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

e^3 is ~20, so for large n you get 95% of success by doing 3n attempts.

Ironically, the even more basic error of probabilistic thinking that people so—painfully—commonly make ("It either happens or doesn't, so it's 50/50") would get closer to the right answer.

Nice.

Similar rule of thumb I find handy is 70 divided by growth rate to get doubling time implied by a growth rate. I find it way easier to think about doubling times than growth rates.

E.g. 3% interest rate means 70/3 ≈ 23 year doubling time.

Curated! A short and sweet post, with a useful rule of thumb for those who live and die by the Fermi. There are some more good ones in the comments, too!

Nice. I have a suggestion how to improve the article. Put a clearly stated theorem somewhere in the middle, in its own block, like in academic math articles.

In case anyone’s wondering, if there’s a 1/n chance of something happening each time (iid), and you try n times (for large n), then it will happen m times with probability $(1/m!)/e$. So 0,1,2,3… hits would be 36.8%, 36.8%, 18.4%, 6.1%, 1.5%, 0.3%, … Nice how it sums to one.

(For the general formula, i.e. where the probability is not necessarily 1/n, see: poisson distribution.)

Relatedly, if you perform an experiment n times, and the probability of success is p, and the expected number of total successes kp is much smaller than one, then kp is a reasonable measure of getting at least once success, because the probability of getting more than one success can be neglected. 

For example, if Bob plays the lottery for ten days, and each days has a 1:1000,000 chance of winning, then overall he will have a chance of 100,000 of winning once. 

This is also why micromorts are roughly additive: if travelling by railway has a mortality of one micomort per 10Mm, then travelling for 50Mm will set you back 5 micomort. Only if you leave what I would call the 'Newtonian regime of probability', e.g. by somehow managing to travel 1Tm with the railway, you are required to do proper probability math, because naive addition would tell you that you will surely have a fatal accident (1 mort) in that distance, which is clearly wrong. 

The excellent book Algorithms to Live By has an entire chapter dedicated to this concept, using the secretary problem as an example: https://www.amazon.com/Algorithms-Live-Computer-Science-Decisions/dp/1627790365 

It's worth noting the mathematically fairly simple and obvious but intuitively annoying fact that if you've tried the 10% chance 9 times with no success, you do not have a 63% chance of succeeding on your next attempt.

However it is true that doing something with a 10% success rate 10 times will net you an average of 1 success.

For the easier to work out case of doing something with a 50% success rate 2 times:

• 25% chance of 0 successes
• 50% chance of 1 success
• 25% chance of 2 successes

Gives an average of 1 success.

Of course this only matters for the sort of thing where 2 successes is better than 1 success:

• 10% chance of finding a monogamous partner 10 times yields 0.63 monogamous partners in expectation.
• 10% chance of finding a polyamorous partner 10 times yields 1.00 polyamorous partners in expectation.

EDIT: To clarify, a 10% chance of finding a monogamous partner 10 times yields 1.00 successful dates and 0.63 monogamous partners that you end up with, in expectation.

My guesses at what the spoiler was going to be:

• Ten non-independent trials, a 10% chance each (in the prior state of knowledge, not conditional on previous results,), and only one trial can succeed. You satisfy these conditions with something like "I hid a ball in one of ten boxes", and the chance really is 100% that one is a "success".

• Regardless of whether the trials are independent, the maximum probability that at least one is a success is the sum of the probabilities per trial. In this case that doesn't yield a useful bound because we already know probabilities are below 100%, but in general it's useful.

Yeah, it's cool that "I did n trials, with a 1/n chance each, so the probability of at least one success is... " does have a general answer, even if it's not 100%. Just noting that it's not the only small modification of the title yielding a useful and interesting correct statement.

The ones that came to my mind still involved the sum of the per-trial probabilities. If it was clear that we were looking for something preserving the "n trials with 1/n chance", rather than the summation, I think it would have been more obvious where you're going with this.

Furthermore, the tries must be independent of each other, otherwise the reasoning breaks down completely. If I draw cards from a deck, each one has (a priori) 1/52 chance of being the ace of spades, yet if I draw all 52 I will draw the ace of spades 100% of the time. This is because successive failures increase the posterior probability of drawing a success.

Nitpick: “odds of 63%” sounds to me like it means “odds of 63:100” i.e. “probability of around 39%”. Took me a while to realise this wasn’t what you meant.

Related: https://en.wikipedia.org/wiki/Secretary_problem

Great post, seems like a handy thing to remember.

Years ago when I was hanging out with day traders there was a heuristic they all seemed to hold. If their trading model was producing winning trades two out of three times they thought the model was good and could be used. No one ever suggested why that particular rate was the shared meme/norm -- why not 4 out of 5 or 3 out of 5. I wonder if empirically (or just intuitively over time) they simply approximated the results in this post.

Or maybe just a coincidence, but generally when money is at stake I think the common practices will tend to reflect some fundamental fact of the environment. 

I thought to myself... "That looks familiar..."

It is also very similar to the formula for calculating the compound interest rate. 
Just swap the minus with a plus and the function tends to e: after all, compounding interest rates was how the constant got known in the first place.

love the simplicity of this. nice work.

This post reminds me of another estimate. Using the same mistake as a starting point it can be phrased like: "It's a p chance which I did n times, so it should be np if np<<1." This is because (1-p)^n = 1 - np + n(n-1)/2 p^2 - ..., and since np<<1 this is approximately 1-np.

However, I find this linearity more useful when combining small changes: a 1% increase followed by a 2% increase is approximately a 3% increase, since (1+p)(1+q)=1+p+q+pq and pq can be ignored in an approximation.

I guess wisdom is about understanding, by observation, how "the wheels" role in the machine. You're probably right but you always need to test different solutions. Cybernetics allows you to find out-of-the-box responses on a "plug n play" logic.