Distribution of number of heads, when we keep tossing a coin until we have 4 tails

Question

We toss a fair coin until we've tossed tails exactly 4 times. Let $X$ be the number of tossed heads. What is the distribution of $X$?

My attempt:

We keep tossing the coin until we have registered 4 tails. Suppose that we needed $n$ tosses. The last toss has to be tails, and exactly three of the preceding tosses had to be tails as well:

$$ P(\text{4 tails}|n \text{ tosses})=\frac12\cdot \binom{n-1}{3}\left(\frac 12\right)^3\left(\frac 12\right)^{n-4}=\frac{(n-1)(n-2)(n-3)}{6\cdot 2^n}.$$ This is also the probability of tossing $n-4$ heads, knowing that we needed $n$ tosses to stop the game.

Is this the answer to the question? How do I find 'the distribution' of $X$?

Thanks.

Answer 1

The distribution of discrete random variable is completely described by the set of its possible values and the probabilities corresponding to each value. If $X$ is the number of tossed heads if the coin is tossed until $4$ tails appeared, then $X$ can take values $0,1,2,\ldots$.

For any $k=0,1,2,\ldots$ the event $\{X=k\}$ occures if there are $k+4$ tosses, and the last toss is tail, and exactly three of the preceding $k+3$ tosses are tails as well: $$ \mathbb P(X=k) = \binom{k+3}{3} \cdot\frac{1}{2^{k+4}}. $$ This is Negative binomial distribution.

Answer 2

It's the Negative Binomial distribution. https://en.wikipedia.org/wiki/Negative_binomial_distribution