The Gumball Proposition

In general, let there initially be $n$ gumballs in each machine. We want to find the probability that, after it is first discovered that one of the machines is empty, there are $k$ gumballs still left in the other machine, where $k$ can be any integer value from $0$ to $n.$ (Note that it could be the case that both machines have been emptied, so that when the next person tries to buy a gumball and comes up empty, it is in fact the case that both are empty, i.e., $k$ can equal $0.$ )

Now this "discovery" can happen in one of two ways:

• (i) for the first combined $2n - k$ gumballs bought from the machines, $n$ have been bought from one machine, call it $A$ , and $n - k$ have been bought from the other machine, call it $B$ . On the $(2n - k + 1)$ th purchase, the unfortunate shopper tries to buy a gumball from $A$ only to come up empty;

• (ii) same as (i) except the roles of machines $A$ and $B$ are reversed.

Both scenarios then have a probability of $\dbinom{2n - k}{n} * \dfrac{1}{2^{2n - k}} * \dfrac{1}{2}$ of happening, and so the probability $P(k)$ that the "other" machine has $k$ gumballs left in it is

$P(k) = \dbinom{2n - k}{n} * \dfrac{1}{2^{2n - k}}.$

In this case we have $n = 100$ and $k = 6, 7, ....., 100.$ It will be easier to calculate using the complement, i.e.,

$S = 1 - \displaystyle\sum_{k=0}^{5} \dbinom{200 - k}{100} * \dfrac{1}{2^{200 - k}} = 0.667546....$

Thus $\lfloor 10000*S \rfloor = \boxed{6675}.$