Coin Flipping Game - Wins at 20 Heads

Question

Game Rules

Let's say you have a coin, with $50/50$ chance of lending on Heads or Tails.

You win the game when you get $20$ Heads.

Question

Now, knowing we already threw the coin $50$ times, what are the odds that we have less than $10$ throws left to win.

Answer

The first thing that came to my mind is that to get the answer, I could do the sum of all the possible outcomes from $1$ throw left to win to $9$ throws left to win.

I'm not sure if that is a problem that could involve Cumulative distribution or maybe Negative Binomial distribution.

Answer 1

Note that one throws the coin exactly $n$ more times means that out of total $(50+n)$ throws the last throw has to result in a head, before that there were exactly 19 heads and the rest $(30+n)$ tails.

So the required probability is $\sum_{n=0}^9 {50+(n-1) \choose 19}(\frac{1}{2})^{50+n}$

Answer 2

That's the probability of needing $\le 59$ tosses (for 20 heads), given that we are already in the range $[51, \infty)$.

Answer 3

Using CLT: You have $S_n = \sum_{k=1}^{n} X_k$, with $n=150, \mu=\frac{1}{2}, \sigma^2 = \frac{1}{4}$. You want $$ P(S_n \geq 10) = P(\frac{S_n - n \mu}{\sigma \sqrt{n}} \geq \frac{10-25}{\frac{50}{2}}) \approx 1- \Phi(\frac{10-25}{\frac{\sqrt{50}}{2}}) $$ where $\Phi$ is CDF of standard normal distribution