Goodluck peeps

I changed my mind, this one easy ....

Give 3 and 2 350 bars each and keep 300 bars for 1. The reason as that 5 will always say no since he stands the most to gain and nothing to lose. 4 would have accepted a fair deal since in a head to head vote aginst 5 he would end up being one vote short of a majority. But the fact that 2 and three's vote is enough, it makes sense to cut him off.
Still this puzzle is not objective enough for me.

dorox: Give 3 and 2 350 bars each and keep 300 bars for 1. The reason as that 5 will always say no since he stands the most to gain and nothing to lose. 4 would have accepted a fair deal since in a head to head vote against 5 he would end up being one vote short of a majority. But the fact that 2 and three's vote is enough, it makes sense to cut him off.
Still this puzzle is not objective enough for me.

.....naaah....

Firstly, understand the main question..

What is the most amount of gold bars that agbero #1 (Bola) can keep to himself, and what would his proposal be?

...Although tackling the question from a back-end is best, but you've got to understand it, #1 is the focal point.

Hiroshize:
Agbero #1 : Bola
Agbero #2 : Chukz
Agbero #3 : Zico
Agbero #4 : Musa
Agbero #5 : Damo

Question: What is the most amount of gold bars that Bola can keep to himself, and what would his proposal be?

bola will have at most 201.. And his 1st proposal is as follows:
#1..... 598
#2..... 201
#3..... 201
#4..... 0
#5..... 0
total votes: 3/5 votes (i.e from #1,#2,#3 becos it favors the 3 of them..under normal condition, they are entitled to 200bars each but now, they have more).

#2 will also present his proposal as below;
#1..... 201
#2..... 598
#3..... 201
#4..... 0
#5..... 0
total votes=3valid one at least

#3 also as below;
#1..... 201
#2..... 201
#3..... 598
#4..... 0
#5..... 0
total votes= 3/5 at least.

#4 is also as below;
#1..... 201
#2..... 201
#3..... 201
#4..... 397
#5..... 0
total votes=4/5
this will definitely get the highest vote of at least 4 agberos..

dejt4u:
bola will have at most 201.. And his 1st proposal is as follows:
#1..... 598
#2..... 201
#3..... 201
#4..... 0
#5..... 0
total votes: 3/5 votes (i.e from #1,#2,#3 becos it favors the 3 of them..under normal condition, they are entitled to 200bars each but now, they have more).
.................................................

Bros, this your maths.... lemme follow it and understand you first ..

Ogbeni, i see you peeping here oo @Drniyi4u

dejt4u: Ogbeni, i see you peeping here oo @Drniyi4u

Lol.....m following!

dejt4u

Am sorry, I've got to disagree with your solution...

It's really simple... Give it some more simplistic thinking...

If no correct answer or near correct answer before 9...I will provide the solution in logical steps...

Hiroshize:

Am sorry, I've got to disagree with your solution...

It's really simple... Give it some more simplistic thinking...

If no correct answer or near correct answer before 9...I will provide the solution in logical steps...

ok

Yess

Hiroshize: Five Ojuelegba (Lagos) agberos are trying to split up 1000 gold bars in an EbolaMustGo bag they in a bus from Okokomiako.

Agbero #1 : Bola
Agbero #2 : Chukz
Agbero #3 : Zico
Agbero #3 : Musa
Agbero #4 : Damo

@ d bolded above..pls correct it as thos typo errors are confusing me..

dejt4u: @ d bolded above..pls correct it as thos typo errors are confusing me..

In simple terms, 5 agberos need to share 1000 gold bars.
There's :
Agbero 1
Agbero 2
Agbero 3
Agbero 4
Agbero 5

Hiroshize:

In simple terms, 5 agberos need to share 1000 gold bars.
There's :
Agbero 1
Agbero 2
Agbero 3
Agbero 4
Agbero 5

at dis point, i yield

dejt4u: at dis point, i yield

Hehehehehe.... Let's reveal the answer...

Hiroshize:

Hehehehehe.... Let's reveal the answer...

waiting

let's start from a back end point of view.... Simple then to most complex - -

If there are two agberos left (#4 & #5) for instance , #4 has no options. No matter what he proposes, agbero #5 will disagree, resulting in a 1-1 vote (no majority). #5 will kill #4 and will keep all of the gold bars.

Now say there are 3 agberos left. #4 has to agree with whatever #3 decides, because if he doesn't #3 will be killed (because #5 won't vote for #3's proposal no matter what it is). #3 will just propose that he keep all of the gold and will get a 2-1 vote in his favor.

Now if there are 4 agberos left: #3 won't vote for #2's proposal because if #2's fails, #3 will get all of the gold. #4 and #5 know that they will get nothing if the decision goes to #3, so they will vote for #2's proposal if he gives them one gold piece each. Therefore, #2 would keep 998 gold, and #4 and #5 would each get one gold.

In summary, Agbero #1 needs 2 other votes. He will not get a vote from #2 because #2 will get 998 gold if #1's plan fails.

#1 offers #3 one gold piece to vote for him, which #3 will accept (if it gets to #2's plan, #3 will get nothing).

#1 then offers #4 or #5 (doesn't matter which) two gold pieces, which is more than they would get with #2's plan.

So #1 can end up with 997 gold pieces, with #3 getting one piece and #4 or #5 getting two bars.


Follow it well, you would understand it.... Don't get confused by the # sign, just like saying Agbero number 1 or number 2.

Thanks....feel free to challenge it...

More still to come

m nw here bro..let me attempt the question

Go on..

Hiroshize: Here's another one

There are five people. One of them shot and killed one of the other five.

Who was the murderer and who was killed ?

1. Emeka ran in the Obudu Mountain marathon yesterday with one of the innocent men.

2. Bosco considered being a Kolanut farmer before he moved to the city.

3. Chukwudi is a badass computer consultant and wants to install Mohammed's new computer next week.

4. The murderer had his leg amputated last month.

5. Mohammed met Yinka for the first time six months ago.

6. Yinka has been in seclusion since the crime.

7. Emeka used to drink heavily

8. Mohammed and Chukwudi built their last computers together.

9. The murderer is Yinka's brother. The grew up together in Ikot Ekpene.

10. Op use your brain !!!! Must you cheat

bosco is d murderer and he killed Chukwudi

dejt4u:
bosco is d murderer and he killed Chukwudi

Nice try.....

But NO

Note : Defend your answer with a plausible explanation

Hiroshize:

Nice try.....

But NO

Note : Defend your answer with a plausible explanation

ok..lemme try again

Intelligent questions...Let me see if I can try it.

At this point, i can now conclude that CHUKWUDI is the murderer..

Analysis..
From #9: YINKA is nt d murderer 'cos his brother is the criminal..so Yinka is not the criminal

From #5: MOHAMMAD is not the murderer 'cos he is not Yinka's brother (they only met for the 1st time 6months ago)

From #4: EMEKA is not the criminal 'cos he is nt amputated (he ran just y'day from #1)

From #2: BOSCO is not the criminal 'cos he is now living in the city (yinka nd his bro grew up 2geda in Ikot-Ekpene)

definitely, CHUKWUDI IS THE MURDERER

AND BOSCO IS THE PERSON KILLED

adebayour26: Intelligent questions...Let me see if I can try it.

you are welcome here bro

dejt4u: At this point, i can now conclude that CHUKWUDI is the murderer..

Analysis..
From #9: YINKA is nt d murderer 'cos his brother is the criminal..so Yinka is not the criminal



AND BOSCO IS THE PERSON KILLED

......pending validation ..

Hold on..

dejt4u

Thats correct

Another question coming shortly....

will give the full explanation for this answer.....

Hiroshize:

Thats correct

Another question coming shortly....

will give the full explanation for this answer.....

i'm loving this bro...keep it coming.. I need to exercise my brain

dejt4u:
i'm loving this bro...keep it coming.. I need to exercise my brain

Thunbs up!!!

Well done

The next question will be tricky

.....