Zoom International Math League: Happy MAR10 Day! (Weekly Brain Potion Discussion)

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Happy MAR10 Day! (Weekly Brain Potion Discussion)

This week's brain potion is a version of a classic puzzle adapted to honor Mario Day (MAR10, i.e. March 10th).

The seven Koopalings (youngest to oldest) are named:

Larry
Morton
Wendy
Iggy
Roy
Lemmy
Ludwig

They were tired of working for Bowser without getting paid, so they decided to steal $100$ of Mario's coins and escape in their airship.

They agreed that they would divide the coins as follows:

Ludwig, as the oldest, would suggest a number of coins to give to everyone.
The others (not including Ludwig) would then vote on whether or not to accept the proposed distribution of the coins.
If $\geq 50\%$ of them vote to accept, then they give out the coins according to the disposal.
If $< 50\%$ of them vote to accept, they throw Ludwig from the airship and then repeat the procedure with the next oldest suggesting a distribution.

Assume that each of the Koopalings is rational, good at logical thinking, and, in order of importance,

Would not like to be thrown overboard.
Would like as many coins as possible.
Enjoys throwing someone else overboard.

How many coins can Ludwig get? Explain your reasoning and include how many coins each of the Koopalings will receive.

Please share any thoughts or questions you have below. We'll monitor the responses and give our thoughts as well!

You copied this from Wait but Why didn't you?

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Versions of this puzzle can be found in many places! But I haven't heard of Wait buy Why before. In fact, Wait but Why gives credit to 538's weekly puzzles in their related post :)

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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.

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You're assuming that the Koopalings would choose to throw each other off-board, knowing that they would get thrown if this strategy were to be enacted. Remember that they're perfectly logical! 1 coin is better than none, and none is better than being thrown off. Therefore, Ludwig could give anyone (who would otherwise get none) 1 and not get thrown off. We can start with your scenario of Wendy, Larry, and Morton. First of all, why would Morton vote yes? If Wendy is overthrown, Morton is the only one voting and keeps everything. Larry has to vote yes for Wendy because if not, Larry would be thrown off. Wendy can keep everything for herself. Iggy, realizing this, would offer 1 coin to Larry and Morton, more than otherwise. The vote is 2-1, meaning Iggy keeps 98. Continuing, Roy knows that Wendy might receive nothing. Roy would give 1 coin to Wendy and 2 to either Morton or Larry, not both, resulting in a 2-2 tie. Lemmy needs 3 votes, so he gives 1 to Iggy, etc...

This solution is not correct, but it's an example of how every Koopaling has to think of the future and the present.

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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.

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keep in mind that the proposer does not vote :l

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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

Iggy:2

Wendy:3

Roy:1

Lemmy:96

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

0-0-2-3-1-0-94

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Good discussion this week!

It seems like people have the right idea that the key to the problem is working backwards, but Ludwig can do even better than $94$ coins!

Starting by what happens with $1$ Koopaling (Larry takes everything!) and then $2$ Koopalings (Larry will vote no to whatever Morton suggests because he wants to take everything and throw Morton overboard). Note then for $3$ Koopalings Wendy can take everthing, because Morton knows that if it gets down to $2$ Koopalings he will be thrown overboard. We continue working our way up to $7$ Koopalings, giving us the proposals (that pass) in the table below.

Total Koopalings	Larry	Morton	Wendy	Iggy	Roy	Lemmy	Ludwig
1	100
2	100	0
3	0	0	100
4	1	1	0	98
5	0	2	1	0	97
6	1	0	2	1	0	96
7	2	1	0	0	1	0	96

For example, with $4$ Koopalings, Iggy needs $2$ votes, so he can "pay off" Larry and Morton with $1$ more coin than they will get if the vote fails. The rest of the table follows similar reasoning. (Note if there are $5$ Koopalings for example, it doesn't matter if Larry or Morton gets the $2$ coins, either will work and lead to the same final answer.)

I'm happy to explain any of the steps in more detail, just let me know!

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There's a version of this on TedEd. Did you get it from there?

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