Khelil said:
Dear Members I am very bad at probability, so please I hope I can find help here. My problem is that I can't see how to use n and m in this problem
Suppose a coin-flipping game between two players, A and B. The probability that the coin lands heads up is p (0<p<1). In case a head appears, player A gets one point and player B loses one point. In case a tail appears, player B gets one point and player A loses one point. When either one of the players lose all their points, the game ends and the player having points becomes the winner. When the initial points of A and B are m and n respectively, calculate the probability that player A wins. Note that m and n are positive integers.
This is not an easy problem, and it does not have a simple solution. Start by noticing that whenever A wins a point, B loses a point; and vice versa, whenever A loses a point, B wins a point. So the total number of points of A and B combined is always $m+n$. Thus A will lose if his number of points ever goes down to $0$, and he will win if his number of points ever goes up to $m+n$.
Denote by $w_k$ the probability that A will win if he starts with $k$ points, and let $q=1-p$. If A wins the first flip (with probability $p$) then he will have $k+1$ points, and his probability of winning becomes $w_{k+1}.$ If he loses the first flip (with probability $q$) then he will have $k-1$ points, and his probability of winning becomes $w_{k-1}.$ Therefore $$w_k = pw_{k+1} + qw_{k-1}.\qquad(*)$$ Also, $w_0=0$ (corresponding to A losing the game), and $w_{m+n} = 1$ (corresponding to A winning the game).
Put $k=1$ in (*) to see that $w_1 = pw_2$, so $w_2 = \dfrac1pw_1$. Now put $k=2$ in (*): $w_2 = pw_3 + qw_1 = pw_3 + pqw_2$, from which $w_3 = \dfrac{(1-pq)}pw_2 = \dfrac{(1-pq)}{p^2}w_1.$ Continue in this way. The next step will be to put $k=3$ in (*), giving $w_4 = \dfrac{1-2pq}{p(1-pq)}w_3 = \dfrac{1-2pq}{p^2}w_1.$ If you go further, you will find that $w_5 = \dfrac{1-3pq + p^2q^2}{p^4}w_1$ and $w_6 = \dfrac{1-4pq + 3p^2q^2}{p^5}w_1$. Eventually, you get to a formula for $w_{m+n}$ as a multiple of $w_1$. And since you know that $w_{m+n}=1$ you then have an expression for $w_1$ in terms of $p$ and $q$.
Suppose for example that $m+n=6$. Putting $w_6=1$ you find that $$w_1 = \frac{p^5}{1-4pq + 3p^2q^2},$$ $$w_2 = \frac{p^4}{1-4pq + 3p^2q^2},$$ $$w_3 = \frac{(1-pq)p^3}{1-4pq + 3p^2q^2},$$ $$w_4 = \frac{(1-2pq)p^2}{1-4pq + 3p^2q^2},$$ $$w_5 = \frac{(1-3pq + p^2q^2)p}{1-4pq + 3p^2q^2}.$$ That gives A's probabilities of winning a game in which $m+n=6$, in the cases where he starts with 1,2,3,4 or 5 points.
You can see that for larger values of $m$ and $n$ the algebra gets rapidly more complicated. If you know something about solving recurrence relations, you could get a formula for the polynomials in $pq$ that occur in the general solution, but it would not look pretty.
In the case of a fair coin, where $p=q=1/2$, the calculations become very much simpler, and $w_k$ increases linearly from $0$ to $1$ as $k$ goes from $0$ to $m+n$. So when $m+n=6$ you would then find that $w_1 = 1/6$, $w_2 = 1/3$, $w_3 = 1/2$, $w_4 = 2/3$ and $w_5 = 5/6$.