Imagine a row of 10 pirates, all very intelligent and all very greedy. We'll label the pirates with numbers 1 through 10.
The pirates are about to divide 100 gold coins among them in the following manner: Pirate #1 proposes how to divide the coins and his proposal is put to a vote (the proposing pirate also votes). If the vote passes, (tied vote also passes) then the coins are divided as per his proposal. Otherwise, he is thrown overboard and pirate #2 makes a new proposal, and so on.
How will the coins be divided? (Yes, there is a logical clear-cut, no-doubt-about-it solution)
2007-03-14 00:33:20 · 2 answers · asked by blighmaster 3 in Science & Mathematics ➔ Mathematics
If pirates 9 and 10 are left, then pirate 9 will take it all, so if pirates 8,9 and 10 are left, #8 will propose 99 for himself and 1 for #10 (#10 will vote in favor since 1 coin is better than none)
If pirates 7,8,9 and 10 are left, #7 will propose 99 for himself and 1 for #9 (same logic), and if pirates 6,7,8,9 and 10 are left, then #6 will propose 98 for himself, 1 for #8 and 1 for #10. #8 and #10 will vote in favor since otherwise they'll get nothing. And so on and so on, so....
Pirate #1 will propose 96 for himself, and 1 each for pirates 3,5,7 and 9 and the vote will pass!
Examine my solution, OZ - thoroughly! And you will see it is correct.
2007-03-14 00:35:23 · answer #1 · answered by pjjuster 2 · 0⤊ 1⤋
Pirate 10 and 9 will share the coins under pirates 9 proposal because the pirates will all vote against the proposal until the last two remain and thus a tied vote so passing the proposal of pirate 9.
if pirate 9 proposes 100% division to him then pirate 9 will take all the coins
2007-03-14 07:38:06 · answer #2 · answered by Oz 4 · 0⤊ 1⤋