 2024-11-18T01:14:27.738Z · LW · GW · 5 comments

Contents

  The math:
  Hold on a sec, that formula looks familiar...
    So, if something is a 1/n chance, and I do it n times, the odds should be... 63%.
  What I'm NOT saying:
None
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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 (incorrect) 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 do 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 $63\%$ figure 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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