TAOCP 3.3.1 Exercise 2

Let the two dice be labeled die A and die B.

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Section 3.3.1: General Test Procedures for Studying Random Data

Exercise 2. [20] If two dice are "loaded" so that, on one die, the value 1 will turn up exactly twice as often as any of the other values, and the other die is similarly biased towards 6, compute the probability $p_s$ that a total of exactly $s$ will appear on the two dice, for $2 \le s \le 12$.

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Solution

Let the two dice be labeled die A and die B. Let $X$ denote the outcome of die A and $Y$ the outcome of die B. We are given that die A is loaded so that 1 appears twice as often as any other face, and die B is loaded so that 6 appears twice as often as any other face. Denote by $p_i^{(A)}$ the probability that die A shows $i$, and by $p_j^{(B)}$ the probability that die B shows $j$, with $1 \le i,j \le 6$.

Since die A has faces $1,2,3,4,5,6$, the total probability must sum to 1:

$p_1^{(A)} + p_2^{(A)} + p_3^{(A)} + p_4^{(A)} + p_5^{(A)} + p_6^{(A)} = 1.$

Let $q$ denote the probability of each face $2,3,4,5,6$ on die A. Then $p_1^{(A)} = 2q$. Therefore,

$2q + 5q = 7q = 1 \implies q = \frac{1}{7}.$

Hence,

$p_1^{(A)} = \frac{2}{7}, \quad p_i^{(A)} = \frac{1}{7} \text{ for } i = 2,3,4,5,6.$

Similarly, for die B, let $r$ denote the probability of each face $1,2,3,4,5$, and $p_6^{(B)} = 2r$. Then

$5r + 2r = 7r = 1 \implies r = \frac{1}{7}.$

Thus,

$p_6^{(B)} = \frac{2}{7}, \quad p_j^{(B)} = \frac{1}{7} \text{ for } j = 1,2,3,4,5.$

The total $s$ on the two dice is $s = X + Y$. The probability $p_s$ is computed by convolution:

$p_s = \sum_{i=1}^{6} p_i^{(A)} p_{s-i}^{(B)},$

where $p_{s-i}^{(B)} = 0$ if $s-i \notin {1,2,3,4,5,6}$.

We compute $p_s$ for each $s$:

  • $s = 2$: only $(X,Y) = (1,1)$ contributes. Then

$p_2 = p_1^{(A)} p_1^{(B)} = \frac{2}{7} \cdot \frac{1}{7} = \frac{2}{49}.$

  • $s = 3$: $(X,Y) = (1,2), (2,1)$. Then

$p_3 = p_1^{(A)} p_2^{(B)} + p_2^{(A)} p_1^{(B)} = \frac{2}{7}\cdot \frac{1}{7} + \frac{1}{7}\cdot \frac{1}{7} = \frac{3}{49}.$

  • $s = 4$: $(1,3), (2,2), (3,1)$

$p_4 = \frac{2}{7}\cdot\frac{1}{7} + \frac{1}{7}\cdot\frac{1}{7} + \frac{1}{7}\cdot\frac{1}{7} = \frac{4}{49}.$

  • $s = 5$: $(1,4),(2,3),(3,2),(4,1)$

$p_5 = \frac{2}{7}\cdot\frac{1}{7} + 3\cdot \frac{1}{7}\cdot\frac{1}{7} = \frac{5}{49}.$

  • $s = 6$: $(1,5),(2,4),(3,3),(4,2),(5,1)$

$p_6 = \frac{2}{7}\cdot\frac{1}{7} + 4\cdot \frac{1}{7}\cdot\frac{1}{7} = \frac{6}{49}.$

  • $s = 7$: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$

$p_7 = \frac{2}{7}\cdot\frac{2}{7} + 5 \cdot \frac{1}{7}\cdot\frac{1}{7} = \frac{4}{49} + \frac{5}{49} = \frac{9}{49}.$

  • $s = 8$: $(2,6),(3,5),(4,4),(5,3),(6,2)$

$p_8 = 5 \cdot \frac{1}{7}\cdot\frac{2}{7} = \frac{10}{49}?$

Check carefully:

  • $(2,6) \to \frac{1}{7} \cdot \frac{2}{7} = \frac{2}{49}$
  • $(3,5) \to \frac{1}{7}\cdot\frac{1}{7} = \frac{1}{49}$
  • $(4,4) \to \frac{1}{7}\cdot\frac{1}{7} = \frac{1}{49}$
  • $(5,3) \to \frac{1}{7}\cdot\frac{1}{7} = \frac{1}{49}$
  • $(6,2) \to \frac{1}{7}\cdot\frac{1}{7} = \frac{1}{49}$

Sum: $2+1+1+1+1=6 \implies \frac{6}{49}$.

  • $s = 9$: $(3,6),(4,5),(5,4),(6,3)$

$(3,6)=\frac{1}{7}\cdot \frac{2}{7} = \frac{2}{49}$

$(4,5)=\frac{1}{7}\cdot\frac{1}{7}=\frac{1}{49}$

$(5,4)=\frac{1}{7}\cdot\frac{1}{7}=\frac{1}{49}$

$(6,3)=\frac{1}{7}\cdot\frac{1}{7}=\frac{1}{49}$

Sum: $2+1+1+1=5 \implies \frac{5}{49}$

  • $s = 10$: $(4,6),(5,5),(6,4)$

$(4,6)=\frac{1}{7}\cdot \frac{2}{7}=\frac{2}{49}$

$(5,5)=\frac{1}{7}\cdot\frac{1}{7}=\frac{1}{49}$

$(6,4)=\frac{1}{7}\cdot\frac{1}{7}=\frac{1}{49}$

Sum: $2+1+1=4 \implies \frac{4}{49}$

  • $s = 11$: $(5,6),(6,5)$

$(5,6)=\frac{1}{7}\cdot\frac{2}{7}=\frac{2}{49}$

$(6,5)=\frac{1}{7}\cdot\frac{1}{7}=\frac{1}{49}$

Sum: $2+1=3 \implies \frac{3}{49}$

  • $s = 12$: $(6,6) \to \frac{1}{7}\cdot\frac{2}{7}=\frac{2}{49}$

All probabilities sum to:

$2+3+4+5+6+9+6+5+4+3+2=49 \implies \sum_{s=2}^{12} p_s = \frac{49}{49} = 1.$

Hence, the probabilities are

$$ \begin{aligned} p_2 &= \frac{2}{49}, \quad p_3 = \frac{3}{49}, \quad p_4 = \frac{4}{49}, \quad p_5 = \frac{5}{49}, \quad p_6 = \frac{6}{49}, \ p_7 &= \frac{9}{49}, \quad p_8 = \frac{6}{49}, \quad p_9 = \frac{5}{49}, \quad p_{10} = \frac{4}{49}, \quad p_{11} = \frac{3}{49}, \quad p_{12} = \frac{2}{49}. \end{aligned} $$

This completes the solution.

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