Wikipedia:Reference desk/Archives/Mathematics/2014 November 30

= November 30 =

Permutations of N unique numbers subject to constraints

 * When V = N - 1, the answer isn't N - 1 but the V th triangular number. The iteration is shown in the material of sequence A130534 in OEIS. X(0, 0) = 1, X(N, V) = 0 if V > N, X(N, V) = X(N - 1, V - 1) + (N - 1) * X(N - 1, V) →86.171.209.142 (talk) 21:26, 30 November 2014 (UTC)
 * Thank you so much! WinterWall (talk) 23:58, 30 November 2014 (UTC)

Probability question involving simulations of picking balls from a bag
I’m working on a chemistry problem, which essentially translates to finding the answer to a related probability problem. However, my knowledge in probability is very limited and I'd be grateful if someone could help me out with it. The following is the problem:-

Suppose I have a bag containing $$70$$ red balls and $$30$$ blue balls. For the purpose of illustration, let’s call them $$R$$s (red balls) and $$B$$s (blue balls). Now, I am going to pick one ball at a time from this bag, without replacement. I define a run to be a sequence of consecutive $$R$$s (or alternately, $$B$$s) picked, along with the first $$B$$ (or $$R$$) that is picked. And I define a red (or blue) run length to be the number of consecutive $$R$$s (or $$B$$s) I pick in a run, before I encounter a $$B$$ (or $$R$$) or until the number of balls run out.

As examples, $$RRRRRRB$$ is a run (for simplicity, let me denote it by $$R_6$$ in shorthand) with red run length $$6$$, $$RB$$ is a run (denoted by $$R_1$$) with red run length $$1$$, $$BBBR$$ is a run (denoted by $$B_3$$) with blue run length $$3$$.

In each simulation, I keep doing runs until all the $$100$$ balls are picked out (since the balls are picked without replacement, the number of runs and the red/blue run lengths are both finite).

Let’s look at a typical simulation of ball-picking: $$R_{50} R_{10} B_{28} R_9$$. In this simulation, there are $$4$$ runs. The first run consists of $$50$$ consecutive red balls, until a blue ball is encountered. The second run consists of $$10$$ consecutive red balls until a ball is encountered. The third run consists of $$28$$ consecutive blue balls until a red ball is encountered. And the last run consists of $$9$$ consecutive red balls, and the simulation ends as there are no more balls to be picked. It is easy to see that the minimum possible number of runs is $$2$$ (attained by $$R_{70}$$ followed by $$B_{29}$$, or $$B_{30}$$ followed by $$R_{69}$$) and the maximum possible number of runs is $$31$$ (attained by $$R_1$$ $$30$$ times followed by $$R_{40}$$, or $$B_1$$ $$30$$ times followed by $$R_{70}$$).

Also, the maximum possible value of red run length is $$70$$ and that of blue run length is $$30$$.

Now, I’m interested in knowing the probability distribution of the red and blue run lengths. For this, I believe that I must first find the expected value of the number of runs in a simulation. But I’m not sure how to proceed from here. So to sum up, the following are my questions:-

$$1$$. How do I find the expected value of the number of runs in a simulation?

$$2$$. For that expected value, how do I calculate the probability distribution of red and blue run lengths?

--Jobsism10 (talk) 10:29, 30 November 2014 (UTC)
 * I don't think your suggested method of first finding the expectation of total number of runs, then use that to find the value for each type of run, will work.
 * Instead, I'd define $$f(b,r,X)$$ as the expected number of times that run X will appear in a simulation with b blue balls and r red balls. Then you have:
 * $$f(b,r,R_i) = \frac{(b+r-i-1)!}{(b+r)!(r-i)!(b-1)!}+\sum_{j=1}^r\frac{(b+r-j-1)!}{(b+r)!(r-j)!(b-1)!}f(b-1,r-j,R_i) + \sum_{j=1}^b\frac{(b+r-j-1)!}{(b+r)!(r-1)!(b-j)!}f(b-j,r-1,R_i)$$
 * With some edge cases, and likewise for blue runs. Then I'd solve this using recursion (with memoization).
 * If you then want the expected number of runs, you can sum up these values over all kinds of runs.
 * (I haven't spent a lot of time verifying the above so there could be errors.) -- Meni Rosenfeld (talk) 15:21, 30 November 2014 (UTC)