Probability of 4 balls in one of 80 bins after 26 throws

Question

This relates to a stock problem: I have a product with 80 sizes, the stock level of each size is 3, I forecast 26 customers per week so what is the probability I will run out of stock of at least one size ie. have an order for at east 4 of at least one size. From this I would like to be able to try various combinations of stock level and forecast customers.

Answer 1

For practical purposes, the probability that this happens for more than one size is so low that you can just take the expected number of sold-out sizes as an excellent approximation of the probability that at least one size is sold out. With $n=80$, $m=26$, $k=4$, this is

\begin{eqnarray*} n\left(1-n^{-m}\sum_{j=0}^{k-1}\binom mj(n-1)^{m-j}\right) &=& 80\left(1-\frac{79^{26}+26\cdot79^{25}+\binom{26}2\cdot79^{24}+\binom{26}3\cdot79^{23}}{80^{26}}\right) \\ &\approx& 0.0234\;. \end{eqnarray*}

If you want to know the exact probability, you can find it using inclusion–exclusion , but the calculation would be tedious since you'd have to sum over many possibilities for each of the sold-out sizes. Another way to find the exact probability would be to model this as a Markov chain, where the states would be characterized by the number of sizes that have $0$, $1$, $2$ and $3$ items left; the number of states would be the number $\binom{80+3}3=91881$ of ways to distribute the $80$ sizes over the $4$ stock values, so this could easily be modeled exactly on a computer.