**The Problem of the Points 計點問題**

**Example***(The Problem of the Points)*- Independent trials, resulting in a success with probability
*p*and a failure with probability 1-*p*, are performed. What is the probability that*n*successes occur before*m*failures? If we think of*A*and*B*as playing a game such that*A*gains 1 point when a success occurs and*B*gains 1 point when a failure occurs, then the desired probability is the probability that*A*would win if the game were to be continued in a position where*A*needed*n*and*B*needed*m*more points to win. *Solution:*- Here, we present two solutions.

**(i)**

Let*P*_{n,m}the probability that*n*successes occur before*m*failures. By conditioning on the outcome of the first trial we obtain:*P*_{n,0}=0,*P*_{0,m}=1, these equations can be solved for*P*_{n,m}.

**(ii)**

For*n*successes to occur before*m*failures, it is necessary and sufficient that there be at least*n*successes in the first*m*+*n*-1 trials. (Even if the game were to end before a total of*m*+*n*-1 trials were completed, we could still imagine that the necessary additional trials were performed.) This is true, for if there are at least*n*successes in the first*m*+*n*-1 trials, there could be at most*m*-1 failures.

On the other hand, if there were fewer than*n*successes in the first*m*+*n*-1trials, there were fewer than*n*successes in the first*m*+*n*-1 trials, there would have to be at least*m*failures in that same number of trials; thus*n*successes would not occur before*m*failures.

Hence, as the probability of exactly*k*successes in*m*+*n*-1 trials is, , we see that the desired probability of*n*successes before*m*failures is**Example**

As an illustration of the problem of the points suppose that 2 players each put up and each of them has an equal chance of winning each point (*p*=1/2). If*n*points are required to win and the first player has 1 point and the second none, then the first player is entitled to*i*=2*n*-2-*k*.Thus