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I encountered the following Pólya's Urn setup:

Suppose originally we have $1$ red ball, $1$ blue ball, and $k$ green balls. Each turn we draw a ball from the urn uniformly at random and replace it with an additional ball of the same color. After $N$ rounds, there are $(2+k+N)$ balls in total, we draw two balls out at once.

What is the probability that these two balls are exactly one red and one blue?

If it's not possible to write out a clean formula for this probability, is it possible to find the size (i.e. "sharp" upper and lower bounds) of this probability in terms of $k$ and $N$?

The motivation behind the problem is that I am unsure of how to draw up probabilistic bounds on problems with multiple unknowns. I am aware of how to possibly ascertain an exact solution but this seems very complicated. Therefore, I am interested to find techniques for simplifying such a problem in order to establish strong bounds (rather than an exact solution) with the trade off being a simpler solution.

However, I am unsure how to progress with this idea.

FD_bfa
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MikeG
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1 Answers1

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We can establish a lower bound and upper bound as follows:

Lower bound

The worst case scenario is where after $N$ rounds, we have picked no red balls or blue balls (and so there are still one of each in the urn). Therefore, the probability of selecting a $\color{red}{ \text{red}}$ ball is $$\frac{1}{2+k+N}$$ and this probability will be the same for selecting a $\color{blue}{ \text{blue}}$ ball by symmetry.

Therefore, the probability that we draw exactly one red and one blue ball (in no particular order) is $$ \fbox{$2 \times \Big{(} \frac{1}{2+k+N} \Big{)} \times \Big{(}\frac{1}{(2+k+N)+1} \Big{)}$}$$

Note$_1$: the reason we multiply by $2$ in the above formula is to account for choosing "red first, then blue" and "blue first, then red". The reason why we add one to the denominator for the second pick is to account for the additional ball that is added after our first pick.

Upper Bound

For the upper bound we want to find the optimal number of picks of red and blue balls such that the probability of selecting one red and one blue is maximized. Let's begin by considering the problem in generality.

In the best case scenario, out of our $N$ picks, we do not select any green balls (since this clearly would only reduce the probability of selecting red and blue. Let's assume that we pick a red ball $A$ times and pick a blue ball $B$ times. Then we know that $A+B = N$.

The probability of picking a red ball first in this scenario is:

$$ \color{red}{\frac{A+1}{2+k+N}}$$

and picking a blue ball first has probability:

$$\color{blue}{\frac{B+1}{2+k+N}}$$

Therefore the probability that we pick a red first or a blue first will be:

$$ \color{red}{\Big{(}\frac{A+1}{2+k+N} \Big{)}} \color{blue}{ \Big{(} \frac{B+1}{(2+k+N)+1} \Big{)}} + \color{blue}{\Big{(} \frac{B+1}{2+k+N} \Big{)}} \color{red}{ \Big{(} \frac{A+1}{(2+k+N)+1} \Big{)}}$$

To find an upper bound on this, we will need to pick the values of $A,B < N$ such that this sum is optimized.

If we treat the above as a function of $A,B$, then the constants in the denominator do not affect which values of $A,B$ maximize the function. Therefore, this is equivalent to maximizing the simplified formula:

$$ \Big{(}\frac{A+1}{N} \Big{)} \Big{(} \frac{B+1}{N} \Big{)} + \Big{(} \frac{B+1}{N} \Big{)} \Big{(} \frac{A+1}{N} \Big{)}$$ $$ = $$

$$ \Big{(}\frac{A+1}{A+B} \Big{)} \Big{(} \frac{B+1}{A+B} \Big{)} + \Big{(} \frac{B+1}{A+B} \Big{)} \Big{(} \frac{A+1}{A+B} \Big{)} $$

$$ = $$

$$ 2 \times \Big{(}\frac{(A+1)(B+1)}{(A+B)^2} \Big{)} $$

Again, if we want to find the values of $A,B$ that maximize this function, the constant multiple $2$ does not affect which values of $A,B$ maximize this, so we can again simplify the above to consider the following maximization problem under the constraint that $A+B=N$:

$$ \Big{(}\frac{(A+1)(B+1)}{(A+B)^2} \Big{)} $$

One way to tackle the above would be using the method of Lagrange Multipliers to tackle this constrained optimization problem for a given value of $N$.

Using this approach, we find that $A = B = \frac{N}{2}$ is the optimal choice of red and blue balls. This now allows us to return to the original problem of establishing our upper bound. We know that the upper bound on the probability is the maximum value of the following:

$$ \Big{(}\frac{A+1}{2+k+N} \Big{)} \Big{(} \frac{B+1}{(2+k+N)+1} \Big{)} + \Big{(} \frac{B+1}{2+k+N} \Big{)} \Big{(} \frac{A+1}{(2+k+N)+1} \Big{)}$$

Where we can now substitute in $A=B= \frac{N}{2}$ to give us:

$$ \fbox{$\frac{2\big{(}\frac{N}{2}+1\big{)}^2}{(2+k+N)(2+k+N+1)}$} $$

Conclusion

Therefore, if we let $P$ denote the probability that we draw exactly one red and exactly one blue after $N$ rounds, then we get the following inequality by combining the bounds established above:

$$ \fbox{$\bbox[#E0FFE0]{\frac{2}{(2+k+N)(2+k+N+1)} \leq P \leq \frac{2\big{(}\frac{N}{2}+1\big{)}^2}{(2+k+N)(2+k+N+1)}}$}$$

Note$_2$: the bounds found above assume that $k \neq 0$. If we allow for $k=0$, then we can improve the bounds further, however, this should be fairly simple to calculate as it reduces down to the setting where we only have red and blue balls.

FD_bfa
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