# Ridiculous numbers of pirates

That's right: three days of ocean pictures, and I don't even like the beach.

We gave an answer to the pirate riddle yesterday, but it was somewhat unsatisfying for a mathematician.  I mean, the pirates will run out of gold eventually, and then what?  Well, notice that yesterday we established, for n < 102:

n pirates– #(n-2) gets 1 PoE, ceil((n-3)/2) of the bottom n-3 pirates gets 2 PoE (so each expects something like 1 PoE).

With a crew of size 101, the pirate captain is just barely able to survive:

101 pirates- #99 gets 1 PoE, and then he needs to pay 2 PoE to 49 of the remaining 98 pirates, meaning he’ll have 1 PoE for himself.

But then adding one more crew member means certain death for #102: #100 gets 1 PoE, and it will cost 100 more PoE to get the required 51 votes!  This creates an interesting dynamic, since #102 will definitely vote for #103 (if it saves his own skin!), and #103 has to compete with the offer that #101 made:

I thought about using this one for mappings of the plane to the plane...

103 pirates- #100 gets 1 PoE, #102 gets 0 PoE, and 49 of the remaining pirates get 2 PoE.  #103 keeps 1 for himself, and has 51.5 votes for, 51 against.  Notice that in this situation, there are 100 pirates (1-99, and 101) whose expectation is 98/100 PoE.  Things just got cheaper, since these pirates can now be bought for only 1 PoE.
104 pirates– Any pirate except #100 and #103 can be bought for just 1 PoE.  Then #104 will buy himself 52 votes for 1PoE each, and the expectation for the pirate payouts is only 52/101.  He keeps 48 for himself.
105 pirates–  Now any pirate except #104 can be bought for 1 PoE.  So he buys 52 votes (expected payout is 52/103).

We have a new pattern, for 105 < n < 202:
n pirates- Any pirate except #n-1 can be bought for 1 PoE.  So the pirate will buy floor(n/2) votes for 1 PoE each.

There is a new problem with 202 pirates on the ship: #202 can only buy 100 votes, plus his own makes 100.5 for, 101 against. But here again is a new pattern: each pirate from now on will have 100 votes, plus the next pirates who will die if he dies.  So #203 gets 100 votes, plus #202, plus his own makes 101.5 for, 101 against, and lives.  #204 gets only 100.5 votes, #205 can only get 101.5, #206 can only get 102.5, but #207 will get 103.5 (vs. 103) and live.  In fact, we get the following pattern for arbitrarily large pirate ships:

For n > 200, the pirate #n can live only if  $n = 200+(2^k-1)$ for some k = 1,2,…

As an example in lieu of proof: with 263 pirates, there will be 100 votes, plus the previous 31 pirates who’ll die if this one dies for 131.5 voting for, 131 against.