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All of your stuffed animals are assembled, anxiously waiting to learn who will be be invited as the Guest of Honor to your next tea party.

This is to be an exclusive party, after all, as there is to be only one Guest of Honor. This is by necessity, because you have only two strawberry pastries, and you obviously intend to eat one of those yourself. (Cook was most annoyed with you for taking so many infinite bags of pastries yesterday, and has since cut off your supply). All the other guests at the tea party shall only receive tea.

So now you must decide which of your many guests will be the Guest of Honor, who receives the coveted strawberry pastry.

Your first instinct, of course, is to invite the valiant Mr. Bun to be the Guest of Honor. After all, he's brave and noble and always tells the truth. But then you think of poor insidious Duke Froggington II, who admittedly is dastardly and always lies, but in your heart you know that these character flaws are only due to his tragic childhood. You see, the perfidious Duke Froggington II grew up all alone in a dank 30-meter well, trying to leap his way to light and freedom. You know what that's like, and your heart aches for him. And there are similar reasons why you might choose almost any of the six dozen other stuffed animals that pack every available surface in your bedroom.

No, you find that you're completely unable to decide which stuffed animal should be the Guest of Honor at your tea party, and so you instead devise a game for the stuffed animals to play, to determine the Guest of Honor fairly.

The game works in this way:

First, you choose Mr. Bun as the initial nominee as Guest of Honor. He's the obvious first candidate, as the noblest of all your stuffed animals.

Then, starting with the nefarious Duke Froggington II and descending through the other stuffed animals in order of seniority and ending with Peanut (an ugly mouse given to you by your annoying older sister at your last birthday fête, and usually tucked far back in your dresser behind your latin coursebooks), each stuffed animal still in attendance is asked if they wish to declare themselves to be the new nominee as Guest of Honor. If they do, then they become the new nominee, and the previous nominee is returned to the toy chest, where they must remain until after the tea party.

Whenever the nominee changes, the whole process starts again; Every animal (in order of seniority, and starting with the most senior remaining animal) is given a fresh opportunity to nominate themselves, even if they had previously chosen not to. As before, if an animal chooses to nominate themselves, the current nominee is returned to the toy chest, and may not attend the tea party.

This process continues until no further stuffed animals wish to declare themselves as the nominee.

Once no more stuffed animals wish to make themselves the nominee, whoever remains as the nominee shall become the Guest of Honor and will receive a strawberry pastry at the tea party. The other remaining stuffed animals shall receive tea. Any stuffed animals in the toy chest get nothing. (At least, not until they're brought out again for the next tea party (which will likely be in ten minutes or so))

A few obvious points which you know instinctively are:

  • All stuffed animals love strawberry pastries (as does everyone of refinement and distinction. Peanut would probably love strawberry pastries too, despite having neither refinement nor distinction, though you've certainly never wasted one on her, even yesterday when you had several infinite supplies of them). This means that each stuffed animal dearly wishes to be the Guest of Honor.
  • All stuffed animals like tea, so none of them wish to completely miss the party by being put back into the toy chest.
  • All stuffed animals are perfect logicians (I mean, obviously. Do I even need to mention this?)
  • Stuffed animals don't hold grudges against one another; they will remain fast friends even if one sends another to the toy chest. On the contrary, they will recognise that it was the only logical thing to do, and will probably laugh about it at your next tea party, later this afternoon.
  • While the treacherous Duke Froggington II always lies, you've long since adapted to that, and aren't ever confused any more about what he means when he tells you which door leads to certain doom and which leads to your afternoon Latin classes (though you're still not convinced that there's actually much of a difference between the two). Similarly, there will be no confusion on your part about whether or not he wishes to become the new nominee, regardless of what lies he might say.
  • For the sake of clarity, you have six dozen and two stuffed animals in total. With Mr. Bun as your favourite, the despicable Duke Froggington II as your second favourite, and Peanut as your least favourite. Other stuffed animals may be identified by number, with the next animal after the treacherous Duke Froggington II being number 3.

Of your stuffed animals, how many will attend your tea party, and which will be the Guest of Honor?

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    $\begingroup$ Confession: I wrote this almost entirely as an excuse to use the line "All stuffed animals are perfect logicians", which is now probably my favourite line of prose of all time. $\endgroup$ – Trevor Powell May 17 '15 at 16:02
  • $\begingroup$ Similar feel to the pirates and gold coins puzzle. $\endgroup$ – Julian Rosen May 17 '15 at 17:15
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    $\begingroup$ leoll2 and I had different interpretations of the rules (see the comments below his answer). Can you clarify which interpretation you intended? $\endgroup$ – Julian Rosen May 17 '15 at 17:17
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    $\begingroup$ I honestly have no idea how I'm going to approach this puzzle, but +1 for the absolutely fantastic setup. $\endgroup$ – VictorHenry May 17 '15 at 21:34
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    $\begingroup$ Who shall come to the tea party? I will, of course! (Sorry.) $\endgroup$ – MadHatter May 19 '15 at 7:17
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All but Mr. Bun will attend the tea party, with Duke Froggington II as the guest of honor. The total number of animals who will be attending the party is $(6*12+2)-1=73$.

Looking at some specific cases:

  1. One animal left: he will definitely nominate himself.

  2. Two animals left: each animal knows that if he nominates himself, there will only be one left, and that one will definitely nominate himself. So neither will nominate himself.

  3. Three animals left: each animal knows that if he nominates himself, there will be two animals left, and neither will nominate himself. So the first animal to be asked will nominate himself.

  4. Four animals left: each animal knows that if he nominates himself, there will be three animals left, and one will nominate himself. So none will nominate himself.

So using backward induction, if there are $n$ animals left, the strategy taken will be:

  • If $n$ is even, no one will nominate himself.
  • If $n$ is odd, the first one asked will nominate himself.

When Duke Froggington II takes his turn, there are 73 animals left (including himself). As he is a perfect logician, he will nominate himself as the guest of honor. No one else will nominate himself after him. So poor Mr. Bun gets sent to the toy box, while everyone else enjoys the tea party and Duke Froggington II hogs the strawberry pastry.

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    $\begingroup$ Great answer. I now know how to respond if I ever get invited to a tea party with only two strawberry pastries in it. $\endgroup$ – Mark Gabriel May 18 '15 at 2:44
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I am not myself a perfect logician, but here the behaviour I observed at my own tea parties, and the explanation my own favourite nounours gave as years passed.

Party of one

This issue was quickly solved before I got my second plush. Nounours was the single invitee, and behaved as well as a bear does at a party.

Party of two

When Dr. Tortue joined us some time later, I observed the following:

  1. Nounours was nominated by default
  2. Dr. Tortue nominated himself, much to Nounours’s chagrin.

But Nounours, being the noble being that he is, quickly excused the good Doctor behaviour. “In her stead, I would have done the same thing: as there were no other candidate for Guest of Honour, she was certain to eat the delicious sugar pie1.”

Party of three

When Johnny McCanasson joined our little assembly, I expected Nounours and perhaps Dr. Tortue to be excluded from the table again. Nounours, however, quickly reassured me: “No, see, if either Dr. Tortue or Mr. McCanasson were to nominate themselves for Guest of Honour, they would be much in the same situation as I last time: the current nominee with only one candidate. This candidate, being sure to win the coveted position if they nominated themself, would without doubt do so. Thus, the first candidate to nominate themself would end up enjoying neither the pie nor the chocolate2. Even that dreaded McCanasson would rather enjoy your company than deprive both him and myself of it.”

Party of four

The next year saw the coming of the Lapin brothers, Mercredi and Dimanche. Unfortunately3, my younger brother requested Mr. McCanasson for his own company, and neither Mr. McCanasson4 nor I had the heart of refusing him.

I was expecting the party to go as it had previously, with Nounours remaining the Guest of Honour nominate throughout the process. It didn’t happen.

“See”, explained Nounours once again, “we are in the very same position as we were when only Dr. Tortue and myself were your guests. The first person to nominate themself for Guest of Honour would be in the same situation than previously, that is the nominee in a party of three. As we have seen last time, this is an excellent position, as it guaranties that you will win the honour and the pie that goes with it. The doctor, being the intelligent tortoise she always is, did not fail to notice, and didn’t let either of the Lapin brothers the opportunity to enjoy this position: she nominated herself as soon as she was offered the chance.”

Party of n, especially when n equals 74

Since then I observed the same behaviour each year: whenever there were an odd number of guests, Mr. Nounours, as first nominee, would remain guest of honour; when the guests were in even number, Dr. Tortue would nominate herself for Guest of Honour and eventually win the position, leaving the poor Mr. Nounours on the side.

As for what will happen with a party of 74, I have no idea as I not yet enjoy the pleasure of such a large company. Nounours, however, insists that Mr. Bun will be put aside as Duke Froggington II will take the seat next to yours, claiming that “the same patterns keep repeating” and using words such as “induction”. But, really, I can’t think of a reason why changing stove would change the issue.

I do however sometimes wonder what would happen if I were to cook additional pies…


  1. I just now notice that my proposed answer is for tea parties with sugar pies. I’ll post it anyways, but feel free to delete if necessary.
  2. Ah! Another possibly important detail.
  3. Although I can hear Nounours mumbling that he doesn’t mind so much.
  4. Don’t listen to Nounours, Mr. McCanasson is gruff, but very kind indeed. He often takes the time to visit.
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  • $\begingroup$ I love this answer! (It looks like you're missing the end of the Party of Four section, though; it just stops in mid-sentence) $\endgroup$ – Trevor Powell May 18 '15 at 1:30
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My answer is

All but Mr. Bun will attend the tea party and the Guest of Honor is Peanut

Reason:

Of course, Peanut declares that he wants to be the Guest of Honor (GoH), since he is the last and nobody can dethrone him.
Everybody else knows that asking to be the GoH will cause Peanut (or someone else) to claim the throne, resulting in a "no tea" situation.
Consequently, everybody except Peanut refuses the GoH.
Mr.Bun, who was forced to accept the throne, is bound to the toy box.

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    $\begingroup$ If I understand the setup correctly, when Peanut (or any other stuffed animal) nominates herself as GoH, the process repeats, so each of the remaining stuffed animals has the opportunity to dethrone Peanut. $\endgroup$ – Julian Rosen May 17 '15 at 16:50
  • $\begingroup$ @JulianRosen the problem says "If the nominee is changed in this way, the process is then repeated, with each stuffed animal still in attendance being offered (again, in order of seniority)". Peanut is the last in the order, so nobody can follow him. $\endgroup$ – leoll2 May 17 '15 at 16:51
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    $\begingroup$ Also, they have to answer "Yes" or "No" when they're asked, they cannot skip the question. Who has already answered cannot be asked again, else he could be incoherent. $\endgroup$ – leoll2 May 17 '15 at 16:53
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    $\begingroup$ Hmm...to me it sounded like a stuffed animal who has declined to declare themself the nomine will have another opportunity to do so if the nominee changes. Perhaps @TrevorPowell can clarify. $\endgroup$ – Julian Rosen May 17 '15 at 17:07
  • $\begingroup$ Apologies! My intention was that each time the nominee changes, every animal is given another opportunity to nominate themselves again, even if they had been asked before. (This was the paragraph beginning: "If the nominee is changed in this way, the process is then repeated") I'll think about how that rule can be clarified. $\endgroup$ – Trevor Powell May 18 '15 at 0:40
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Alternate Universe Solution:

If we ignore the 5th 'instinctive knowledge' statement about Duke Froggington (taking him at his word, rather than reading his intentions) we get a happier solution.

Setup:

Duke Froggington always lies and is a perfect logician. He knows that if he declares himself the guest of honor, he will partake of the strawberry pastry. (By mmKings correct logic, Duke Froggington will not be dethroned if he declares himself the Guest of Honor right away.) So when asked "Do you wish to declare yourself the guest of honor?" Duke Froggington must say "no," because the answer is "yes" and he always lies. The phrasing of the question means that he must always work against his own interests.

So here's what happens:

Duke Froggington responds "no." The other animals all know that Duke Froggington will dethrone the next guest of honor if and only if it results in he himself being dethroned at some future point. If one of the other animals declares itself the Guest of Honor, Duke Froggington will replace it, secure in his knowledge that the next animal will bump him off in turn (due to the odd/even math presented in mmKing's solution). Therefore no one will declare themselves the Guest of Honor first.

Answer:

Everyone enjoys the tea party, and Mr. Bun remains the guest of honor, thanks to Duke Froggington's perverse nature keeping the other animals at bay.

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  • $\begingroup$ This really does feel like the intended solution to me. I can see that someone else got the checkmark already, but can anyone tell me why this isn't the answer? $\endgroup$ – Solocutor Mar 8 '16 at 22:42
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    $\begingroup$ "While the treacherous Duke Froggington II always lies, you've long since adapted to that... there will be no confusion on your part about whether or not he wishes to become the new nominee, regardless of what lies he might say." $\endgroup$ – f'' Mar 8 '16 at 22:46
  • $\begingroup$ Thank you for the quick response. I guess I glossed over the end a little, and read that (incorrectly) as indicating that all the little perfect logicians will be able to understand the Duke's motivations and predict his lies. $\endgroup$ – Solocutor Mar 8 '16 at 23:13
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    $\begingroup$ The diabolical Duke Froggington II's "always lies" trait was a joking reference to the classic Knights & Knaves puzzle. (I particularly liked giving him the classic "frog in well" puzzle as backstory, as a traumatic experience which caused his Knave-like behaviour in later life). It's all just a bit of fun that's completely irrelevant to the logic of the puzzle, as per the line @f'' quoted. When stripped of its theme, this is actually a variant of the classic Lions & Sheep puzzle. $\endgroup$ – Trevor Powell Mar 8 '16 at 23:15
  • $\begingroup$ Thank you for the correction. I feel a little silly, but I'll leave it up as an alternate universe version where Duke Froggington's lies actually save the day for Mr. Bun. $\endgroup$ – Solocutor Mar 8 '16 at 23:16
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Clearly,

Peanut (or possibly Mr. Bun)

will be the sole member of your tea party.

Here's why:

All 74 attendants of the party should know that you have access to infinite bags of pastries. So, they would ask if you have more than 1 to spare. When you explain that you do not, then the pastry becomes a precious commodity, while the chance at a tea party comes multiple times in a day.

Faced with this realization, each animal would take the risk at the pastry, knowing that, while they may miss this party, there will be another later that day. Since you enraged the cook, who knows how long it will take to get another delicious pastry.

Peanut, being the last in line, is unable to be overthrown, unless you give Mr. Bun the chance to overthrow him. You see, Mr. Bun did not have a choice in his being given the pastry or not, while there is no one to come after Peanut. The only fair solution to this would be to give Mr. Bun a chance to recover the pastry from Peanut.

Ok, so maybe that last bit is a copout, but have you seen Peanut? That guy is just creepy.

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    $\begingroup$ "Would take the risk" is an irrational strategy for a logician. Also, the OP explicitly says "none of them wish to completely miss the party" $\endgroup$ – leoll2 May 17 '15 at 18:26
  • $\begingroup$ @leoll2 While it's true none of them want to completely miss the party, once they realize that they have ready access to parties but only one chance at a strawberry tart, the tart becomes a much more valuable commodity. Couple that with the fact that the tart is of higher base value than tea, the logical approach would be to try and be the person who gets the most valuable commodity. $\endgroup$ – tfitzger May 19 '15 at 13:09
  • $\begingroup$ Also, if all the animals, save the least favorite, are in the box, they can have their own party and socialize, and then get a second party with tea. Win Win. $\endgroup$ – tfitzger May 19 '15 at 13:10

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