My reasoning is based on these assumptions:
Each banker values his own life more than anything. I assume they would choose the death of others over money, because you mentioned the thirst for blood before the lack of generosity, and said:

That "bloodthirsty" means "will kill if given the opportunity" -> yes

If there are two bankers banker #1 (the last one making the proposal) wins. He simply rejects the proposal of banker #2 who will then be assassinated, leaving all the money to banker #1.

If there are three bankers they know what will happen if the first proposal is rejected. That is why banker #2 must accept the proposal of banker #3. Banker #3 will of course take all the money.

If there are four bankers, #3 gets the best outcome if he rejects #4's proposal. Since #1 and #2 are so bloodthirsty, they'll also reject #4's proposal no matter how much he offers them. So #4 knows he's screwed.

Which leads to five bankers. #4 will want to live so he'll accept any proposal. #5 will now have to persuade only one banker to accept his proposal. But he can't do that because the bankers' thirst for blood is greater than their greed. Therefore #5 and #4 will die no matter what and #3 will get all the money.

If I'm wrong, could you say whether it's my reasoning or my assumptions.

hmmmm

The first proposal is that they split the money equally because, otherwise 4 of them end up dead and one takes the entire sum.

If that proposal were rejected, then the next could be rejected and the next.
Only by agreeing that they split the money equally will they agree, but only by agreeing do the first 4 live.

Banker one is so smart, he has hired a sniper. He proposes he gets all the money and signals to the sniper to kill off 3, 4 and 5.
Banker two will be scared and runs off. Two will spread the word about how mighty and ruthless One is, earning One the spot for the national mob boss and gathering more and more wealth, until one day Two infiltrates into his inner circle and shoots him under the table.

Banker Two wins.

I'm smarter than all of you.

Banker #5 declines to make the first proposal... the riddles says, "may make" not "must make".

If he must make a proposal... and assuming the proposal order is 54321 and $1 denominations.

4) nothing to #4 because #1 and #2 have to accept #4s non-zero offer or get nothing.
3) nothing to #3 because #3 gets everything if #4 and #5s proposals gets rejected.
2) $2 to #2 because #4 need only give him $1.
1) $2 to #1 because #4 need only give him $1.

#3 will vote for nobody's proposal.
#3 is an ass since #2 has to agree with whatever he proposes or die, and so #3 can easily take it all.
#2 has to vote for whatever #4 proposes or he'll get $0 by #3
#1 has to vote for whatever #4 proposes or he'll get $0 by #3 too

Banker 5 says that only bankers 1 and 2 get any of the money, so that it passes 3 - 2. The rest of the Bankers then fail to agree on how the money is split between 1 and 2 and get picked off.

My take on it:
I assume that their desire to kill is less than their greed, if only because the opposite leads to Finch's solution. I would think that being smart means that they value their bloodlust below greed, but that's my interpretation.

In the case of 2 bankers, #1 will always vote against #2. There is no situation where it is advantageous for him to do anything else, no matter his order of motivations.

In the case of 3 bankers, #1 will always vote against, since a failure will lead to the 2 banker situation. #3 must convince #2; either he does so by allowing him to live, or offering him $1 if he knows #2 will not accept only his life. We do not know whether the bankers prefer to kill someone, or live themselves. So let's examine the outcome of both scenarios:


In the case that they would rather live than kill:
3 bankers: #1=0, #2=0, #3=all
(#2 prefers his life over killing)

4 bankers: #1=1, #2=1, #3=0, #4=all-2.
(The logic here is clear. #4 cannot win #3s vote no matter what. #1 and #2 know they will get nothing if this vote fails, so they will take any other amount.)

5 bankers: #1=2, #2=0, #3=1, #4=0, #5=all-3 OR #1=0, #2=2, #3=1, #4=0, #5=all-3
(#4 will demand all-1 or more, and #5 will not pay this. Thus, #5 will need two votes. #3 can be bought for 1, and either of #1 or #2 can be bought for 2, the other getting 0.)



In the case that they would rather kill than live.
3 bankers: #1=0, #2=1, #3=all
(Same as before, but #2 must be paid something, or he will prefer to kill.)

4 bankers: #1=1, #2=2, #3=0, #4=all-3
(As before, #1 and #2 are bought for 1 more than in the 3 banker situation.)

5 bankers: #1=2, #2=0, #3=1, #4=0, #5=all-3
(Same solution as before, but now #1 and #2 are non-redundant, and #1 is cheaper to buy.)


My answer is therefore that #5 offers #1 $2, and #3 $1, and keeps the rest, and the motion passes.

Undeath wrote:My answer is therefore that #5 offers #1 $2, and #3 $1, and keeps the rest, and the motion passes.

I agree with this... I was thinking the dollar savings while getting groceries.

Offer 1 xor 2 $2 and offer 3 $1. Assuming Survival always wins over Greed and Killing & Assuming Greed always wins over Killing.

If the initial proposal were that some number of the bankers got an unequal share, then by the majority rejecting that proposal, the bankers all know they can reduce the number of bankers by one, thereby increasing the inequality. They also know that this same process will remove 4 of them.
Their only logical recourse so far as I can see is to split the money equally knowing that any other proposal will result in one banker ending up with the entire sum and 4 of them will be dead.

Each knows that by not accepting such a proposal initially that any proposal they themselves make will result in their own death.
Only by accepting the initial proposal do they avoid death.
The only one who would disagree would be the last man standing.
If the order in which they make a proposal is fixed then that banker may disagree knowing that he stands to gain the entire sum, he will be in the minority though.
If the order is not fixed then each knows that 4 would die, but not which 4.

To preserve their lives that majority will understand that only a proposal acceptable to all initially guarantees their own survival.
Since the banker making the initial proposal knows this, he understands that the proposal he makes must be persuasive of not being rejected and the first banker to propose equitable division would then survive. Since the banker making the pro9posal knows this and knows that each of the others know this, he will propose equity.

If the order of suggestions is not fixed but selected at random, then this becomes a different problem. #5 needs to buy two votes at regret cost; which with five people is 25%, possibly adding in another dollar. #5 should propose that 25% go to two other bankers and the rest to himself.

SchrodingersFinch wrote:My reasoning is based on these assumptions:
Each banker values his own life more than anything. I assume they would choose the death of others over money, because you mentioned the thirst for blood before the lack of generosity, and said:

That "bloodthirsty" means "will kill if given the opportunity" -> yes

If there are two bankers banker #1 (the last one making the proposal) wins. He simply rejects the proposal of banker #2 who will then be assassinated, leaving all the money to banker #1.

If there are three bankers they know what will happen if the first proposal is rejected. That is why banker #2 must accept the proposal of banker #3. Banker #3 will of course take all the money.

If there are four bankers, #3 gets the best outcome if he rejects #4's proposal. Since #1 and #2 are so bloodthirsty, they'll also reject #4's proposal no matter how much he offers them. So #4 knows he's screwed.

Which leads to five bankers. #4 will want to live so he'll accept any proposal. #5 will now have to persuade only one banker to accept his proposal. But he can't do that because the bankers' thirst for blood is greater than their greed. Therefore #5 and #4 will die no matter what and #3 will get all the money.

If I'm wrong, could you say whether it's my reasoning or my assumptions.

On the other hand, if greed is worth more then killing, then #5 can buy off #1 and #2 for 2 cents each. They know they get nothing if #3 is reached, and #4 only has to give them 1 cent.

ArthurWilborn wrote:If the order of suggestions is not fixed but selected at random, then this becomes a different problem. #5 needs to buy two votes at regret cost; which with five people is 25%, possibly adding in another dollar. #5 should propose that 25% go to two other bankers and the rest to himself.

I actually think this variant of the problem is interesting... but I think we need assumptions on the gambling nature of the bankers.


And as your subsequent post indicated for the original variant... if we work with 1 cent increments of cash, then

Offer 1 xor 2 two cents and offer 3 one cent. Assuming Survival always wins over Greed and Killing & Assuming Greed always wins over Killing.

Here's a secret code:
2, 99027402891791685081075611082139, 103247, 83355839641

Hint:
20307105945254617

(You might want to use more than just a pen and paper.)

^Dude... is that really ALL we get to work with?

Each number is a word, so there are 5 words, including the hint.

Is it in another language that then translates into English?
[is that close?]


<----feels very intellectually lazy.

UrbanMasque wrote:Is it in another language that then translates into English?
[is that close?]


<----feels very intellectually lazy.

It does translate to English... I think.
a=2 -> z = 101 in order, first 26 primes.

"Hint" appears to be off and I am at a loss. It is evenly divisible by 23, 43, 71, (I, N, T) but not evenly by 19 (H)

The first word appears to be "a"
The third "is"
The forth "you"

The second appears to be evenly divisible by each letter a multitude of times.

Very good, you pretty much got it.

DeathofSpeech wrote:"Hint" appears to be off and I am at a loss. It is evenly divisible by 23, 43, 71, (I, N, T) but not evenly by 19 (H)

20307105945254617 is divisible by 19. It's 1068795049750243.

SchrodingersFinch wrote:Very good, you pretty much got it.

DeathofSpeech wrote:"Hint" appears to be off and I am at a loss. It is evenly divisible by 23, 43, 71, (I, N, T) but not evenly by 19 (H)

20307105945254617 is divisible by 19. It's 1068795049750243.

DOH... my bad, it is.

The first letter is found by the first iteration in which the remaining code is no longer divisible by the prime representing that letter...

so, 20307105945254617 is equal to 19^1*23^2*43^3*71^4

First word 2^1... a
Second word is a pain in my ass 83^1* 23^2 *43^3* 43^4* 11^5* 61^6... winner
Third word 23^1*67^2... is
Forth word 97^1*47^2*73^3... you

Just use WolframAlpha.

And congratulations!

SchrodingersFinch wrote:Just use WolframAlpha.

And congratulations!

ROFL... I finally hashed it out and you beat me to the congrats...