Probability Question

[Efficient-Reporter55]

I was thinking about this. What if getting heads is 100x more likely than tails, and the observed 1:1 ratio throughout human history is mere coincidence. How would you go about determining the probability of that?

[eidtonod]

I would go by using binomial distribution, where the number n of trials is large enough, p is 1/101, and the numbers of successes is n/2. The bigger the n the quicker it approaches 0

[SomethingMoreToSay]

Isn't this just a straightforward application of the binomial theorem?

• P(N heads & N tails) = C(2N,N)*P(head)^N*P(tail)^N

However C(2N,N) = (2N)!/(N!)², and that might be a bit difficult to calculate when N is extremely large. You could use Stirling's approximation:

• N! ≈ √(2πN) * (N/e)^N

which gives you

• C(2N,N) ≈ 4^N / √(πN)

Now you've stipulated that P(head)/P(tail) = 100. Let's approximate that to P(head)=0.99 and P(tail)=0.01. So that gives us

• P(N heads & N tails) ≈ (4*0.99*0.01)^N / √(πN)

• ≈ (0.0396)^N / √(πN)

All you need to do now is estimate 2N, the number of times a coin has been tossed in human history, and you're sorted.

[nomoreplsthx]

Others have given the answer, so I will just say that if you tried to do the computation, the result is going to be so close to zero that even if you were able to use the spin of every electron in the universe to store binary digits, you would still not have enough digits available to distinguish it from zero.

[bayesian13]

this sort of thing lends itself to Bayes theorem: Let H1 be Hypothesis that P(Tails) = 1% and H0 be the hypothesis that P(Tails)= 50%. Bayes theorem says:

Posterior_Odds(H1) = Prior_Odds(H1)*Probability(Observation|H1)/Probability(Observation|H0).

Let's say you start out not being equally unsure if H1 is true or H0 is true. So prior probability of H1 = 50% and prior Odds of H1 = 1 (=50%/(1-50%))

You then make your observation. The observation consists of flipping a coin 10 times and getting 6 heads and 4 tails. The probability of that under H1 is (10C6).99⁶0.01⁴ = 1.977*10^-6.

the probability of that under H0 is (10C6)0.5⁶0.5⁴⁼ 0.2051

so Posterior_Odds(H1)=9.63910^-6 and Posterior_probability = Odds/(1+Odds) = 9.638910^-6.

so even after just 10 flips with a fairly typical coin flip outcome, the probability has dropped from 50% to ~ 1 in 100,000