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.

Skip to statement of Problem 8.

[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 \(q=1-p\).

Review statement of Problem 4.

Solution 2: Noting that the game must end in no more than \(l+m-1\) rounds, we assume that a player does not stop immediately upon achieving one of the appropriate numbers of winning rounds but continues to play until exactly \(l+m-1\) rounds are played.

Under these assumptions, the probability that player \(L\) wins the game is equal to the probability of winning, by that same player \(L\), no fewer than \(l\) rounds out of all \(l+m-1\) rounds.

In this case, if the game is won by player \(L\), then the number of rounds won by him reaches the value \(l\) and the subsequent rounds can only increase this number, or leave it unchanged.

And conversely, if player \(L\) wins no fewer than \(l\) rounds out of \(l+m-1\) rounds, then the number of rounds won by player \(M\) will be less than \(m\); hence, in that case, it follows that player \(L\) wins \(l\) rounds before player \(M\) manages to win \(m\) rounds and, in this way, the game will be won by player \(L\).

On the other hand, the probability that player \(L\) wins no fewer than \(l\) rounds out of \(l+m-1\) rounds coincides with the probability that, in \(l+m-1\) independent experiments, an event whose probability for each experiment is \(p\) appears no fewer than \(l\) times.

The previous probability is expressed by the known sum of products \[\frac{1\cdot 2\cdot \cdots\cdot (l+m-1)}{1\cdot 2\cdot \cdots\cdot (l+i) \cdot 1\cdot 2\cdot \cdots\cdot (m-i+1)} p^{l+i} q^{m-i+1}\]
where \(i= 0,1,2,\dots,m-1\).

Then \[(L) = \frac{(l+m-1)\cdot\cdots\cdot m}{1\cdot2 \cdot\cdots\cdot l} p^l q^{m-1} \left\lbrace 1 + \frac{m-1}{l+1} \frac{p}{q} +  \frac{m-1}{l+1}\frac{m-2}{l+2}\frac{p^2}{q^2} + \cdots \right\rbrace;\] also, we find \[(M) = \frac{(l+m-1)\cdot\cdots\cdot l}{1\cdot2 \cdot\cdots\cdot m} p^{l-1} q^{m} \left\lbrace 1 + \frac{l-1}{m+1}\frac{q}{p} +  \frac{l-1}{m+1} \frac{l-2}{m+2} \frac{q^2}{p^2} + \cdots \right\rbrace.\]

It is not hard to see that these new expressions for \((L)\) and \((M)\) are equal to those found previously.¹⁶

Continue to Markov's numerical example for Problem 4.

Skip to statement of Problem 8.

Review statement of Problem 4.

Numerical examples:

\(\mathrm{1)}\ \ p = q = \dfrac{1}{2},\ l=1,\ m=2\).

\[ \begin{align} (L) &= p(1+q) = 2pq\left(1+ \dfrac{1}{2}\dfrac{p}{q}\right) = \dfrac{3}{4}, \\
(M) &= q^2 = \dfrac{1}{4}. \end{align} \]

\(\mathrm{2)}\ \ p=\dfrac{2}{5},\ q= \dfrac{3}{5},\ l=2,\ m=3\).

\[ \begin{align} (L) &= p^2\left\lbrace 1+ 2q + 3q^2\right\rbrace = 6p^2q^2\left\lbrace1 + \dfrac{2}{3}\dfrac{p}{q} + \dfrac{1}{6}\dfrac{p^2}{q^2}\right\rbrace = \dfrac{328}{625}. \\ (M) &= q^3 \left\lbrace 1+ 3p \right\rbrace = 4q^3  p\left\lbrace 1 + \dfrac{1}{4}\dfrac{q}{p}\right\rbrace  = \dfrac{297}{625}. \end{align} \]

Continue to Markov's statement of Problem 8.