To win 6 out of 10 rounds, choose a stone opposite to the majority in the bag, and alternate your strategy after 5 rounds. For 7 out of 11, agree on a fixed sequence and adjust the 11th round based on prior rounds and the dealer’s choices.

For the second puzzle:

To find the murderer in the fewest gatherings, divide the 70 people into two groups and invite one after the other. If the witness remains silent in the first group, they are in the second; otherwise, question both groups after the first gathering Check Vine Density Calculator

My program is not given the question “are 6 out of 9 wins possible?”. Instead, it’s given the question, “how much do we need to constrain the dealer to make 6 out of 9 possible?”. In other words, “what is the maximum size of the family of binary sequences of length 9 that we allow the dealer to choose from in order to make scoring 6 out of 9 always possible?”. If the answer is 512 (all sequences), then a strategy exists, if the answer is smaller, then it doesn’t.

My program does not compute the exact answer, instead it computes an upper bound. This is because it doesn’t check that all the sequences in the family of sequences are distinct. Instead (for tractability), it just assumes that if we take a family of size A and union it with a family of size B, the union will have A + B elements (modulo some rudimentary checks for overlaps that are not comprehensive).

In this case, for this blog’s version of the “6 out of 9”, my program outputs that we need to restrict the dealer to at most 464 sequences out of the 512 possible. So unless there’s a bug in the reasoning or in the implementation, this should mean that the actual number of sequences is at most 464, which would imply that this blog’s version of “6 out of 9” has no solution. For the stack exchange version, my program says 512 out of 512 sequences are possible, which is consistent with the strategy for “6 out of 9” existing (which we know it does from other sources, but we can’t infer that from the output of my program alone).

]]>I should have referenced Sperner’s Theorem as part of the canonical numerical convenience above in Puzzle 2.

]]>While the difference is small, I believe it may be crucial for the “6 out of 9” case. For example, the condition “If you played the same value as the Casino, but Leader didn’t” is not testable in this blog’s version of the puzzle. Of course, there may be another strategy that still guarantees 6 out of 9, so I wrote a short C++ program that attempts to estimate the maximum number of wins that can be guaranteed for a given number of rounds. (The program is not perfect, as enumerating all possible strategies seems prohibitively expensive, but the results seem plausible. For example, by backtracking the computation, I was able to convince myself that a “3 out of 5” strategy actually exists.)

https://gist.github.com/zielaj/f130fe15d36340538fa0743c0bd4c351

Here are a few results. In each row, the first number is the number of rounds, the second the maximum number wins in the stackexchange version, and the third is the maximum number of wins in this blog’s version:

1 0 0

2 1 1

3 1 1

4 2 2

5 3 3

6 3 3

7 4 4

8 5 5

9 6 5 ***

10 6 6

11 7 7

12 8 8

13 9 8 ***

…

1000 806 768

The value for 1000 rounds also is also mildly encouraging because it is not that far from the asymptotic value derived in “Online Matching Pennies” (0.810710…). This also suggests that this blog’s version is asymptotically harder.

https://oeis.org/A320084

Still, at 1000 rounds I’d expect a better agreement than 0.806 ≈ 0.810710…, so it may be that my program misses some strategies.

]]>We will call each person into the office for 4 out of the 8 gatherings. Since 8C4 = 70, we can assign each person a different 4 gatherings. So the murder and the witness are called in the same number of times, but not on exactly the same occasions. So there must be one time when the witness is in and the murderer is not.

To show that it can’t be done in 7, note that since 70 > 2^6 there must be some pair of people who are called into the office on exactly the same gatherings out of the first 6. If these people are the witness and the murderer in some order, then one further gathering isn’t enough to deduce which is which.

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