I found this solution while solving an equivalent problem (it was about apartments and residents , instead of pirates and golds , respectively.) We begin with a claim . Key claim. This process ends in at most $15$ controls. Proof. FTSC assume it doesn 't necessarily stop , i.e. we can find a pirate with $\ge 15$ pieces at $15^{\text{th}}$ control . Clearly if a pirate with $\ge 15$ pieces (let's call them greedy) is found ,he loses all his pieces and only after $\ge 15$ steps he can be greedy again $(\clubsuit)$ , because we check the pirates in the following way : check them until we find a greedy pirate , then apply the given process (take all from him) and then begin to check again , from the first pirate and check until we find another greedy one ,so he can get at most one piece at each checking , etc. Now $( \clubsuit)$ is obvious and all these $15$ pirates are distinct ,say $P_1,P_2,\dots,P_{15}$ , now wlog assume $P_1$ has $\ge 15$ pieces of gold , then $P_2$ will have $\ge 14$ pieces , $\dots$ $P_{15}$ will have $\ge 1$ pieces and so the total number of pieces that are distributed among them is at least $1+2+\dots+15=120$ while we have only $119$ pieces , contradiction. $\blacksquare$