Probability of Rolling a Pairwise Seven rolling $4$ Dice

Question

My friend posed this problem to me. He wrote a computer program to generate all possible combinations of dice and counted the cases with pairwise $7$'s. I thought that the problem had a simpler mathematical answer. My theory was:

1. There are $6$ Possible combinations among four dice (A,B,C,D).
2. Each pair of dice has a probability of $5/6$ of NOT totaling $7$.
3. The probability that none of the Six pairs will total $7$ is $(5/6)^6 \approx 0.334897977$.
4. The probability of rolling one or more pairs of $7$'s is the complement or =$0.665102023$.

The problem is that we had a discrepancy of $.01$ in our calculations. I don't really feel like debugging his computer program. Could you please comment on my mathematical approach to this?

thanks, Bob K

Answer 1

the probabilities in step (3) are not independent. So probability of the compound event is not the product of the simple events' probabilities...

Answer 2

Let us imagine momentarily that the dice are all different color so that we may think of these in a specific order, say $\color{red}{red},\color{blue}{blue},\color{green}{green},$ and $\text{black}$.

The number of possible ways of rolling these dice will be $6^4$.

The number of possible ways of rolling these dice while avoiding any pairwise seven can be counted as follows:

• Pick the number for the $\color{red}{red}$ die: $6$ possibilities.

• Pick the number for the $\color{blue}{blue}$ die: It may be any of the available digits except $7-\color{red}{red}$: $4+1$ possibilities.

Unfortunately, from here, we have a different amount of possibilities for the $\color{green}{green}$ die depending on whether the $\color{red}{red}$ die's result matched the $\color{blue}{blue}$ result. Break into cases:

• In the case that $\color{red}{red}\neq \color{blue}{blue}$, the $\color{green}{green}$ die may be any of the remaining results except $7-\color{red}{red}$ or $7-\color{blue}{blue}$: $2+2$ options

• In the case that the $\color{green}{green}$ die matched either red or blue, then there will be $4$ options for black. In the case that green did not match, then there will be $3$ options for black.

  • for a total of $6\cdot 4\cdot (2\cdot 4 + 2\cdot 3)=336$ in this subcase

• In the case that $\color{red}{red}=\color{blue}{blue}$, the $\color{green}{green}$ die may be any of the results except $7-\color{red}{red}$ for a total of $4+1$ options.

• If $\color{green}{green}=\color{red}{red}$ then there will be five options available for black. If $\color{green}{green}\neq \color{red}{red}$ there will be four options for black.
  • for a total of $6\cdot 1\cdot 4\cdot 4 + 6\cdot 1\cdot 1\cdot 5 = 126$ in this subcase

This gives me a grand total of $336+126=462$ possible ways to have a subset of two of the dice which totals seven.

This gives a probability of $\frac{462}{6^4}=\frac{77}{216}=0.35648\overline{148}$.

The chance then of getting at least one subset totalling seven is $\approx 0.643512$

Answer 3

We will compute the negative (avoiding a $7$), and subtract from $1$. Without loss of generality (wlog), assume the first roll is a $1$.

Case I. With probability $1/6$, the second roll is also a $1$. There are two subcases:

• The third roll is also a $1$, with probability $1/6$. The last roll then avoids the $6$ with probability $5/6$, and the probability contribution of avoiding a $7$ is $1/6 \times 1/6 \times 5/6 = 5/216$.

• The third roll is $2$ through $5$, with probability $2/3$. Wlog, assume it is a $2$. Then the last roll avoids a $5$ or $6$ with probability $2/3$, and the probability contribution is $1/6 \times 2/3 \times 2/3 = 2/27$.

Case II. With probability $2/3$, the second roll is a $2$ through $5$. Wlog, assume it is a $2$. There are two subcases:

• The third roll is a $1$ or $2$, with probability $1/3$. Then the last roll avoids a $5$ or $6$ with probability $2/3$, and the probability contribution is $2/3 \times 1/3 \times 2/3 = 4/27$.

• The third roll is a $3$ or $4$, with probability $1/3$. Wlog, assume it is a $3$. Then the last roll avoids a $4$, $5$, or $6$ with probability $1/2$, and the probability contribution is $2/3 \times 1/3 \times 1/2 = 1/9$.

The total probability of obtaining no $7$ is thus

$$ \frac{5}{216}+\frac{2}{27}+\frac{4}{27}+\frac{1}{9} = \frac{77}{216} $$

which means that the probability of obtaining a $7$ (from two dice) is

$$ 1 - \frac{77}{216} = \frac{139}{216} \doteq 0.64352 $$