A person decides to flip a coin until it lands on heads.
Part 1:
What is the probability that the first time it lands on heads is on the 4th flip (that is, it lands on 3 tails before it lands on heads for the first time).
I understand this question. Using the geometric distribution, I solve for P(x=$4$):
P(Heads) = $0.5$
P(Tails) = $0.5$
P(X=$4$) = $0.5^3$ x $0.5^1$ = $0.5^4$ = $0.0625$
Part 2:
If their first $3$ flips are tails, what is the probability that the next flip will be heads?
I'm not sure how where to start? To me, this sounds exactly like Part 1. Any help is appreciated.
Thanks
Best Answer
Each toss is independent of the last. If I'm not mistaken, the idea that the previous tosses impart "probabilistic weight" on subsequent tosses is known as the gambler's fallacy.
Even if my last 100000 tosses yielded tails, my next flip still has a 50% probability.
The reason this might not seem intuitive is because in real life, if your last few thousand tosses were tails, the coin probably isn't fair. But with this sort of calculation, we assume that the coin is fair, thus implying that our 100000-tail streak is a lovely coincidence.