Naive Significance Statements and Bitcoin (or Anything)

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Avatar for Beakerboy
3 years ago
Topics: Math, Bitcoincash

I saw the following image on noise.cash and felt I had to respond.

The image shows a table of month and years, and the percent return on bitcoin for that period. Gains are highlighted in green and losses are highlighted in red. The heading contains the text “March is time to buy. 8/10 is a bad month for bitcoin” and you can see that the month of March has eight years where it experienced losses and only two years where there were gains.

People make statements like this all the time in bitcoin, general life advice, etc. that some seemingly rare event has happened so you should avoid putting yourself in that situation. What most people don’t realize is exactly how common seemingly unlikely events are when things are repeated. Hopefully, by the end of this short article, you will gain a little intuitive sense into this.

Let’s assume we are speaking about coin flips, and ask, how likely is it, that if I flip a coin for 12 groups of 10 flips, that at least one of the groups has only two tails? There are 1024 permutations of two in ten, 45 of which have two tails. 45/1024 is .044, which is about 1 in 23 odds. Now, the naive response would be to say “well, we only have 12 groups, not 23, so it’s obviously not going to happen”. However, this is looking at the next step backwards. In order for something to never happen, it has to “not happen each time”. Stop for a second and read that sentence again. For something to happen it only has to happen once but for something to never happen it has to not happen each and every time. What this means in math is we have repeat the result of not getting 2 in 10, roughly 22/23 odds or 96%, 12 times by raising it to the power of 12. This equals 58%. What this tells us is that the odds of NEVER getting 2 heads in 10 flips, when repeated 12 times is only 58%, or, there is a 42% chance of getting “at least one instance” of 2 tails in 10, when repeated 12 times. This event is nowhere near remote enough to make any sort of investing choice, or offer any advice. This type of analysis was formalized by Bernoulli...and Fisher liked to say that something has to be outside the 97.5% and 2.5% range to be significant. Even a 1 tails in 10 has a 10% chance of happening at least once in this chart, so if we saw one of those, you could still call it random chance.

There are a whole family of paradoxes or conundrums that confuse us due to most people’s difficulty in wrapping their head around these concepts. The Birthday Problem being the most famous.

Another related question is, given an n-sided die, how likely is it to never roll a 1 when rolled n times. The answer is ((n-1)/n)ⁿ and is around 1/3, so the odds of getting at least one “1” is 2/3....pretty likely. The funny thing is, this is true for all n! My question to you is, what is the answer when n is infinity! What are the odds of never rolling a 1 when an infinite sided die is rolled an infinite number of times. First person to give me an answer and how they got it will get a tip.

Edit:Fixed typo in equation.

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$ 0.80
$ 0.50 from @Omar
$ 0.30 from @lerkfrend
Avatar for Beakerboy
3 years ago
Topics: Math, Bitcoincash

Comments

The answer is ((1-n)/n))ⁿ

You mean the answer is ((n-1)/n))ⁿ. The limit of that as n --> infinity is 1/e as explained nicely here. https://www.youtube.com/watch?v=c7VOC5U1Et4

$ 0.50
3 years ago

Actually, that video does a pretty shitty job at explaining anything. Here’s a better one. https://youtu.be/hpCenDJgy7w

$ 0.00
3 years ago

You found a typo as well...excellent! Now, would you have been able to figure it out without YouTube?

$ 0.00
3 years ago

Now, would you have been able to figure it out without YouTube?

Ha, good question. As t--> infinity... maybe?

$ 0.00
3 years ago

I didn't do any complex math so this is a guess, but the odds are 0 because infinite doesn't end, so you are bound to roll a 1.

$ 0.00
3 years ago

No, you are not bound to roll a one. It is certainly possible to roll the number 2 each and every time. Very unlikely, but it’s there...so it’s not zero.

$ 0.00
3 years ago

very nice article, more of these please! as for the puzzle, I tried, had a headache, gave up.
I am pretty sure it is one of the following though: zero, 1/3, infinity

$ 0.00
3 years ago

Nope, none of those three. The problem can be solved with advanced High School level maths. Also, probabilities are bounded by zero and 1, but it’s not one either.

$ 0.00
3 years ago