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by furyofantares 1612 days ago
They specified that in the post, they were able to constrain to a strategy that guarantees 6 or fewer and then optimize for EV to get 3.55, but could get down to 3.52 by allowing some small number of words to take more than 6.

I would be interested to know if 6 is the lowest you can constrain it. Can you guarantee a solution in 5, and if so what is the best EV with that constraint?
1 comments
npinsker 1612 days ago
Yes, you can -- I believe the EV was around 3.47 for that. The "greedy" strategy (minimizing the worst-case size of the largest set) also ends up using at most 5 words.

Perhaps unsurprisingly, you can't guarantee at most 4 words.

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Sesse__ 1612 days ago
You can get EV of 3.46393 with a guaranteed maximum of five tries. (I don't know whether this bound is tight.)
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pxx 1612 days ago
This bound is not tight. http://sonorouschocolate.com/notes/index.php?title=The_best_...
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Sesse__ 1612 days ago
Good to know.

Another possible question is: What is the best strategy minimizing (at any point) _first_ the maximum number of tries, and second the EV (average number of tries over all possible words)? That's subtly different from either answer.

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