# Probability of the number of tosses to get heads dependent?

I understand if a fair coin is tossed repeatedly, the probability of getting a head during each toss does not change and is independent from that of other tosses.

What about the number of tosses required to get heads - are they dependent?

For example let X be the random variable representing number of tosses required to get the first head, and Y the random variable representing the number of tosses to get the first two heads. Are the two variables dependent?

Intuitively it seems that Y depends on X. What is a more 'formal' reasoning than mere intuition?

• Not sure I follow. As it stands , of course they are dependent. If it takes you $20$ tosses to get the first head, then it is impossible that you got two heads in the first $5$ tosses. Is that what you meant? – lulu Jul 13 at 13:52
• The number of tosses has a negative binomial distribution. The questions is whether X (number of tosses getting the 1st head) and Y (number of tosses getting the first two heads) are independent. – Cee Cee Jul 13 at 13:56
• And as I remarked, they are obviously dependent. Do you think $P(X=3, Y=2)\,=\,P(X=3)\times P(Y=2)$? Or am I misunderstanding the question? – lulu Jul 13 at 13:58
• I do not think P(X=x, Y= y) = P((X = x) * P(Y = y). What I am asking is a more formal "proof". I have the distribution function for P(X) and P(Y) respectively, both follow the negative binomial distribution. But I cannot figure out the joint distribution P(X, Y).... – Cee Cee Jul 13 at 14:03
• What I wrote is a formal proof. Just compute both sides and verify that they are not equal. – lulu Jul 13 at 14:03

We have $$Pr(X=3) >0$$ and $$Pr(Y=2)>0$$
But $$Pr(X=3, Y=2) =0 \ne Pr(X=3)Pr(Y=2)$$
We know the joint distribution requireds $$X < Y$$ to be positive. But $$X$$ follows a geometric distribution while $$Y$$ follows the sum of two geometric distributions.