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 nsuccesses 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 Pn,m the probability that n successes occur before m failures. By conditioning on the outcome of the first trial we obtain:
By using the obvious boundary conditions Pn,0=0, P0,m=1, these equations can be solved for Pn,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 nsuccesses 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

Now,
where the last identity follows from the substitution i=2n-2-k.Thus
and so the first player is entitled to