What is the probability that I throw a bigger number with $4$ dice than with $8$?

Question

That is: Alice throws $4$ dice, Bob throws $8$ dice. What is the probability that the sum of Alice’s points is bigger than Bob’s?

Answer 1

Given $n\geq1$ denote by $p_n(k)$ the probability that in $n$ throws of a fair die we obtain a sum of exactly $k$ points. For the purposes of this problem I consider $p_n$ as a function $$p_n:\quad{\mathbb Z}\to{\mathbb R}_{\geq0},\qquad k\mapsto p_n(k)\ .$$Then $$p_1(k)={1\over6}\quad(1\leq k\leq 6),\quad {\rm and} =0\ {\rm otherwise.}$$ We only need $p_4$ and $p_8$. These can be obtained via $$p_2=p_1*p_1,\quad p_4=p_2*p_2,\quad p_8=p_4*p_4\ ,$$ whereby the $*$ denotes convolution. Of course there is also the recursion $$p_{n+1}(k)={1\over6}\sum_{j=1}^6 p_n(k-j)\qquad(k\in{\mathbb Z})\ .$$ When $p_4$ and $p_8$ have been computed the requested probability $p$ is given by $$p=\sum_{k=9}^{24} \left( p_4(k) \sum_{j=8}^{k-1}p_8(j)\right)\ .$$ The computation gives $$p={771\,485\over120\,932\,352}=0.00637948\ .$$

Answer 2

Let's rephrase our problem. We now have $n_1=4$ ordinary dice and $n_2=8$ «negative dice» with negative numbers $-1$, $-2$,... $-6$. We throw all of them and want to see if the sum is positive. Next, we can notice that throwing a negative die is the same as throwing an ordinary die and adding $-7$ to the result (that is because the probability of having $k$ on ordinary die and $k-7$ on negative die are the same). So we can throw 12 ordinary dice subtract from the sum $8\times7=56$ and see whether the result is positive. Finally, we want to answer the question: what is the probability to throw more than $M=7n_2=56$ on $n=n_1+n_2=12$ ordinary dice.

To find the exact answer is a hard thing to do without a computer. So we can make a good estimation. The number of dice $n=12$ is large enough for the application of Central Limit Theorem.

Estimation

Let $X$ be a random value of the one die and $Y=X_1+\dots+X_{12}$ the sum of $n=12$ dice. Then $$ \mu=E[Y] = 12E(X) = 12\times 3.5 = 42 \\ \sigma^2 = E\left[ (Y-E[Y])^2\right] = 12 E[(X-E[X])^2] = 12\times\frac{2}{6}(0.5^2+1.5^2+2.5^2)=35 $$

According to CLT, the value $(X-\mu)/\sigma$ is distributed closely to standard normal distribution $N(0,1)$. So we are asking what is the probability of random value $Z\sim N(0,1)$ to be greater than $z_0=({56.5-42})/{\sqrt{35}}=2.37$ (we take 56.5 here for a better estimation, as it is an average value between 56, which isn't allowed, and 57, which is). That probability can be found either in tables or calculated: $$ P(Z>z_0) = 0.007 $$

Exact answer

If we can use WolframAlpha or other software we can find the exact answer with the help of generating functions. The generating function of our distribution $Y$ is: $$ g(x) = \frac1{6^{12}}(x+x^2+x^3+x^4+x^5+x^6)^{12} $$

The coefficient at $x^k$ is the probability to get sum of $k$ in the throw. Thus, we need to sum the coefficients $k>56$. With the help of WolframAlpha we get first coefficients of the expansion: $$ 6^{12}g(x) = x^{72} + 12 x^{71} + 78 x^{70} + 364 x^{69} + 1365 x^{68} + 4368 x^{67} + 12364 x^{66} + 31680 x^{65} + 74646 x^{64} + 163592 x^{63} + 336336 x^{62} + 653016 x^{61} + 1203632 x^{60} + 2115048 x^{59} + 3555564 x^{58} + 5734664 x^{57} + \dots $$

Summing these up, we get the answer: $$ p = \frac{13\,886\,730}{6^{12}} = 0.00638 $$