Pirates

[bugzpodder]

Two pirates A and B stole 100 gold coins. They split it as follows: A takes a pile of n coins (left over from the undistributed coins) and B chooses who gets the pile. If there are still coins left over (not distributed) after either A or B gets nine piles, then whoever did not get nine piles take the leftover coins. suppose A and B do this in an optimum ways (such as to maximize their profits), what is the number of coins A gets? justify.

[sql_lall]

OK, so the first time, A picks the pile, and B decides who gets it.
- ***** -
Then, once one of them gets 9 piles, the other gets all the rest.

My question is, what happens in the -*****- bit??

Obviously, after the initial pile is distributed, anywhere between 8 and 16 more piles need to be given out, so what happens for these? Do A&B alternate picking&distributing, or is it always A?



Underpants gnomes:
Step 1: steal underpants
Step 2: ?
Step 3: $$$ Profit $$$
Now all we need is a step 2.

[DiGiTaIErRoR]

I don't know pirate A. How would I know how many coins he got?

But I'd bet it's 50 or less(if A's stupid), but definately more than 8. Maybe A gets more than 50... if B is stupid.

B should take the pile if it has 6 or more coins, or if the sum the coins in the piles he has is greater than number of piles * 5.5.

This insures B gets 50 or more coins.

[bugzpodder]

Originally posted by sql_lall
OK, so the first time, A picks the pile, and B decides who gets it.
- ***** -
Then, once one of them gets 9 piles, the other gets all the rest.

My question is, what happens in the -*****- bit??

Obviously, after the initial pile is distributed, anywhere between 8 and 16 more piles need to be given out, so what happens for these? Do A&B alternate picking&distributing, or is it always A?

they repeat the action of A choosing a pile, and B gets to decide who gets it, until one of them gets nine piles and the other gets all thats left over -- or until the coins are all distributed before somebody gets nine piles, whichever comes first.

DE: if you were make that bet, you'd lose horribly

[DiGiTaIErRoR]

I justified my answer.

Perhaps I just don't share your capitalist perspective.

[bugzpodder]

your justification was again, BS

[sql_lall]

No need for that here.
Stop you bickering

DE is right though, if B always assigns groups <=5 to A, and >=6 to B, then they get more than 50.

If B gets 9 groups first, they have 54, A gets 46. If A gets 9 groups first, they have 45, and B gets 55. So the best A can hope for is 46, by offering 6 coins each time.

However, i guess the question is optimal, so i'm not sure exactly how that effects DE's reasoning. I don't think A can expect to get any more than 46. If they do, which means B has 9 groups and <54 coins, at least one of these groups must have been <6, in which case those <6 coins could have been given to A, and B would have gotten extra at the end.

[DiGiTaIErRoR]

Sql:

Basically my reasoning would produce:

offer01: 5 A gets it because it's < 5.6
offer02: 6 B gets it because 6 > 5.6
offer03: 5 A gets it because the would-be sum of B's coin, 11, would be less than 11.1(piles * 5.555...)
offer04: 6 B gets it, because 12 > 11.1
offer05: 5 B gets it, because 17 > 16.7
offer06: 6 B gets it, because 23 > 22.2
offer07: 5 B gets it, because 28 > 27.8
offer08: 6 B gets it, because 34 > 33.3
offer09: 5 B gets it, because 39 > 38.8
offer10: 6 B gets it, 45 > 44.4
offer11: 5 B gets it, and they have split it 50/50

Deviation from this would most likely result in A coming out short.

As ultimately B has the power to make himself more money, but only if A allows it (A can only screw himself).

[sql_lall]

I think you can guarentee B getting at least 54, as suggested, splitting groups <=5 to A, >=6 to B

[DiGiTaIErRoR]

Originally posted by sql_lall
I think you can guarentee B getting at least 54, as suggested, splitting groups <=5 to A, >=6 to B

Yup.