Puzzle 1: score 19. Edit: that was for score = sum of numbers used. The product score for the same solution is 198.
Puzzle 2: 1.0030 expected rolls.
Edit: Scott pointed out that the primes had to be below 2021. For this I have a solution with exactly 2 rolls.
Edit, no, I’m still wrong, there are 2022 people to choose among, not 2021.
My latest attempt gets 2.000000579 expected rolls.
Puzzle 3: 4 coins (and at most 14 flips).
Full solutions:
Puzzle 1:
211−33 scores 19.
Edit: for the problem as originally stated, i.e. the score is the sum of the numbers used. For score = product, the score is 198.
Puzzle 2:
Use a die with 2027 faces (the smallest prime above 2021). Roll to choose; if the result is above 2021 roll again. The expected number of rolls is 2027/2021 = 1.0030.
Edit: I missed the condition that the primes had to be below 2021. Since 2021=43×47, use one roll each of a 43-sided and a 47-sided die.
Edit, no, I’m still wrong, there are 2022 people to choose among, not 2021. So I don’t have a solution to puzzle 2 yet.
New attempt: I used some computational assistance in finding this solution. Roll one die of 1811 sides and one of 1907. The product of these is 3453577 = 1708*2022 + 1. In 3453576 out of 3453577 cases this gives you your choice, otherwise roll both again.
Use six coins, with probabilities 1⁄2, 1⁄3, 1⁄5, 1024/2021, 729⁄997, and 243⁄268.
Flip 1024/2021 to divide the people into groups of 1024 and 997. Choose from the 1024 group with 10 flips of 1⁄2.
For the 997 group, use the 729⁄997 to get groups of 729 and 268.
The 729 group can be chosen from with the 1⁄2 and 1⁄3 coins in at most 12 rolls. (Use 1⁄3 to cut off one third of the group, and 1⁄2 to split the remaining 2⁄3 into two thirds. Do this 6 times.)
For the 268 group, use the 243⁄268 to split it into groups of 243 and 25.
These can both be chosen from with the 1⁄2, 1⁄3, and 1⁄5 coins, the group of 243 with at most 10 flips, the group of 25 with at most 6.
In the worst case 14 flips are needed.
Better solution with five coins: 2000/2021, 1⁄2, 1⁄3, 1⁄5, and 4⁄7.
Use 2000/2021 to divide the group into 2000 and 21. The 1⁄2 and 1⁄5 will choose from 2000 in at most 13 rolls. Use 1⁄3 and 1⁄2 to divide the 21 into three groups of 7. Use the 4⁄7 to split 7 into 4 and 3, which can be chosen from with the 1⁄2 and 1⁄3.
Further improvement with four coins: 2000/2021, 1⁄2, 1⁄5, and 20⁄21.
Use 2000/2021 to divide the group into 2000 and 21. 2000 is as before, using the 1⁄2 and 1⁄5. For the group of 21, use 20⁄21, then the group of 20 can be solved with the 1⁄2 and 1⁄5.
Considering the way these solutions all work, I doubt if there is one with three coins along these lines. UnexpectedValues claims to do it with just one coin, so he must be taking a completely different approach. I want to think about that before looking at his solution.
Summary results (without derivations):
Puzzle 1: score 19. Edit: that was for score = sum of numbers used. The product score for the same solution is 198.
Puzzle 2: 1.0030 expected rolls.
Edit: Scott pointed out that the primes had to be below 2021. For this I have a solution with exactly 2 rolls.
Edit, no, I’m still wrong, there are 2022 people to choose among, not 2021.
My latest attempt gets 2.000000579 expected rolls.
Puzzle 3: 4 coins (and at most 14 flips).
Full solutions:
Puzzle 1:
211−33 scores 19.
Edit: for the problem as originally stated, i.e. the score is the sum of the numbers used. For score = product, the score is 198.
Puzzle 2:
Use a die with 2027 faces (the smallest prime above 2021). Roll to choose; if the result is above 2021 roll again. The expected number of rolls is 2027/2021 = 1.0030.
Edit: I missed the condition that the primes had to be below 2021. Since 2021=43×47, use one roll each of a 43-sided and a 47-sided die.
Edit, no, I’m still wrong, there are 2022 people to choose among, not 2021. So I don’t have a solution to puzzle 2 yet.
New attempt: I used some computational assistance in finding this solution. Roll one die of 1811 sides and one of 1907. The product of these is 3453577 = 1708*2022 + 1. In 3453576 out of 3453577 cases this gives you your choice, otherwise roll both again.
Expected rolls = 2*(3453577/3453576) = 2.000000579.
Puzzle 3:
Use six coins, with probabilities 1⁄2, 1⁄3, 1⁄5, 1024/2021, 729⁄997, and 243⁄268.
Flip 1024/2021 to divide the people into groups of 1024 and 997. Choose from the 1024 group with 10 flips of 1⁄2.
For the 997 group, use the 729⁄997 to get groups of 729 and 268.
The 729 group can be chosen from with the 1⁄2 and 1⁄3 coins in at most 12 rolls. (Use 1⁄3 to cut off one third of the group, and 1⁄2 to split the remaining 2⁄3 into two thirds. Do this 6 times.)
For the 268 group, use the 243⁄268 to split it into groups of 243 and 25.
These can both be chosen from with the 1⁄2, 1⁄3, and 1⁄5 coins, the group of 243 with at most 10 flips, the group of 25 with at most 6.
In the worst case 14 flips are needed.
Better solution with five coins: 2000/2021, 1⁄2, 1⁄3, 1⁄5, and 4⁄7.
Use 2000/2021 to divide the group into 2000 and 21. The 1⁄2 and 1⁄5 will choose from 2000 in at most 13 rolls. Use 1⁄3 and 1⁄2 to divide the 21 into three groups of 7. Use the 4⁄7 to split 7 into 4 and 3, which can be chosen from with the 1⁄2 and 1⁄3.
Further improvement with four coins: 2000/2021, 1⁄2, 1⁄5, and 20⁄21.
Use 2000/2021 to divide the group into 2000 and 21. 2000 is as before, using the 1⁄2 and 1⁄5. For the group of 21, use 20⁄21, then the group of 20 can be solved with the 1⁄2 and 1⁄5.
Considering the way these solutions all work, I doubt if there is one with three coins along these lines. UnexpectedValues claims to do it with just one coin, so he must be taking a completely different approach. I want to think about that before looking at his solution.
Puzzle 2 does not allow primes over 2021
I changed the scoring, so now 211−33 scores 198.