10

My question has two parts:

  1. Am I approaching the problem correctly, resulting in a reasonable formula?
  2. How do I do the final calculation for numbers as large and as small as these. I put my formula into google and it just gives an answer of "1", which surely isn't the right answer.

My approach:

  • Lets start with an assumption, that the average person experiences 225 coin flips in a lifetime (3/year for 75 years)

  • The odds of an individual having the extraordinary results of "only heads" or "only tails" for all 225 of their flips would be 0.5^224 (the first flip can be either so we only say 224).

  • Then we can say that the odds that someone goes through life without these extraordinary results would be 1-(0.5^224).

  • The odds then of nobody experiencing these extraordinary results will be to take the previous calculation multiplied by itself for each member of the human population, so we raise it all to the power of 7.8 billion.

  • The final formula for this calculation is then (1-(0.5^224))^7800000000.

Edit: Given the insight from Stacker's answer its clear that I missed a step, I only "calculated the probability that everyone will not have such an extraordinary event in their lifetime. To find the probability that at least one person will, subtract that number from 1 (the complement rule)."

The final formula for this calculation is actually 1-((1-(0.5^224))^7800000000)

P.S. If your interested in some context, I'm trying to do this calculation to explore the idea of perspective. To better understand how likely it is that there are people that will experience the world in a way that is entirely different from reality.

user1937279
  • 103
  • 2
    I think the more likely way for this to happen is for someone to not experience any coinflips :) After all, 225 is just an average. – Misha Lavrov Apr 21 '21 at 14:44
  • 4
    Someone who has only ever had coins flip one way for years is likely to think there is something odd going on, and start flipping far more coins than three a year to investigate. cf. https://daily.jstor.org/statistics-of-coin-tosses-theater-geeks/ – armb Apr 21 '21 at 15:52

2 Answers2

11

Your final formula of $(1-(0.5^{224}))^{7800000000}$ correctly gives the probability that 0 people in a group of 7.8 billion, each of whom flip 225 coins, flips either "all heads" or "all tails". So if you want the chances that at least one person DOES flip all heads or all tails, you have to subtract that quantity from 1.

This is a reasonable formula. There may be some clever simplification, but this is just going to be a tough problem for a calculator. You can put it in Wolfram Alpha and it gives a solution: https://www.wolframalpha.com/input/?i=%281+-+.5%5E224%29%5E7800000000

But you may get some intuition from the following limit:

$$e^{-1} = lim_{n \rightarrow \infty} (1 - 1/n)^n$$

Heuristically, we can say that $(1 - 1/(2^{224}))^{2^{224}} \approx e^{-1} \approx .368$. Then the question is, how much bigger is $2^{224}$ than 7,800,000,000? And the answer is, like, way bigger. It's pretty easy to check that $7800000000 < 2^{33}$ (A good rule of thumb: $2^{10} \approx 1000$). So then we can write:

$$ ((1 - 1/(2^{224}))^{7800000000})^{2^{191}} > ((1 - 1/(2^{224}))^{2^{33}})^{2^{191}}) $$

$$ = (1 - 1/(2^{224}))^{2^{224}} \approx e^{-1} \approx .368 $$

And then, rearranging the beginning and the end of the inequality, we get that our quantity of interest $ 1 > (1 - 1/(2^{224}))^{7800000000}> .368^{(\frac{1}{2^{191}})}$, which is very, very, very close to 1.

So ultimately, pretty unlikely that even in this big group of people, anyone gets 225 H or T in a row.

housed_off_space
  • 384
  • Follow up: So 225 is pretty unlikely, what is the highest N such that it can be reasonably expected (>50% chance) that one person in the world flips N times and gets all heads or all tails? – Darrel Hoffman Apr 21 '21 at 18:21
  • Let's find the $N$ such that there is exactly 50% chance that there is someone that flips $N$ heads/tails, or equivalently, there is exactly 50% chance that there is nobody that flips $N$ heads/tails. If $x$ is 7.8 billion, then we want $\left(1-2^{1-N}\right)^x=\frac12\Rightarrow1-2^{1-N}=\frac1{\sqrt[x]2}\Rightarrow1-\frac1{\sqrt[x]2}=2^{1-N}\Rightarrow\log_2\left(1-\frac1{\sqrt[x]2}\right)=1-N\Rightarrow N=1-\log_2\left(1-\frac1{\sqrt[x]2}\right)$. – Kyan Cheung Apr 22 '21 at 16:00
  • Using Wolfram Alpha, we find $N\approx34.3$. As long as $N\leq34$, there is at least 50% chance that someone gets $N$ heads/tails. So your answer is 34. – Kyan Cheung Apr 22 '21 at 16:00
  • Just as a (very belated) follow-up: The reason $N \leq 34$ appears in Kyky 's astute answer is tightly related to the reason $2^33$ shows up in my heuristic explanation: 7.8 billion is between $2^33$ and $2^34$, and we should expect the probability to have just crossed over $1 - e^{-1} \approx .63$ (on its way downward) with $N = 34$ flips...As Kyky's calculation shows, shortly after that it passes $.5$. – housed_off_space Apr 27 '22 at 19:05
3

You calculated the probability that everyone will not have such an extraordinary event in their lifetime. To find the probability that at least one person will, subtract that number from 1 (the complement rule).

I don’t think anyone would experience the world completely different from reality unless they died extremely early on (very few coin flips). But then would you stipulate that you have to have a certain number of coin tosses/live to a certain age?

Vons
  • 11,004
  • 2
    Thanks for the sanity check on my calculation. Also it appears that wolfram alpha agrees with you that this scenario is unlikely. It gives it a probability of zero. – user1937279 Apr 21 '21 at 05:29
  • For simplicity sake I was assuming everyone lived a full-ish life. But its worth noting that the odds of someone having 10ish years and having an unrealistic life experience is not that unlikely so its very reasonable for there to be delusional 10 year olds running around. – user1937279 Apr 21 '21 at 05:31
  • No problem. If you search “extremely accurate calculator” and use the first one (Keiseo), the probability is about 2.89(10)^-58, or basically 0 as wolfram said. – Vons Apr 21 '21 at 05:33