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?
Added: Thanks to Henry's great answer below, the above questions have mostly been answered, though I'd appreciate it if anyone has an answer that adds more to it. I'm interested in finding something else now;
Let's say there are $r$ red balls and $b$ blue balls. How do I find
the probability that the length of the longest red (or blue) run is
$M$, where $M \in \{1,2,…r\}$? Essentially, I'm trying to find the
probability distribution of the lengths of the longest run.
I've read Schilling's classic MAA paper on the subject and tried to extend it, but this seems to be a much harder problem, as the probability of a red (or blue) ball keeps changing with the trial. Could someone help me out with this?
Best Answer
All ${100 \choose 30}$ patterns are equally likely. There is probably a better approach to your first question, but this produces a nice answer:
Suppose that $R(r,b)$ is the expected number of runs given that the first ball is red and $B(r,b)$ is the expected number of runs given that the first ball is blue, in each case when there are $r$ red balls and $b$ blue balls. Then $R(r,0)=1$ and $B(0,b)=1$ and by symmetry $R(r,b)=B(b,r)$. More generally we have the recurrence $$R(r,b)=\frac{r-1}{r+b-1} R(r-1,b) + \frac{b}{r+b-1} (1+R(b,r-1))$$ which has the solution $R(r,b)= \dfrac{2rb+r-1}{r+b-1}$ with $R(1,0)=1$. So the expected number of runs overall when there are $r$ red balls and $b$ blue balls is $$\frac{r}{r+b}R(r,b)+\frac{b}{r+b}R(b,r)=1+\frac{2rb}{r+b}.$$
In your question you had $r=70$ and $b=30$, giving an expected number of runs of $43$.
It is slightly unfortunate that $43$ is odd, as it complicates the distribution of runs (e.g. there might be $21$ or $22$ red runs with probabilities $0.3$ and $0.7$).
Added: In general, if there $n$ balls of a particular colour and they are distributed between $k$ positive runs, then a simple stars and bars argument says that the probability that a given run is of length $M$ is $$\Pr(M=m|n,k)= \dfrac{\displaystyle {n-m-1 \choose k-2}}{\displaystyle {n-1 \choose k-1}}$$ the each of the $k$ runs has a identical (but not independent) distribution. So for example if you have $70$ red balls in $22$ runs or $30$ blue balls in $21$ runs, the probability distribution for the length $M$ in a red run or of a blue run would be about
But we can go further than that, and see the distribution of runs unconstrained by how many runs there are. As an example consider two red balls and one blue ball: the patterns $R_2B_1$, $R_1B_1R_1$, and $B_1R_2$ are equally likely and so with repeated experiments over time you expect to see as many runs of $R_2$ as $R_1$ in this particular case. More generally, if there are $r$ balls and $b$ blue balls, the expected proportion of red runs which would be of length $m$ in that sense, is
$$ \dfrac{b}{r} \dfrac{ {r \choose m} }{ {r+b-1 \choose m}}$$
where reversing $r$ and $b$ would give the expected proportion of blue runs of length $m$ in that sense. With $r=70$ and $b=30$ this would give about
which is not that far away from the constrained case.