I'm pretty proud of myself for having taken this many photos with boats/oceans/pelicans.

Yesterday’s post had a riddle about 10 very logical pirates trying to fairly split up 100 Pieces of Eight (PoE). Today we give the solution, which is a great example of backwards induction. What we do is first imagine an easier situation, then solve for more complicated solutions using that. Specifically, we’ll first assume we have 1 pirate, then 2, and so on, up to 10 (or however many we have on board!) For ease of notation, we’ll name the lowest ranking pirate on the ship #1, the second lowest #2, and so on. So if there are 8 pirates, #8 is the captain (and only gets 1/2 of a vote).

**One pirate**– This pirate would get 100 PoE, and be pretty happy about that.

**Two pirates**– #2 is doomed. He can offer #1 all 100 PoE, but pirate logic says #1 is happier with 100 PoE *and *a dead crewmate.

**Three pirates- **#3 is in a great situation. If he dies, so does #2, so he can keep all 100 PoE for himself, get 1.5 votes (#2 would rather live with 0 PoE than die with 0 PoE), and live.

**Four pirates- **#4 needs two votes. There’s no way #3 will vote his way, but if he offers #1 and #2 a single PoE each, he’ll make it with 2.5 votes, and 98 PoE.

All of these are from Corpus Christi and some surrounding areas.

**Five pirates- **Now we’re cooking. #5 also needs two votes. The cheapest pirate vote he can buy is #3 with 1 PoE. Then he can offer *either *#1 or #2 two PoE. This is tricky because the riddle is very sensitive to initial conditions (more on that later). Let’s say that both #1 and #2 have an *expectation *of 1 PoE. In any case, #5 gets to keep 97 PoE.

**Six pirates**– This guy needs to buy 3 votes. Again, #5 is useless, #4 only costs 1 PoE , and we need to buy 2 of the remaining three pirates, each of which costs 2PoE. Again, using expectations, each of the pirates #1,2,3 expects 4/3 PoE. #6 keeps 95 PoE.

**Seven pirates**– We give #5 a single PoE, and two of the bottom 4 pirates 2 PoE (so again, they expect 1 PoE), keeping 95 for himself.

**Eight pirates***– *#6 gets 1 PoE, 3 of the bottom 5 pirates gets 2PoE (so the expectation is 6/5 PoE), #8 keeps 93 PoE.

**Nine pirates- **#7 gets 1 PoE, 3 of the bottom 6 pirates gets 2PoE (so the expectation is 1 PoE), #9 keeps 93 PoE.

**Ten pirates- **#8 gets 1 PoE, 4 of the bottom 7 pirates gets 2 PoE (expectation is 8/7 PoE), #10 keeps 91 PoE.

This is a sculpture *near* the ocean, and reminds me of creepy pirates.

So there’s our answer, but tomorrow we come back with ridiculous numbers of pirates.

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The computation of expected values is unconvincing. Can we assume that everyone is equally likely to be bribed? This does not follow from the rules. Maybe the natural choice is to bribe the higher-ranked pirates, other things being equal.

Yes, that would certainly make it better. This is another aspect of this riddle that I enjoy: it is very sensitive to initial conditions, by which I mean that changing any one rule (i.e., maybe the captain gets 1.5 votes) can change the optimal strategy significantly. I’ll be honest in that I did not work out the solution when I wrote down the rules, and probably would have included a change like yours if I had!