There are 7 choose 3 possibilities, so you can't do better than having up to 17 possibilities (actually 18, but that doesn't change the result) after the first guess, and up to 8 after the second guess, so I'd say no.
Right, but I'm thinking of those coin-weighing problems where you can extract information about other possibilities without guessing them. For instance, you can get info about nine coins by weighing 3 vs 3, splitting the candidates into three groups rather than getting just a split in half.
The extra credit reminds me a lot of this
https://jblblog.wordpress.com/2021/07/05/finding-the-magic-coins/
but in that case you're trying to find a set that contains exactly 2 cards from one group (rather than identify the whole group).
"when you thought about it rationally" har har har :-p
I try. :-)
Making the *Rounds* ....
"l'une" décembre
Hmm. I found (after the deadline) a method of finding the quartet with six steps. I'll be rather interested to see if a 5-step method exists.
There are 7 choose 3 possibilities, so you can't do better than having up to 17 possibilities (actually 18, but that doesn't change the result) after the first guess, and up to 8 after the second guess, so I'd say no.
Right, but I'm thinking of those coin-weighing problems where you can extract information about other possibilities without guessing them. For instance, you can get info about nine coins by weighing 3 vs 3, splitting the candidates into three groups rather than getting just a split in half.
In regards to this weeks connections, does getting all 4 from the same set count as a special case of only one left or is it excluded from that count.
For last week's Fiddler, I derived an expression for the probabilities p(k, n) of landing on the nth space when rolling k dice.
Let the solution set E(k) = {6^k = ((x^5 + x^4 + x^3 + x^2 + x + 1)/(x^6))^k}.
Then,
p(k, n) = 1/k sum_(x∈E(k)) ((x^5 + x^4 + x^3 + x^2 + x + 1) x^n)/(x (x (x (x (x+2)+3)+4)+5)+6).
https://i.imgur.com/uqiXur8.png
For example,
p(2, n) = 1/2 (2/7 + (0.157048 + 0.0734772 i) (0.610859 + 0.535522 i)^n + 1/7 (0.294195 + 0.668367 i)^n + (0.171101 + 0.0255587 i) (-0.0802299 + 0.718983 i)^n + 1/7 (-0.375695 + 0.570175 i)^n + (0.171851 + 0.00709332 i) (-0.613963 + 0.324952 i)^n + 1/7 (-0.670332)^n + (0.171851 - 0.00709332 i) (-0.613963 - 0.324952 i)^n + 1/7 (-0.375695 - 0.570175 i)^n + (0.171101 - 0.0255587 i) (-0.0802299 - 0.718983 i)^n + 1/7 (0.294195 - 0.668367 i)^n + (0.157048 - 0.0734772 i) (0.610859 - 0.535522 i)^n),
or using trigonometric function,
p(2, n) = 1/14 (2 + 2.42741×0.812363^n cos(0.719775 n + 0.43761) + 2×0.73025^n cos(1.15615 n) + 2.42199×0.723446^n cos(1.68192 n + 0.148281) + 2×0.682823^n cos(2.15341 n) + 2.40797×0.694654^n cos(2.6548 n + 0.0412525) + (-0.670332)^n).