game of throwing 200 die [see desc.]

Question

Here is dice game question about probability.

Play a game with $2$ die. What is the probability of getting a sum greater than $7$?

I know how the probability for this one is easy, $\cfrac{1+2+3+4+5}{36}=\cfrac 5{12}$.

I don't know how to solve the follow-up question:

Play a game with $200$ die. What is the probability of getting a sum greater than $700$?

Answer 1

Really, the probability for two die to get a sum greater than 7 is $$\cfrac{1+2+3+4+5}{36}=\cfrac 5{12}.$$ For $200$ die it is too hard to calculate the required probaility exactly. A reasonable way is to choose Central Limit Theorem and obtain the approximate answer using normal approximation to the distribution of the sum.

Let $X_i$ be the points we get at $i$-th throw. Then $\mathbb P(X_i=k)=1/6$, $k=1,2,3,4,5,6$ and $X_1,X_2,\ldots,X_{200}$ are independent and identically distributed r.v's with mean $\mathbb E[X_1]=3.5$ and variance $\text{Var}[X_1]=35/12$. By CLT for $S_n=X_1+\ldots+X_n$, $$ \mathbb P\left(S_{200}>700\right) = \mathbb P\Biggl(\frac{S_{200}-200\cdot \mathbb E[X_1]}{\sqrt{200 \text{Var}[X_1]}}>\underbrace{\frac{700-200\cdot 3.5}{\sqrt{200\cdot\frac{35}{12}}}}_{0}\Biggr) \approx \mathbb P(Z >0)=0.5 $$ where $Z$ has standard normal distribution.

You can also give the same approximate answer to this question without CLT. It it obvious that the distribution of $S_{200}$ is symmetric w.r.t. $700$: it is equally probable to get sum $200$ and $1200$, $201$ and $1199$, ..., $699$ and $701$. Therefore $$\mathbb P(S_{200}>700)=\mathbb P(S_{200}<700)=\frac{1-\mathbb P(S_{200}=700)}2$$ And the probabililty $\mathbb P(S_{200}=700)$ is sufficiently small, so $\mathbb P(S_{200}>700)$ is close to $0.5$

Answer 2

I would not suggest that one should attempt the computation by hand, and I'm pretty sure this is not what you are asked to do, but the following MATLAB code returns

$$ 0.49174766008476854489594153068254 $$

for the probability of rolling a number greater than $700$. It does so in less than 5 s.

poly1 = poly2sym(sym([ones(1,6), 0]) / sym(6));
poly200 = poly1^200;
c200 = coeffs(poly200, 'All'); % from highest to lowest degree
pmf = c200(end:-1:1);          % from lowest to highest degree
pr = vpa(sum(pmf(702:end)));   % pmf(i+1) == Pr(X=i)

The computation is carried out exactly until the call to vpa returns a result accurate to 32 digits (the number of digits may be increased). Numerical convolution (with conv) is much faster, of course, and while it does not boast the same accuracy, it's not too shabby:

$$ 0.491747660084763 \enspace.$$

A look at the graph of the exact and approximate PDFs confirms that the normal approximation suggested by @NCh is more than adequate for most uses.

Indeed, if we apply a continuity correction:

1-normcdf(0.5/sqrt(200*35/12))

we obtain $0.491741700046665$.