# A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 8 – Solution, Numerical Example

Author(s):
Alan Levine (Franklin and Marshall College)

Review statement of Problem 8.

For example, for $m = 6$ and $n = 3$, we find:

$216 P_{18} = 216 P_3 = 1, \ \ \ 216 P_{17} = 216 P_4 = 3,$
$216 P_{16}= 216 P_5 = \frac{3\cdot 4}{1 \cdot 2} = 6, \ \ \ 216 P_{15}= 216 P_6 = \frac{3\cdot 4\cdot 5}{1 \cdot 2\cdot 3} = 10,$
$216 P_{14}= 216 P_7 = \frac{3\cdot 4\cdot 5\cdot 6}{1 \cdot 2\cdot 3 \cdot 4} = 15,$
$216 P_{13}= 216 P_8 = \frac{3\cdot 4\cdot 5\cdot 6 \cdot 7}{1 \cdot 2\cdot 3\cdot 4 \cdot 5} = 21,$
$216 P_{12}= 216 P_9 = \frac{3\cdot 4\cdot 5\cdot 6 \cdot 7 \cdot 8}{1 \cdot 2\cdot 3\cdot 4 \cdot 5 \cdot 6} - 3 = 25,$
$216 P_{11}= 216 P_{10} = \frac{3\cdot 4\cdot 5\cdot 6 \cdot 7 \cdot 8 \cdot 9}{1 \cdot 2\cdot 3\cdot 4 \cdot 5 \cdot 6 \cdot 7} - 3 \cdot 3 = 27,$

Three ordinary six-sided dice, whose sides have numbers 1, 2, 3, 4, 5, 6 can serve to illustrate this example.

If these three dice are thown on a surface and if $X_1, X_2, X_3$ denote the numbers on the upper sides, then $\{1, 2, 3, 4, 5, 6\}$ will be the exhaustive and equally likely values of $X_1$, as well as of $X_2$ and $X_3$.

Corresponding to these, the numbers $P_3, P_4, P_5, \dots P_{18}$ represent the probabilities of different assumptions on the sum of the numbers,18 revealed on three ordinary throws of the dice.

And the equation $P_3 + P_4 + P_5 + \cdots + P_{10} = P_{11} + P_{12} + P_{13} + \cdots + P_{18}$ shows the equal probability of the assumptions that this sum does not exceed 10 and, conversely, that it is bigger than 10.

Figure 7. Three dice made of bone or ivory, ca 9th–10th century CE, excavated from Nishapur, Iran.
Metropolitan Museum of Art, Rogers Fund, 1938, public domain.

Continue to the second part of Markov's solution of Problem 8.

[18] The text has the last entry in this list as $P_{21}$, which clearly is incorrect. The sum of the faces on three dice cannot exceed 18.