# A Selection of Problems from A.A. Markov’s Calculus of Probabilities: Problem 4 – Solution 2

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

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.

[16] Note that the second solution demonstrates a nice connection between the binomial and negative binomial distributions. This argument is essentially the following theorem: If $Y$ has a binomial distribution with parameters $n$ and $p$, and $T$ has a negative binomial distribution with parameters $r$ and $p$, where $n\geq r$, then $P(Y \geq r) = P(T \leq n)$. In other words, if it takes no more than $n$ trials to get $r$ successes ($T \leq n$), then there must have been at least $r$ successes in the first $n$ trials ($Y \geq r$).