Happy MAR10 Day! (Weekly Brain Potion Discussion)
I don’t think Ludwig can give any number of coins that would not result in his being thrown overboard, and here’s why.
Ludwig would have to play in favor of everyone else because priority states that he first would not want to be thrown overboard. So let’s take the most extreme case: he gives all 100 coins to the other six Koopalings while he gets nothing. Therefore, everyone would have an average of 100/6 or approximately 14.28 coins. However, why would most of the other Koopalings accept this value when they could just throw Ludwig overboard and instead have an average of 100/5 or 20 coins when the next Koopaling does the distributing. The only person that would vote in favor of this proposal is Lemmy because he would be the one to distribute the coins next, ending with potentially nothing, and by a vote of 1 in favor, 5 against, this motion fails, and Ludwig is thrown overboard.
This pattern would continue on: Lemmy gives all coins to the other Koopalings, he gets thrown overboard by a vote of 1 for, 4 against, then Roy is thrown overboard by a vote of 1 for, 3 against, then Iggy: 1 for, 2 against.
Then, Wendy would be the one to distribute. Unlike the previous Koopalings, a single vote of this distribution will pass the proposal. In addition, she can keep more coins because Morton has to vote yes or else he gets nothing when he divides up the coins after Wendy is thrown overboard. Specifically, Wendy can keep up to 98 coins (because I don’t think that the coins are specifically given to people. If so, the number goes up to 99). The remaining 2 coins would be divided between Larry and Morton, with Larry voting no, and Morton having to vote yes to avoid getting nothing. So the final distribution would be Wendy: 98 coins, Larry: 1 coin, Morton: 1 coin.
This solution is not correct, but it's an example of how every Koopaling has to think of the future and the present.
Very interesting rebuttal. And upon further inspection, I think there is a way to prove that Ludwig’s initial distribution of 94-1-1-1-1-1-1 will not fail.
Going to work backwards for this one. Obviously when we have just Larry and Morton, Larry will always get all 100 coins because he can just veto any other planned distribution by Morton. When we throw in Wendy, she can just give one coin to everyone else, so Morton will have to vote yes to prevent the lock situation while Larry votes no to encourage the lock, so the motion passes 1-1. Throwing in Iggy this time, Wendy will want the situation described above, so she will vote yes, and the others vote as per Wendy’s situation, so the vote fails, 1-2. When Roy distributes, Iggy knows the next situation will fail, so Iggy will vote yes, allowing his distribution to pass 2-2.
Now I could just continue on the pattern and say that Ludwig’s initial motion will pass with a vote of 3-3, allowing that distribution to hold true, but I want to see if I can prove a general case. The proof will come.... in another post.
Ludwig can have at max 94 coins
this is why:
if all of the Koppalings aside from Larry and Morton have been thrown off, then Morton will be thrown off by Larry no matter what he does. when the youngest 3 are alive, Morton will vote for Wendy no matter what, so Wendy gets 100. Meanwhile, when 4 are alive, Morton will take 1 coin, and Larry will also be fine with 1 coin, so Iggy would keep 98 coins. When 5 are alive, Roy will need to bribe 2 of the other koppalings, 1 for Wendy, and 2 for either Morton or Larry. When only Ludwig is gone, Lemmy needs to bribe 3 of them. Iggy will take 1, Wendy takes 2, and 3 for Morton and Larry, I think 2 coins would work, as if the one that is chosen doesn't accept, then he might get 2 coins, or he get 0, so 2 should be worth it for him. However, if he worked out a deal with Roy to give him 2 coins should he throw off Lemmy, then Lemmy start a bidding war on who will survive, I'm going to go with case 1, as in case 2, Roy has a total of 99, and Lemmy has a total of 97. However, Roy could trick Morton/Larry, and therefore they would most likely just accept the 2 coins. Now so Ludwig knows this, and so he needs to bribe 3 people. Here are the costs for bribing each person
Larry:? 3 or 1
Morton:? 3 or 1
Larry and Morton have questions marks because of the previous situation with a bidding war, but since they can either get 0 or 2 in the other situation, they might only accept 3, in which case Wendy would be a safer bet, at the same cost. 100-2-3-1=94
his proposition would be