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)
2007-03-13 23:41:24 · 10 answers · asked by pjjuster 2 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!
2007-03-14 00:32:01 · answer #1 · answered by blighmaster 3 · 0⤊ 0⤋
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 06:45:31 · answer #2 · answered by Oz 4 · 2⤊ 0⤋
Pirates 5,6,7,8,9 and 10 all vote against the proposals of pirates 1 to 4 (the last 6 have a majority over the first 4)
At this stage pirates 7,8,9 and 10 vote against pirates 5 and 6 proposals.
Pirates 8,9 and 10 vote against pirate 7 and pirates 9 and 10 vote against pirate 8.
Thus the two last remaining pirates share the 100 gold coins.
50 each.
2007-03-14 07:03:40 · answer #3 · answered by Mikem 3 · 0⤊ 0⤋
pirates 10, 9, 8, 7, 6, 5, 4, 3, and 2 are all out voted because they are all greedy and want the most for themselves. this leaves only pirate 1 who takes all the 100 gold coins for himself
2007-03-14 08:02:29 · answer #4 · answered by captainblyton 2 · 0⤊ 0⤋
are they going to have the vote not pass and each pirate thrown overboard until there's only one pirate who gets the whole lot?
2007-03-14 06:46:07 · answer #5 · answered by mhm 3 · 0⤊ 0⤋
The answer is that pirates 8,9 and 10 will divide the remaining coins because 1,2,3,4,5,6,7 will be trown aboard. ..
2007-03-14 07:36:59 · answer #6 · answered by richard 2 · 0⤊ 0⤋
pirate no.1
will vote :
96 for himself and 1 coin for each pirate of numbers(3,5,7,8)
2007-03-15 08:32:40 · answer #7 · answered by hasan 2 · 0⤊ 0⤋
they can give me the money and ill throw them all over board.. im 100 coins richer than you
stupid pirates
2007-03-14 06:49:21 · answer #8 · answered by Anonymous · 0⤊ 0⤋
100/10? lol sorry
2007-03-14 06:44:00 · answer #9 · answered by Konjo Nashi Pirate™ 5 · 0⤊ 0⤋
My brain hurts....
2007-03-14 06:51:06 · answer #10 · answered by Anonymous · 0⤊ 0⤋