SOLUTION: If you could set p twice, two flips would be enough: Set p = 1/3 and on Heads give the widget to Alice, else reset to 1/2 and use the result to decide between Bob and Charlie. Since you can only set p once, it might make sense to try to design a scheme first, then pick p to make the scheme work.
For example, you could flip your coin three times and if the flips are not all the same, you could use the "different" flip to decide to whom to award the widget (e.g., "HTT" means Alice gets it). But if you get all heads or all tails, you'll have to do it again, and maybe yet again — so, it might take more than 10 flips.
But now suppose we flip the coin four times. There are four ways to get one head, six to get two, four to get three: Even numbers all, so anytime you get non-uniform results you can use the flips to decide between Alice and Bob. If the coin is fair, the remaining probability — that is, the probability q = p4 + (1 - p)4 of "HHHH or TTTT" — is only 1/8, too small for Charlie.
As you reduce p, however, q changes continuously and approaches 1. Thus, by the Intermediate Value Theorem, there is some p for which the probability of "HHHH or TTTT" is exactly the 1/3 you need.
[The actual value of p works out to 1/2 − sqrt(sqrt(2/3)−3/4), which is about 0.24213069021. The "other root," p = 1 − 0.24213069021, also works. Since these numbers are close to 1/4 and 3/4, you could do a pretty good job with 8 flips of a fair coin; think of each pair of 50-50 flips as one flip of a p-coin, e.g., HH is interpreted as H and HT or TH or TT as T. The result is "fair" between Alice and Bob but gives the widget to Charlie with probability (1/4)4 + (3/4)4 = 0.3203125, a bit short of his just desserts.]
