Probability that there exists a person who will "only flip heads" or "only flip tails" in their lifetime?

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)

Solution 1:

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.

Solution 2:

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?