I throw 750 red and 250 blue balls into 10 bins. What is the probability no bin is over 1/3 blue?

Question

We have $1000$ balls, $750$ red and $250$ blue. We distribute them uniformly at random into $10$ bins. What is the probability none of these bins is at least $1/3$ blue?

More generally I'd like to consider the case where you have $n$ balls and $k$ bins, such that $3/4$ of the balls are red and $1/4$th are blue. What is the probability no bin has $1/3$ blue?

Approximations are also desirable, mostly ones that give a probability strictly bounded below the true probability.

Answer 1

Let's distribute the 250 blue balls randomly in the 10 bins. So we will choose 1 to 10 bins out of the 10 bins and distribute the blue balls to them randomly, so that each set contains at least one ball, which is $$\sum_{i=1}^{10} {{10}\choose{i}}\cdot{{250 - 1}\choose{i-1}}$$ if the (250 - 1) term seems odd to you, check the stars and bars method.

So now for 500 out of the 750 red balls we know exactly where they should go (namely for each blue ball we have to assign two red balls in order to keep on the one third condition.

we still have 250 red balls we can distribute randomly, so it's again $$\sum_{i=1}^{10} {{10}\choose{i}}\cdot{{250 - 1}\choose{i-1}}$$ for the 250 rest red balls. Hence the total answer should be: $$\left( \sum_{i=1}^{10} {{10}\choose{i}}\cdot{{250 - 1}\choose{i-1}} \right)^2$$

Edit Instead of this sum you can use the Formula: $${250 + 10 - 1}\choose{10 - 1}$$ for which there are many explanations, one of them is, the problem is equivalent of having 250 + 10 balls distributing them to 10 slots but every slot must have a ball at least. Another explanation is found in the second theorem of stars and bars. The total answer in this case will be: $$\left({250 + 10 - 1}\choose{10 - 1}\right)^2$$

Answer 2

This looks quite difficult, I doubt that there is a closed form solution.

For the case of large $n$ balls ($ \alpha n$ red, $(1-\alpha)n$ blue), and $k$ bins, here's a simple asympotic, for large $n,k$, and $t=n/k$ approx. constant.

The amount of red balls in each bin can be approximated by $k$ iid Poisson variables $X_i$ with mean $\lambda_x= \alpha n/k=\alpha t$.

Same for blue balls, $Y_i$ with mean $\lambda_y= (1-\alpha) t$ (also $X_i, Y_j$ are independent).

We are interested in the event $E = \cap_i E_i=\cap_i [X_i > b Y_i]. $ Then:

$$P(E_i)=\sum_{x=0}^{\infty} \sum_{y=0}^{\lceil x/b \rceil-1} e^{-\alpha t} \frac{(\alpha t)^x}{x!} e^{-(1-\alpha) t} \frac{((1-\alpha) t)^y}{y!} $$

and $$P(E)= \left(e^{-t}\sum_{x=0}^{\infty}\sum_{y=0}^{\lceil x/b \rceil-1} \frac{(\alpha t)^x}{x!} \frac{((1-\alpha) t)^y}{y!}\right)^k$$

This is not very nice, and I'm afraid it's not even easy to approximate (even for $b=1$, see here). But, at least, the term inside the parenthesis only depends on $\alpha, \beta, t$. In our case, of course, $\alpha=\frac34$ and $b=2$.