2 Really Hard Questions

[muchspl2]

no correct answers in this thread

[Chevy]

Pirates:

Vote 1= yourself
Vote 2= Give just one coin to pirate #1 will guarentee his vote,
Vote 3= Give just one coin to pirate #3 to get his vote (I think)


Races:

Play it as a knockout competition knocking out the last 2 each time
5 races of 5 runners leaving 15
3 races of 5 runners leaving 9
1 race of 5 runners and a race of 4 runners leaving 6
1 race of 5 runners (1 entrant gets a bye) leaving 3 and the bye
the final 4 determine the top 3
=12 (not to sure about this answer)

edit: pirates - pirate 1 is always the top pirate's guarenteed vote with one coin as if it gets down to 2 he'll get nothing. Pirate 3 can be bought with only one coin as if it gets down to 4 (he doesn't vote now) pirate 4 can just buy pirate 1 in the same way that you are.

[Snee]

But if the pirates are "maximally greedy" wont they keep voting against the highest numbered pirate until there are only two left at which point the highest numbered pirate left will take all the money, and vote for his own suggestion.

[sexfem]

The 5th pirate votes

5 = 32
4 = 32
3 = 32
2 = 2
1 = 2

Guaranteed carried vote. B)

[DanB]

Originally posted by sexfem@24 April 2004 - 21:50
The 5th pirate votes

5 = 32
4 = 32
3 = 32
2 = 2
1 = 2

Guaranteed carried vote. B)

Wouldnt they just fight and the winner ie the person still alive, take all?

[muchspl2]

Originally posted by muchspl2@24 April 2004 - 15:34
no correct answers in this thread

[Chevy]

Was sure I had the pirate one right (see the edit) but for the races one, instead of a cup style knockout tournament (takes 12 races), it could be a race of 5 repeated with the top 3 staying in and the bottom 2 being knocked out.

Race 1 would leave 23 left etc.., with 11 races needed to end up with a final 3.

[muchspl2]

nope, here are some hints:

1. regarding question 1 for those of you with no reading comprehension skills you are the 5th numbered pirate i.e. you decide the first allocation and you try to make the allocation such that the other 4 pirates wouldn't kill you for it, and you end up with the maximum number of gold coins.

please also remember all the pirates are logical, they don't want to die either and are intelligent.

2. regarding question 2 none of the answers given before this post is correct. btw if you mention an answer please show how you get to that number.

[hobbes]

ok, read the clarification. Have to think some more.

As the first distributor, I would give 50 to #4 and 50 to #3 and escape with my life.

If 4 does not vote for it, he will get 0 when he is chosen to distribute, the maximum 3 can get is 50 anyway, so he will vote for it.

1, 50, 49, 0, 0

y, y, n, n, n,-you die

then
50, 50, 0, 0

y, y, n, n

So the max that 3 and 4 can get is 50. You will either get none and live or be killed.

[Chevy]

If there are 2 pirates left #2 can just keep the money as he can give 50% of the vote

If there are 3 pirates left, #3 can just give 1 coin to pirate one to get the vote as if he's killed pirate one will get nothing (scenario above)

If their are 4 pirate left, #4 can give one coin to pirate number 2 as pirate number 2 will be aware of the scenaio above.

Solution, keep 98 coins, one coin to pirate 1 and one to pirate 3. (as my edit on the other page)