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I found Joel David Hamkins' riddle quite fun and wanted to construct one for a larger ordinal. I removed the fraction aspect since I think that would add too much difficulty to this version. However, I'm not 100% sure that the puzzle is unambiguous/correct as stated so I'm hoping people can point out if there's an error. I'll spoiler the answer in case anyone wants to think about it independently first.

Cheryl is throwing a birthday party, and as an excellent host decides to give each guest a gift. As a not so great host, her gift to each of them consists of one of the two following items:

  • A month of the year
  • A pair consisting of a month and a finite length, non-empty sequence of non-negative integers

Two gifts may be compared for "size" as follows:

  • If both gifts are just months, the later month is larger
  • If one gift is a month and the other is a pair, the month is larger
  • If both are pairs and one has a later month, the pair with the later month is larger
  • If both months are equal, sequences are compared first by length (longer sequences are bigger), then lexicographically

Albert and Bernard arrive and receive distinct gifts, which they each keep hidden.

Cheryl: Which of you do you think has the larger gift?

Albert: I don't know

Bernard: Neither do I

Albert: I still don't know

Bernard: And still either do I

Cheryl: Well, no matter how long you two go back and forth like that you'll never know who has the bigger gift

Albert: Despite that hint, I still don't know

Bernard: Neither do I

Albert: I still don't know

Bernard: And still neither do I

Cheryl: Let me call that last statement $S_0$, and say that no matter how many times we went back and forth with you two saying you don't know, and me saying $S_0$, you still wouldn't know who has the bigger gift

Albert: Despite that hint I still don't know, and also I find that subscript a bit ominous

Bernard: I also don't know

Cheryl: Call my last statement $S_1$, and now I will tell you that no matter how many times we alternate with you two saying you don't know, and myself saying $S_0$ or $S_1$, you still will not know who has the larger number

Albert: And yet I still don't know

Bernard: Me neither

Cheryl: Let me now define $S_n$ inductively to be the statement that no matter how many times we alternate with you two saying you don't know, and myself saying any $S_k$ for $k < n$, you two will still not know who has the larger gift. I now tell you that all statements $S_n$ are true.

Albert: Even still, I don't know who received the bigger gift

Bernard: Me neither

Albert: Ah I now know who has the bigger gift!

Bernard: Oh! Now I know as well, and also I know what gift you got

Albert: Then I know what gift you got as well

What gift did they each receive?

I believe that with this set of utterances, the solution is that

Albert received a pair of February, and a one element sequence: $[1]$, and Bernard received a pair of February and $[2]$

My intention is that the gifts are ordered with order type

$\omega^\omega \cdot 12 + 12$. The $+ 12$ is added as a confounder, since I think the obvious wrong answer would be that all pairs are ruled out so they both must each have only a month.

Edit: Let me also add my reasoning so that someone can comment directly on whether it is correct:

I believe that each "I don't know" rules out the lowest available final number of a finite-sequence for the speaker, so the initial I don't knows are stepping through (January, $[1]$), (January, $[2]$) etc. They also eliminate the month-only gifts in descending order.

Cheryl asserting the truth of $S_n$ increments the lowest possible $n$-th element of the sequence in a (January, sequence) pair by 1 (and $S_0$ rules out all month-only gifts). Therefore, since all $S_n$ are true, exactly the pairs with January as their month are ruled out, so we're left with the just the version on the natural numbers (the 1 element sequences)

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    $\begingroup$ Can the integers in a sequence be negative? $\endgroup$ Commented Feb 6, 2023 at 0:28
  • $\begingroup$ Ah great catch, they should be natural numbers let me edit that. $\endgroup$ Commented Feb 6, 2023 at 0:30

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