nprofile1qy2hwumn8ghj7un9d3shjtnyd968gmewwp6kyqpq00fuygn2kuesukt4egm42rgde8x7a4g6kuuhx58q30x930uxdvqqp0vhyf (nprofile…vhyf) I can prove you need at least five checks. This is because you have six variable, and you already know obey one equation (that the probabilities sum to one). Each check gives you the additional equation that three of the variables sum to the same as the other three. So with fewer than five checks we would have an underdetermined linear system. And since we already have one solution (all probabilities equal) we would therefore have a family of solutions with dimension at least one. Any solution near the fair solution would also be a valid set of probabilities.
So we need at least five checks.
Furthermore, it's clear that some number of checks will eventually work. Any alternative set of probabilities will be ruled out by the check that compares the three largest probabilities to the three smallest. From this I think we can conclude that it can be done in exactly five. Just keep picking a check that's linearly independent from the information you already have.
Oscar Cunningham
npub1my…dtj8l
2026-01-18 12:36:56 UTC
in reply to nevent1q…mheg
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npub1my7d7x8gs4a96erlhhj9hn2lstz9qrdhex4hhuhgq9kqe5z06k6qldtj8lSeen on
wss://relay.ditto.pubPublished at
2026-01-18 12:36:56 UTCEvent JSON
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