The CubeCast 2.0 - Montage Released!

penguinz7 said:
Awesome! The fact that I don't even know what RSS is(Programming Language?), means I probably can't help.
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Really Simple Syndication. You publish something, the RSS "feed" updates automatically, and RSS "readers" check the feed every so often. If there's an update, they notify you. It's a really fantastic way of seeing new posts without having to go check all the time.

These are the things I know of RSS.
 
The puzzler was harder than I expected.

I had to use the hint that you're asking for "the" optimal split, meaning there's just one. That clarifies the ambiguity of whether they have to use full dollars or can use cents (must thus be cents) as well as what to do in case of uncertainties. If someone knows they'll get either 0¢ or 2¢ if they vote against me, clearly I can buy their vote by offering them 2¢ (they can't get more and it beats the expected value), but 1¢ might be enough (matches the expected value and beats the minimum). Fortunately, using the former rule leads to many optimal splits (John gives 1¢ to the third-youngest and 2¢ to each of three of the five oldest and keeps the rest) while using the latter rule leads to just one, so that must be it.

John gives 1¢ each to the third-youngest, the fifth-youngest, and the two oldest, and keeps the rest.
 
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Stefan said:
The puzzler was harder than I expected.

I had to use the hint that you're asking for "the" optimal split, meaning there's just one. That clarifies the ambiguity of whether they have to use full dollars or can use cents (must thus be cents) as well as what to do in case of uncertainties. If someone knows they'll get either 0¢ or 2¢ if they vote against me, clearly I can buy their vote by offering them 2¢ (they can't get more and it beats the expected value), but 1¢ might be enough (matches the expected value and beats the minimum). Fortunately, using the former rule leads to many optimal splits (John gives 1¢ to the third-youngest and 2¢ to each of three of the five oldest and keeps the rest) while using the latter rule leads to just one, so that must be it.

John gives 1¢ each to the third-youngest, the fifth-youngest, and the two oldest, and keeps the rest.
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There are actually two correct answers, after doing the simulation again. This isn't quite right.
 
Puzzler Solution

So do the other 7 people vote? So you need 4/7 to be majority? Or what?

EDIT:

After about an hour, this is my solution, assuming that John votes and you need 4/8 votes to win and you can use cents:

Note: Shafted = pay 0 cents.

Suppose that the robbers are named by letters, with the oldest robber in each scenario being Robber A. I'll list the robbers in order of youngest to oldest for each case:

In a two-robber case (John, Robber A), John always keeps the cash. Robber A has no leverage whatsoever, since he'll never have superior force to remove the captain. So, we would say here that Robber A would be shafted because John could win optimally by giving him no money whatsoever.

In a three-robber case (John, Robber B, Robber A), John only needs one vote. Robber B wants to get rid of John so he can be in the advantageous two-robber case. Robber A wants to AVOID that, so the John easily buys Robber A's vote with one cent -- the status quo is maintained, and Robber B gets shafted.

In a four-robber case (John, Robber C, Robber B, Robber A), John still only needs one vote (a tie means no superior force, which means the status quo is maintained and John wins). We again will ask our standard question: "If John loses the vote, then who gets shafted?" -- well, if the captain loses, it's a three-robber case, which means Robber B gets shafted. So the John easily buys Robber B's vote with one gold coin, and the status quo is maintained. Result: Robbers A and C would be shafted, B would be given one cent, and John would get the rest.

In a five-robber case (John, Robber D, Robber C, Robber B, Robber A), John needs TWO votes to avoid losing power. So we ask ourselves, once again: "If John loses the vote, then who gets shafted?" -- Robbers A and C (as determined above, in the four-robber case). So, you’d want to pay Robbers A and C one cent each to make them vote for you. Therefore, robbers B and D would be shafted.

In a six-robber case (John, Robber E, Robber D, Robber C, Robber B, Robber A), John needs two votes to avoid death. So, we ask, "If John loses the vote, then who gets shafted?". Well, as determined above, it would be robbers B and D shafted. So, you can easily win their vote by giving them just one cent. Therefore, Robbers A, C, and E would be shafted.

In a seven-robber case (John, Robber F, Robber E, Robber D, Robber C, Robber B, Robber A), John needs THREE votes to avoid death. So, we ask, “If John loses the vote, then who gets shafted?”. As determined above, A, C, and E would get shafted. So, you’d want to pay A, C, and E one cent each in order to get your vote. Therefore, robbers B, D, and F would not be paid at all, or in other words would be shafted.

In the desired eight-robber case (John, Robber G, Robber F, Robber E, Robber D, Robber C, Robber B, Robber A), John needs three votes, yet again, to avoid death. (Reminder: This is because he will obviously vote for it, making it (1+3)/8 = 4/8, and as described earlier, a tie means no superior force, which means the status quo is maintained and John wins.) So, we would then ask, “If John loses the vote, then who gets shafted?”. As determined above, robbers B, D, and F would be shafted if he lost the vote. So, you can easily get their vote by simply paying them one cent each and John can keep the rest. Therefore: Robbers A (the oldest), C, E, and G would be shafted; robbers B, D, and F would each get one cent each; and John would keep the rest.

So:
John: $7,999,999.97
Next youngest: $0.00
Next youngest: $0.01
Next youngest: $0.00
Next youngest: $0.01
Next youngest: $0.00
Next youngest: $0.01
Oldest: $0.00
 
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You can use cents.

All eight team members vote.

A majority of votes in the proposer's favour is required to enact the proposer's split.

An optimal split is one that guarantees John Johnson keeps his life while John takes as much money as possible.
 
Ranzha said:
You can use cents.

All eight team members vote.

A majority of votes in the proposer's favour is required to enact the proposer's split.

An optimal split is one that guarantees John Johnson keeps his life while John takes as much money as possible.
Click to expand...

So does that mean that you need 5 out of 8 to vote in favor to enact the proposer's split? Because I have been assuming 4 out of 8 need to vote in favor to enact the proposers split... whoops.
 
A is oldest, down to H, Johnny.
Each will vote yes if they will get something better than what they might get if they vote no.
Number denotes how many cents they get. 'X' denotes everything else after the cents are taken away.

If 1 person: AX

If 2 people: AX B0
A will vote no unless he gets everything.

If 3 people: A0 B1 CX
B votes yes or he gets nothing.

If 4 people: A1 B2 C0 DX
A votes yes or he gets nothing. B votes yes or he gets 1.

If 5 people: A2 B0 C1 D0 EX
A votes yes or he gets 1. C votes yes or he gets nothing.

If 6 people: A0 B1 C2 D1 E0 FX
B votes yes or he gets nothing. C votes yes or he gets 1. D votes yes or he gets nothing. Only one best case for F.

If 7 people: A1 B2 C0 D0 E1 F0 GX
A votes yes or he gets nothing. B votes yes or he gets 1. E votes yes or he gets nothing. Note that G can choose either B or D to get 2 and still get majority vote.

If 8 people: A2 B0 C1 D0 E2 F1 G0 HX
A votes yes or he gets 1. C votes yes or he gets nothing. E votes yes or he gets 1. F votes yes or he gets nothing.

This is probably better done in a chart but this should explain it well enough I hope.
So two cents go to A and E each, 1 cent to C and F each, and Johnny gets $7,999,999.94.
 
sneaklyfox said:
A is oldest, down to H, Johnny.
Each will vote yes if they will get something better than what they might get if they vote no.
Number denotes how many cents they get. 'X' denotes everything else after the cents are taken away.

If 1 person: AX

If 2 people: AX B0
A will vote no unless he gets everything.

If 3 people: A0 B1 CX
B votes yes or he gets nothing.

If 4 people: A1 B2 C0 DX
A votes yes or he gets nothing. B votes yes or he gets 1.

If 5 people: A2 B0 C1 D0 EX
A votes yes or he gets 1. C votes yes or he gets nothing.

If 6 people: A0 B1 C2 D1 E0 FX
B votes yes or he gets nothing. C votes yes or he gets 1. D votes yes or he gets nothing. Only one best case for F.

If 7 people: A1 B2 C0 D0 E1 F0 GX
A votes yes or he gets nothing. B votes yes or he gets 1. E votes yes or he gets nothing. Note that G can choose either B or D to get 2 and still get majority vote.

If 8 people: A2 B0 C1 D0 E2 F1 G0 HX
A votes yes or he gets 1. C votes yes or he gets nothing. E votes yes or he gets 1. F votes yes or he gets nothing.

This is probably better done in a chart but this should explain it well enough I hope.
So two cents go to A and E each, 1 cent to C and F each, and Johnny gets $7,999,999.94.
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Ding ding ding, a winner is you!
This was actually the only correct answer; I dun goof'd with the second answer I tried.

PM'd you.
 
Ranzha said:
Ding ding ding, a winner is you!
Click to expand...

Her answer is wrong.

sneaklyfox said:
A is oldest, down to H, Johnny.
Each will vote yes if they will get something better than what they might get if they vote no.
Number denotes how many cents they get. 'X' denotes everything else after the cents are taken away.

If 1 person: AX

If 2 people: AX B0
A will vote no unless he gets everything.
Click to expand...

That's the split that B would propose, but not necessarily the outcome. As B is not offering A more than A would get otherwise, A might vote "no" and B might get killed.

sneaklyfox said:
If 3 people: A0 B1 CX
B votes yes or he gets nothing.[/spoiler]
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C doesn't need to offer 1 to B. Offering 0 is enough, as "0" is better than "0 or death".

Her end result is wrong as well.
 
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The only wrong thing is that actually any two of A, B, D, or E could be offered 2 cents and still vote yes because B or D would get 2 or 0 if Johnny dies so they probably wouldn't take a chance on getting 0.

As for A voting no, causing B's unnecessary death, if this is the kind of premise we have in the problem then it's unsolvable because every odd-oldest person will vote no and every even-oldest person will vote yes. Then H (Johnny) will die and he cannot guarantee his life, even if he offers everything to G, G can still choose to vote no.

In other words, I work on the premise that the vote is yes if you get better than what you expect to get or if you are offered everything. This is his safest bet.
 
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sneaklyfox said:
The only wrong thing is that actually any two of A, B, D, or E could be offered 2 cents and still vote yes because B or D would get 2 or 0 if Johnny dies so they probably wouldn't take a chance on getting 0.
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That's wrong as well, yes. Didn't bother mentioning it because it was irrelevant.

sneaklyfox said:
As for A voting no, causing B's unnecessary death, if this is the kind of premise we have in the problem then it's unsolvable because every odd-oldest person will vote no and every even-oldest person will vote yes. Then H (Johnny) will die and he cannot guarantee his life, even if he offers everything to G, G can still choose to vote no.
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No, it does work. I suggest you try it. Two persons is the only unstable situation.

You yourself btw already did use this everywhere else, by offering people 1 more than they'd get otherwise because you knew that offering them the same wouldn't guarantee their yes.
 
Stefan said:
You yourself btw already did use this everywhere else, by offering people 1 more than they'd get otherwise because you knew that offering them the same wouldn't guarantee their yes.
Click to expand...

Yes, I know this. Offering the same would not guarantee their yes. Offering more does guarantee a yes. We have to assume though that whoever is offered everything will say yes, or Johnny will never live (as I said above). Since we assume that there IS a solution this must be the case or else his safest bet.
 
I did it again now based on the uncertainty that B will live.
A is oldest, down to H, Johnny.
Number denotes how many cents they get. 'X' denotes everything else after the cents are taken away.

On the assumption that there is no guarantee if one is offered the same amount...
If 1 person: AX

If 2 people: AX B0
This is B's safest bet though there is no guarantee he will live.

If 3 people: A0 B0 CX
B votes yes or he risks losing his life at the next turn.

If 4 people: A1 B1 C0 DX
A votes yes or he gets nothing. B votes yes or he gets nothing.

If 5 people: A2 B0 C1 D0 EX
2 can be offered to either A or B who votes yes or gets nothing. C votes yes or he gets nothing.

If 6 people: A2 B0 C2 D1 E0 FX
2 must be offered to either A or B who votes yes or risk getting less. C votes yes or he gets 1. D votes yes or he gets nothing.

If 7 people: A2 B0 C0 D2 E1 F0 GX
2 must be offered to either A or B who votes yes or risk getting less. D votes yes or he gets 1. E votes yes or he gets nothing.

If 8 people: A2 B0 C1 D0 E2 F1 G0 HX
2 must be offered to either A or B who votes yes or risk getting less. C votes yes or he gets nothing. E votes yes or he gets 1. F votes yes or he gets nothing.

This is almost the same answer as before except your options are limited. You give 1 cent to C and F each, 2 cents to E, and 2 cents to either A or B. This will satisfy either assumption. Johnny still keeps $7,999,999.94 for himself.

Edit: So my original answer was still correct.
Edit2: And I still think it's more reasonable to assume that your vote is yes if you are offered everything. No unnecessary loss of life, you know.
 
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