The Dynamics of an Urn Without Replacement

Question

Setup:

Consider an Urn with $N$ balls, where $M\leq N$ are red and $N-M$ are black.

Let $R_t$ and $B_t$ denote the event of drawing a red or black ball at time $t$, respectively. Further, let $p_t=P(B)$ and $q_t=1-p_t=P(R)$ such that $P(R_t+B_t)=p_t+q_t=1$ on each trial.

We assume a frequentialist stance on probabilties and define $p_t=\frac{N_t-M_t}{N_t}$. At each time-step $t$, we draw from the Urn without replacement. This creates the dynamics, as each draw/trial changes the number of red or black balls, and the total amount of balls in the urn, and thus the probability of drawing them.

1. Let $t$ denote the number of draws ${1,2,3,\dots}$
2. We define a $success$ as the event $B$.
3. Time / number of draws before a success is $T$.

Useful definitions/statements:

4. Time of success $T$ is a random variable with density function $f(t)$ cumulative density $F(t)$.
5. The conditional probability of success (conditioned on no success) between $t$ and $t+\triangle t$ is given as $P(t<T\leq t+\triangle t|T\geq t)\approx\frac{f(t)\triangle t}{1-F(t)}$.
6. It seems obvious that, for a finite $N$ and $N-M>0$ then $P(t\leq T)$ (the probability that a success has not yet transpired at time $t$) must necessarily converge to zero. E.g. the Urn is emptied until a black ball is drawn.

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Questions:

Generally, I am interested in the dynamics of the following probabilities. E.g. as in writing up simple one (or $n$) step dynamics.

For some $N$ and $M$ such that $M\leq N$ how does the following identities evolve as time moves forward (e.g. more balls are drawn from the Urn)?

1. $P(t=T)$ (Probability of success in trial $t$)

2. $P(t\leq T)$ (Probability of success before or in trial $t$)

3. $P(t<T\leq t+\triangle t|T\geq t)$ (Probability of success in $t+\triangle t$, given failure up to $t$)

4. Do any of the above identities (or their log-transform) follow simple recursive relations such as $p_{t+\triangle t}=\lambda p_t$? E.g. "The probability of success at time $t + \triangle t$ depends on $p_t$ via. this simple relation ...

5. Can anyone tell me what kind of littrature asks these questions? I've looked shortly at waiting time problem or first passage problems but none of them with easy-to-understand intuition of the Urn.

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Intuition/Answers:

1. $P(T=t)=(1-p)^{t}p$ and thus follows a geometric sequence.

2. $P(t\leq T)\rightarrow0$ for $t\rightarrow\infty$. The speed at which this goes towards zero, must depend on $\frac{N-M}{N}$ - but I can't seem to formulate it rigorously.

3. $P(t<T\leq t+\triangle t|T\geq t)$. Assume at time $t$ that $N=4$ and $N-M=1$ thus $p_t=0.25$. Can I deduce what $p_{t+\triangle t}$ is from definition 5.?

4. Don't really have an answer or know where to begin on this. But my intuition is that there are some easy linearities hidden in this setup. E.g. If I know $N$ and $M$, calculating average (expectations) to future probabilities might be possible or even using Bayes.

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I've tried to be as clear as possible, but I am not a mathmatician, so any calls for clarity are appreciated. Crosspost to math.SE.

Answer 1

Considering your first three questions, such draws follow a hypergeometric distribution if you draw exactly $n$ balls from the urn, to observe $k$ successes. It follows a negative hypergeometric distribution if you draw balls until observing exactly $r$ failures.

Hyprergeometric distribution has the following probability mas function

$$ P(X = k) = \frac{\binom{M}{k} \binom{N - M}{n-k}}{\binom{N}{n}} $$

that has the following resursive property (as described in Wikipedia and in this pdf that refers to Lieberman and Owen, 1961, Tables of the Hypergeometric Probability Distribution):

$$ P(X = k+1) = P(X=k) \frac{(M-k)(N-k)}{(k+1)(N-M-n+k+1)} $$

It's pretty popular probability distribution so you'll easily find lots of references about it.