In a not so distant land, there were 5 selfish, logical protistas, named Omeali, Zygoe, Mortopejia, Bengalii, and Yorz. Though there were part of one gang, there was no brotherhood among them.

One day, they managed to rob 50 pennies from a local shoestore. Each protista wanted all 50 for itself, so... to split up the loot, they decided to play a game. They lined up in alphabetical order; starting from the first protista, each would propose a method of dividing up the money. No protistas can discuss with each other before their turn is up (i.e. no alliances). After a method is proposed, everyone would vote on it. If the majority votes for it, this method is used for dividing up the method. If not, the protista is forced to eat a poison apricot and die. Therefore, if Bengalii proposes a solution that only two protistas (including itself) vote for, he will die. There will be four left to vote on the next solution (only need half to be a majority in the case of even number of voters).

What solution should Bengalii propose such that he can live (of course!) and gain as much of the loot as possible? (Notes: the loot is discrete; can't split into fractions of a penny; these protistas are completely rational and selfish--each wants as much money as possible for itself; if it doesn't think it is getting as much as it possibly can, it will vote down the method.)

Solved by Wiggum1006 for 25 Protista Points. This is his second triumph.

first round: bengali-6 -agree zygoe-0-disgaree omeali-0-disgree morto-22-agree yourz-22-agree 2nd round zygoe and omeali elimanted bengali-5-agree morto-45-agree yourz-0-disagree 3rd round yourz eleminated bengali-50-agree morto-0-disagree bengali wins 50 coins.
—dean on 2003-06-21 11:19:57

answer...

first round:
bengali-6 -agree
zygoe-0-disgaree
omeali-0-disgree
morto-22-agree
yourz-22-agree
2nd round
zygoe and omeali elimanted
bengali-5-agree
morto-45-agree
yourz-0-disagree
3rd round
yourz eleminated
bengali-50-agree
morto-0-disagree
bengali wins 50 coins.
—dean on 2003-06-21 11:24:53

it seems the comments were reset.
re: azn_deto (deleted)
He had proposed that each protista would always vote against the proposed method of division, so that the money would be split among less and less protistas
However, these protistas are completely rational. If the second and third protista know that nomatter what they propose, they will be voted against anyway, why not just agree with the Bengalii's method and save their own lives?

re: dean
A clarification: Only one protista can die each turn, and that is the one that proposes the solution. If it does not die, then its method of splitting up the loot is used and the game is over.
—thelogicfiend on 2003-06-21 02:11:50

Keep posting your guesses until...
I confirm the answer.

re: dean
Also, remember each protista wants as much money as it can possible get for itself. Therefore, looking at round one, why would morto and yourz settle 22 each, when, using the same logic, they can get 24 each. (Likewise, wouldn't Bengalli want to propose 14 for himself?)
—thelogicfiend on 2003-06-21 02:32:58

To split it five ways. Each gets ten pennies.
—Heather on 2003-06-21 05:13:10

I change my answer. Bengali should propose that he gets 16 pennies. and that two other protistas (irrelevant witch ones, perhaps the ones he liked most) gets 17 pennies. The other two would disagree but since the two getting the loot would agree thats three which is the majority and Bengali gets to live.
—Heather on 2003-06-21 05:20:27

re: Heather
Remember each protista desires as much money as possible. Using the same logic, the other 4 protistas will want to kill Bengalii so that the pennies can be split 25/25 among 2 of the 4 remaining protistas.
—theLogicFiend on 2003-06-21 06:18:46

Philhuang.com
bengalli should propose that they split the pennies evenly
—ph23man on 2003-06-21 06:24:16

bengalii must die
the first 3 will die and the fourth will get all of the loot
—ph23man on 2003-06-21 06:27:53

re: ph23man
The question asks for what Bengalii should propose, so that he lives and ends up with as much money as possible. If he dies, your answer is wrong.
—theLogicFiend on 2003-06-21 06:31:38

yo
why do u hurt my feelings?
—ph23man on 2003-06-21 10:32:28

48
If it were to come down to the last two protistas, Yorz and Zygoe, then Yorz would take all 50 for himself, and give none to Zygoe, and Zygoe could not do anything about it.
Therefore, Zygoe will take any offer Omeali will give him, which would be 49 for Omeali, and 1 for Zygoe.
This is obviously unfavorable to YOrz, so Yorz will take any offer by Mortopejia.
Mortopejia would take 49 pennies and offer 1 to Yorz, thus getting two votes on his side.
Now, under Mortopejia's terms, Omeali and Zygoe are getting screwed, so they will take any offer by Bengalii.
Therefore, Bengalii can take 48 pennies for himself and offer Omeali and Zygoe one penny each.

—Wiggum1006 on 2003-06-21 11:41:58

Wiggum wins again!
You have the potential in you to become a logical fiend like myself. MUaHAhHAA.
—theLogicFiend on 2003-06-21 11:45:00

i win again
Yesssss...obviously it takes a selfish, logical protista to know a selfish, logical protista.
—Wiggum1006 on 2003-06-22 12:06:41

Will manage somehow.
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