Probability of rolling more 1's or 2's than 6's on X dice

Question

I'm trying to figure out what the odds are for rolling more 1's and 2's than 6's (on standard dice), according to the number of dice rolled.

For one die it's obviously 33.3%

What about for 2 dice through 10 dice? (Or X dice)

Answer 1

Let $X_n$ be the count for rolling 1 or 2 on $n$ die, and $Y_n$ bethe count for rolling 6 on the same $n$ die. The joint probability mass function is then:

$$\mathsf P(X_n=x, Y_n=y)=\dbinom{n}{x,y,n-y-x}\dfrac{2^x 3^{n-y-x}}{6^n}~\mathbf 1_{\langle x,y\rangle\in\Bbb N^2~:~0\leq x\leq n-y\leq n}$$

So indeed, for 1 die, it is clearly the probability for rolling one of 1 or 2. $$\mathsf P(X_1>Y_1)~{=\mathsf P(X_1=1,Y_1=0)\\=\dfrac{1}{3}}$$

For two dice, it is similarly clear that it is the probability for not rolling any 6 and at least one 1 or 2: $$\mathsf P(X_2>Y_2)~{=\mathsf P(X_2=1,Y_2=0)+\mathsf P(X_2=2,Y_2=0)\\=\dfrac{\dbinom 212^1 3^1+ \dbinom 22 2^23^0}{6^2}\\=\dfrac{4}{9}}$$

Beyond that, things become less straightforward.

$$\mathsf P(X_3>Y_3)~{=\mathsf P(X_3=1,Y_3=0)+\mathsf P(X_3=2,Y_3\in\{0,1\})+\mathsf P(X_3=3,Y_2=0)\\=\dfrac 1{6^3}\left(\dbinom 3 1 2^13^2+\dbinom{3}{2}2^23^1+\dbinom 322^2+\dbinom 332^3\right)\\=\dfrac{55}{108}}$$

And in general.

$$\mathsf P(X_n>Y_n)=\sum_{x=1}^{n}\sum_{y=0}^{\min\{x-1,n-x\}} \dbinom{n}{x,y,n-y-x}\dfrac{2^x 3^{n-y-x}}{6^n}$$

Answer 2

Let $X_n$ be the random variable that assigns the value of $\{0, 1, \dots , n \}$ to the outcome of rolling a die $n$ consecutive times which represents the number of 1's or 2's that are rolled. Let $Y_n$ be the random variable that assigns the value of $\{0, 1, \dots , n \}$ to the outcome of rolling a die $n$ consecutive times which represents the number of 6's that are rolled. The question is: For fixed $n$, what is the probability, $\mathbb{P}(X_n > Y_n)$?

• For $n=1$: $\mathbb{P}(X_1 > Y_1)=\mathbb{P}((X_1=1) \cap (Y_1=0))=\mathbb{P}((X_1=1)\;|\;(Y_1=0))\cdot\mathbb{P}(Y_1=0)= \frac{2}{5}\cdot \frac{5}{6}=\frac{1}{3}$.
• For $n=2$: $\mathbb{P}(X_2 > Y_2)=\mathbb{P}( ((X_1=1) \cap (Y_1=0)) \;\dot\cup\; ((X_1=2) \cap (Y_1=0)) \;\dot\cup\; ((X_1=2) \cap (Y_1=1)) )$ $=\mathbb{P}((X_1=1)\;|\;(Y_1=0))\cdot\mathbb{P}(Y_1=0)$ $+\;\mathbb{P}((X_1=2)\;|\;(Y_1=0))\cdot\mathbb{P}(Y_1=0)$ $+\;\mathbb{P}((X_1=2)\;|\;(Y_1=1))\cdot\mathbb{P}(Y_1=1)$ $= \frac{12}{25} \cdot \frac{25}{36}+\frac{4}{25} \cdot \frac{25}{36}+\frac{0}{10} \cdot \frac{10}{36}$ $=\frac{16}{36}=\frac{4}{9}$.

You should continue this line of reasoning by doing similar calculations for $n=3$ and $n=4$. Then, see if you can observe a pattern that will get you to a formula that will work for the rest of the values for $n=5$ thru $n=10$.