To decide who will do the dishes, we put in a jar tickets with prime numbers up to 111. The one who draws the ticket with a larger number wins.
I look at my ticket: 3. Maybe you have a low number like 11... So I offer to exchange our tickets before showing them.
What is the probability I will do the dishes?

I get 13.
Take one coin of each chest.
a) You get two Gold, one Silver:
Silver is identified, remains to distinguish all_Gold and Gold_Silver.
Take 10 more coins from the first G:
- if they are all gold, you identified all_Gold, the remaining is Gold_Silver
- if you get a silver coin, you identified Gold_Silver, the remaining is all_Gold
b) You get one Gold, two Silver:
Same proceeding, permute Gold/Silver.

There is a strip of N (N>13) squares. In the middle is the SILVER DOLLAR.
Two players alternatively place a penny on an empty square.
Then, at each turn, a player must:
- move one coin one or more squares to the left, observing:
- the coin cannot move out of the strip
- the coin cannot jump on another coin
- the coin cannot jump over another coin
- or -
- pocket the leftmost coin
Best strategy?

@Plasmid I translated your notation into mine, as far as we are gone, we have same results (excepted some doublets).
I thought about interpolation, too, but I did not go this way estimating there is not enough data.

Thanks for finding the problem. I was so sure that if X contradicts Y, they must be in different categories - it did not occur to me that can be both liars.

Do not worry, on my first attempt, I did not manage it to write it clearly enough that I myself could read and understand it. When you asked your question, I checked my solution written about a year ago and wondered whether it would not be easier to start from the beginning. I have rewritten it and found a kind of notation:
P.S. Can someone change in the title ONE in ONCE?
Thanks in advance.

On an island, every statement is true if the islander is aged less than L and false if he is at least L years old. Find their ages.
[1] A: "B is more than 20 years old."
[2] B: "C is more than 18 years old."
[3] C: "D is less than 22 years old."
[4] D: "E is not 17 years old."
[5] E: "A is more than 21 years old."
[6] A: "D is more than 16 years old."
[7] B: "E is less than 20 years old."
[8] C: "A is 19 years old."
[9] D: "B is 20 years old."
[A] E: "C is less than 18 years old."

There be sixty-and-four flowers-de-luce (in a grid 8x8), and the riddle is to show how I may remove six of these so that there may yet be an even number of the flowers in every row and every column.
I am not able to remove 6 of them: interactive version
What am I missing?
Solution. that does not help me.
I got it now, to late to delete.