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Chitchat Sammyboy Streetsmart Training Program

Ice was poisoned .
First girl no die cus ice not melt

2nd drink so slow then all ice melt, cfm die
 
Thick Face Black Heart said:
Five pirates are parting ways after finding a treasure of 100 pieces of gold. The pirates decide to split it based on a vote. Each pirate, from oldest to youngest, gets to propose a plan on how to split the gold.

If at least 50 percent of the other remaining pirates agree on the plan, that is how they will split the gold. If less than 50 percent of the pirates agree, the pirate who came up with the plan will be thrown overboard. Each pirate is smart, greedy, and wants to throw as many others overboard as possible without reducing the amount of gold they get.

What plan can the first (oldest) pirate propose to live and get as much gold as possible?
Click to expand...

Assume Pirates A B C D E (A Oldest, E Youngest)
Need >50% votes to survive and let’s count backwards

Scenario 4: - left with 2 pirates D & E,
E will get all the coin and survive. (100coins)
Reason: D’s vote alone does not exceed 50% (just 1/2)

Scenario 3 - left with 3 pirates C,D,E,
C needs 2/3 votes to survive. Split 99+1
D can never survive when in next Scenario D, so he lan lan got to agree with C.
In Scenario 2, C will propose C-99, D-1, E-0 to survive

Scenario 2 - left with 4 pirates, B,C,D,E,
B needs 3 votes to survive. Split 98+1+1
E will not let Scenario 3 happen because he will get nothing. So he will be low-baller and accept 1 coin.
D will survive Scenario 2, so he will not accept anything less than 98 coins, B is at his mercy
C will win like crazy in next Scenario 3, if B is ousted, so he will not vote for this scenario.
In Scenario B, B will propose B-1, C-0, D-98, E-1 to survive






Scenario 1:
A needs 3 votes to survive (3/5 votes)

B will only get 1 coin in next scenario 2 or else he will die. So 2 coins are enough to buy his vote.
(Asking Price 2)

C has no risk and will win all the way upon Scenario 3. He is the only decision maker who can take the bulk of the coins. So he will never vote for A and his asking price is 99 coins which is impossible (max 98 coins this round)
(Asking Price 99)

D is the king maker in scenario 2 and his asking price is 99 because if he don’t get 98 coins from Scenario 1, he has nothing to lose.
(Asking Price 98)

E knows that he will never survive Scenario 3, his best deal will be one coin from Scenario 2
(Asking Price 2)


So I think: A will offer A-96, B-2 (no better deal ahead), C-0, D-0, E-2 (no better deal ahead),
 
Last edited:
tricky-riddles-3.jpg
 
Kakamora said:

So I think: A will offer A-96, B-2 (no better deal ahead), C-0, D-0, E-2 (no better deal ahead),
Click to expand...

Then knowing that they will get nothing if A lives, C & D will offer to take one coin.
So in the end A will get 98 coins, and 1 coin for C & D each?
 
Kakamora said:
Then knowing that they will get nothing if A lives, C & D will offer to take one coin.
So in the end A will get 98 coins, and 1 coin for C & D each?
Click to expand...

Your method is correct. The question however is a bit ambiguous. It should state that the people voting for a proposal excludes the proposer himself, and if a person is already offered the best deal, he votes no if he knows he can be offered the same deal later on
 
Last edited:
Can have another one? Now I very free.
Hahaahahaha. So funny
 
SNTCK said:
What is the gender of Doctors , I can't see their face?
Can be ah gua not?
Ah gua called Sister or Brother?
Go female or male toilet?
Click to expand...

ya, nobody is lying. the other three are his sisters.
 
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