Probability of rolling a 1 and 2 in 4 rolls?

Question

I was working on a problem to compute the variance of the number of distinct roll values after 4 dice rolls and tried approaching it with indicators. I defined $I_n$ to be the indicator random variable of rolling the value $n$ within the 4 rolls. Thus $X$, the number of distinct roll values would be $\sum_{n = 1}^6 I_n$.

Upon calculating $X^2$, I was having trouble calculating $I_nI_m$ where $n \neq m$. I tried using inclusion-exclusion, but didn't really get anywhere. In the end I just brute force calculated all of the cases since there were only 4 rolls.

I was wondering if there is a more elegant solution to calculating this probability for larger number of rolls?

Answer 1

Hmm... I used a different approach

So, I first simplified this with 2 dice. Since there are only 2 possibilities ${1,2}$ and $2,1$, the probability is $\frac1{18}$.

For each time we add a die, the first roll might be:

• $1$, in which case we only need to get a $2$ in the next rolls. The probability of this happening is:

  • $\frac16$
  • $1-\left(\frac56\right)^{n-1}$ for rolling a $2$ at least once, where $n$ is the number of dice (including the added one)
  • Multiplying those together gives us a probablity of $\frac16(1-\left(\frac56\right)^{n-1})$.

• $2$, which has the same chance of happening as our first scenario: $\frac16(1-\left(\frac56\right)^{n-1})$

• Something else (greater than $2$), and our chance of getting both $1$ and $2$ in the next rolls is:

  • $\frac23$
  • The chance of getting $1$ and $2$ with 1 less die, or $P(n-1)$.

This gives us the following function $P$:
$P(n)=\frac13(1-\left(\frac56\right)^{n-1})+\frac23P(n-1); n\in\Bbb{N}; n\gt2; P(2)=\frac1{18}$

So:
$P(3)=\frac13(1-\frac{25}{36})+\frac23\cdot\frac1{18} = \frac13\cdot\frac{11}{36}+\frac1{27}=\frac5{36}$ $P(4)=\frac13(1-\frac{125}{216})+\frac23\cdot\frac5{36} = \frac13\cdot\frac{91}{216}+\frac5{54}=\frac{151}{648}$

Hopefully that's correct :)