Calculating probability for FUNPROB

问题

Regarding - FUNPROB

The solution is :

int N, M;
while(1) {
    scanf("%d %d", &N, &M);
    if (0 == N && 0 == M) break;

    if (N > M) printf("0.000000\n");
    else {
        double res = (double) (M-N+1) / (M+1);
        printf("%.6f\n", res);
    }   
}

My question is regarding line

res = (M-N+1) / (M+1);

How to arrive at the conclusion that the probability is calculated in this way ?

回答1:

At first it is obvious that if N>M probability is zero.

now I want to use indication on N to prove. consider M>0 I want to prove for every N=<M we have res = (M-N+1) / (M+1) for N=0 it is obvious that probability is 1.

For N=1 put every body we 5$ in a queue in arbitrary order, now for the one person with 10$ you could put him every where except in front of the queue so you have between M+1 places that are available you have M+1-1 choices. so for N=1 you have : res = (M-1+1) / (M+1)

suppose formula is correct for every N=<k I want to prove that if N=k+1 formula is still correct. for that put M people with 5$ and k people with 10$ in a arbitrary queue. we suppose that res = (M-K+1) / (M+1) is the probability of working this queue and every body could have his ticket. consider one of the working queues in this queue if a 10$ person is behind 5$ person remove both of them and do this recursively until there is no 5$ person. this will work because as I said above the first person of the queue is a 10$ one, and also I said that N<M the probability of putting the K+1 person in the queue is choosing one place among M-k+1 because we remove k 10$ person from the queue. and it is like what we said for N=1 so we have probability of putting K+1th 5$ person in the queue is : ((M-k) - 1 +1) / ((M - k) +1)(*) and by indication we have that probability of a working queue for N=k is : (M-k +1) / (M +1)( * ) from () and ( * *) we have probability of putting K+1 people with 10$ and M people with 5$ in a queue with questions condition is :

[((M-k) - 1 +1) / ((M - k) +1)] * [(M-k +1) / (M +1)] = ((M-k) - 1 +1) / (M +1) = (M-(k+1) +1) / (M +1)

And it is end of the proof :).

回答2:

Found the answer. The problem keywords are Dyck words and Catalan number. @Ali's answer is the proof that answer is correct but doesnt explain how we arrive at the number.

Realise that if M < N, probability is 0.
If M >= N, its a necessary condition for having change but not sufficient. You need to have the correct order of people in queue. For eg. M=3, N=2 MMMNN is correct while NNMMM is not correct. So we have to filter out incorrect possible queues.

The problem can be thought of in terms of paths that go up/down in one time step. X axis is time axis, +Y axis is M and -Y axis is N. If we start from (0, 0) traverse up 1 unit for each M and -1 unit for each N then we always end up at (M+N, M-N).

Our answer is: (Total number of paths from (0, 0) to (M+N, M-N) - Paths which go below X axis at least once) / Total number of paths.

enter image description here

Solving for above gives answer as (M-N+1)/(M+1).

*The second term in numerator comes using reflection principle. If you want more details, add a comment and I will update the answer.

来源：https://stackoverflow.com/questions/25281005/calculating-probability-for-funprob

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c++

math

probability