# A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 4 – Solution 1

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

Review statement of Problem 4.

Solution 1: First of all, we note that the game can be won by player $L$ in various numbers of rounds not less than $l$ and not more than $l + m - 1$.

Therefore, by the Theorem of Addition of Probabilities,13 we can represent the desired probability $(L)$ in the form of a sum $(L)_l + (L)_{l+1} + \cdots + (L)_{l+i} + \cdots + (L)_{l+m-1}$, where $(L)_{l+i}$ denotes the total probability that the game is finished in $l+i$ rounds won by player $L$.

And in order for the game to be won by player $L$ in $l+i$ rounds, that player must win the $(l+i)$th round and must win exactly  $l-1$ of the previous $l+i-1$ rounds.

Hence, by the Theorem of Multiplication of Probabilities,14 the value of $(L)_{l+i}$ must be equal to the product of the probability that player $L$ wins the $(l+i)$th round and the probability that player L wins exactly $l-1$ out of $l+i-1$ rounds.

The last probability, of course, coincides with the probability that in $l+i-1$ independent experiments, an event whose probability for each experiment is $p$, will appear exactly $l-1$ times.

The probability that player $L$ wins the $(l+i)$th round is equal to $p$, as is the probability of winning any round.

Then15 $(L)_{l+i} = p \frac{1\cdot 2\cdot \cdots\cdot (l+i-1)}{1\cdot 2\cdot \cdots\cdot i \cdot 1\cdot 2\cdot \cdots\cdot (l-1)} p^{l-1} q^i = \frac{l (l+1)\cdot \cdots\cdot (l+i-1)}{1\cdot 2\cdot \cdots\cdot i} p^l q^i,$ and finally, $(L) = p^l\left\lbrace 1 + \frac{l}{1}q + \frac{l(l+1)}{1\cdot 2}q^2 + \cdots + \frac{l(l+1)\cdots(l+m-2)}{1\cdot 2\cdot \cdots\cdot(m-1)} q^{m-1}\right\rbrace.$
In a similar way, we find $(M) = q^m\left\lbrace 1 + \frac{m}{1}p + \frac{m(m+1)}{1\cdot 2}p^2 + \cdots + \frac{m(m+1)\cdots(m+l-2)}{1\cdot 2\cdot \cdots\cdot(l-1)} p^{l-1}\right\rbrace.$

However, it is sufficient to calculate one of these quantities, since the sum $(L)+ (M)$ must reduce to 1.

Continue to Markov's second solution of Problem 4.

Skip to Markov's numerical example for Problem 4.

[13] This “theorem,” presented in Chapter I, says (in modern notation): If $A$ and $B$ are disjoint, then $P(A \cup B) = P(A) + P(B)$.  We would now consider this an axiom of probability theory. Since Markov considered only experiments with finite, equiprobable sample spaces, he could “prove” this by a simple counting argument.

[14] This “theorem,” also presented in Chapter I, says (in modern notation): $P(A \cap B) = P(A |B) \cdot P(B)$. Nowadays, we would consider this as a definition of conditional probability, a term Markov never used. Again, he “proved” it using a simple counting argument. He then defined the concept of independent events.

[15] The conclusion $(L)_{l+i} = \binom{l+i-1}{i}p^l q^i$ is a variation of the negative binomial distribution; namely, if $X$ represents the number of Bernoulli trials needed to attain $r$ successes, then $P(X=n) = \binom{n-1}{r-1}p^r q^{n-r},\ \ n\geq r$, where $p$ is the probability of success on each trial and $q=1-p$.

Alan Levine (Franklin and Marshall College), "A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 4 – Solution 1," Convergence (November 2023)