Show that the probability of the number of times a *biased* die is rolled to reach a cumulative sum $\ge k$ is normally distributed.

Question

Take a hypothetical game where you need to roll a 6 sided biased die multiple times to reach a cumulative sum of $k$ to win. Once you reach a cumulative sum $\ge k$, you stop rolling the die. The probability of rolling a 6 is $a$, and the probability of rolling any other number in the range $[1,5]$ is $b = (1-a)/5$.

I ran a computer program to simulate the game for $k=100$. This is the histogram of the number times the program needed to roll the die to win the game:

The histogram seems to be normally distributed about the expected value. Is the probability of winning the game in $x$ moves normally distributed? If yes, how can I arrive to this conclusion mathematically?

Answer 1

By the Central Limit Theorem, the score after a fixed but large-ish number of rolls can be approximated by a normal distribution. A single die roll has mean $\mu=a\cdot6+(1-a)\cdot3=3(a+1)$ and we compute its variance as $\sigma^2=-9a^2+7a+2=(9a+2)(1-a)$. So after $N\gg 0$ rolls, the distribution for the sum after $N$ rolls will closely resemble a standard bell curve with mean $N\mu$ and variance $N\sigma^2$, i.e., standard deviation $\sqrt N\sigma$.

However, you are not interested in the random variable $Y_N=$"sum after $N$ rolls", but rather in $X=$"number of rolls until the sum is $\ge k$". How are these related? The probability that $$X\le N$$ equals the probability that $S_N\ge k$. And this latter probability depends on the distance of $k$ from the mean of the bell curve, measured in standard deviations, $$z=\frac{k-N\mu}{\sqrt N \sigma}=\frac k\sigma N^{-\frac12}-\frac \mu\sigma N^{\frac12}.$$ We are interested mainly in the case that we expect to need large $N$, i.e., when $k\gg \mu$, and for $N$ where $z$ is close to $0$, i.e., $N\approx \frac k\mu$. There, the function $N\mapsto z$ looks roughly linear (as is always the case for a differential function), namely there we have $$\tag1\frac {\mathrm d}{\mathrm dN} z= -\frac12\frac k\sigma N^{-\frac32}-\frac12\frac\mu\sigma N^{-\frac12}=-\frac{\mu+k/N}{2\sigma \sqrt N}\approx -\frac{\sqrt\mu}{2\sigma\sqrt{k}}$$ and hence $$ z\approx -\frac{\sqrt\mu}{2\sigma\sqrt{k}}\left(N-\frac k\mu\right)=\frac{N-\mu^*}{\sigma^*}.$$ with $$\tag2\mu^*:=\mu^{-1}\cdot k,\qquad \sigma^*:=\frac{2\sigma}{\sqrt \mu}\cdot\sqrt k.$$ So as long as the "$\approx$" is good enough, we can say that the random variable $X$ behaves approximately like a normally distributed with man and standard deviation given by $(2)$. This is the case when the approximation n $(1)$ holds over a large interval around $\frac k\mu$, which is the case when $k\gg \mu$.

Answer 2

Well it's simple

Make a discrete random variable table

Take out the expected value which you'll get 3+3a

Now divide k by 3+3a to get expected minimum no. of rolls needed to achieve k

To get the probability for number make a binomial distribution

Where mean = np= $\frac {k}{3+3a}a=\frac{ka}{3+3a}$

And variance = npq=$\frac{ka(1-a)}{3+3a}$

For binomial distribution condition that apply before you make the distribution as an approximation to normal are that np>=5 and nq>=5

$\frac{ka}{3+3a}>=5$ and $\frac{k(1-a)}{3+3a}>=5$

$(k-15)a>=15$ and $k-15>=(k+15)a$ and $a=<1$

Now as k gets large enough these inequality holds true

For k=100 we get $85a>=15$ and $85>=115a$ a should be greater than or equal to 1.756 and should be less than or equal to 0.739

As k goes to around 3000, a could take any value between 1 and 0