#thisweeksfiddler
I hit on the below 6x6 prime magic square but unable to find an 8x8. Interested if anyone hit on an 8x8 or larger one...
337 367 307 421 241 353
293 331 383 271 317 431
389 281 347 359 439 211
401 167 313 349 647 149
233 311 17 487 379 599
373 569 659 139 3 283
The below arrangement showing level one circles - which seems to be a possibility based on the wording of the question - results in a lower average score (~3.66).
However, the problem description also seems allow alternative gasket geometries. For example, instead of a large circle centered on the y-axis directly above circles 1 and 2, what if there are two smaller circles (or a less symmetric arrangement of C3, C4, etc.).
I got the same answer with a much less elegant analysis. This aligns with the fractal geometry reflected in the illustration accompanying the problem.
docs.google.com/spreadsheets...
Expected score with no mulligan: 19933/4096 miles
Expected score with mulligan: 21921/4096 miles
#thisweeksfiddler
Miles Probability
3 17/256
3.5 32/256
4 24.75/256
4.5 72/256
5 9.1875/256
5.5 33/256
6 20.0625/256
6.5 48/256
Avg = 3*17/256+3.5*32/256+4*(24+3/4)/256+4.5*72/256+5*(9+3/16)/256+5.5*33/256+6*(20+1/16)/256+6.5*48/256
= 19933/4096 ~ 4.8665 miles
Yes, you're correct the $55 vouchers can be converted to cash if you continue to exactly hedge each bet until all the vouchers are gone. However, this can only be accomplished with certainty by betting cash. The online version of the question elaborated... "you can only wager vouchers (not cash)."
The Fiddler question asks how much $ is guaranteed. There is only a 50% probability of winning $55 all-or-nothing. On the other hand, betting $10 each on both red and black guarantees $10 cash. Second bet put $25 on red, $30 on black to be guaranteed at least $25 additional $ for total $35 minimum.
#ThisWeeksFiddler
EC: W= $90 cash with 50% likelihood from the outset.
First bet the entire $55 on one side. This has 50% probability of winning $55 cash while retaining the $55 of vouchers.
After winning use the same strategy as in Part 1 to add $35 or more cash guaranteed.
#ThisWeeksFiddler
colab.research.google.com/drive/1zobjk...
#thisweeksfiddler
B = 2+sqrt(2) ~ 3.414
colab.research.google.com/drive/1RaUbB...
Another try - not assuming the 2nd cut is along an axis perpendicular to the first...
colab.research.google.com/drive/1jWWlg...
This covers 99.9989% after 10 cuts with several remaining pockets left around the surface. I think 13 total cuts are needed.
colab.research.google.com/drive/12ZqvB...
Our solutions are the same for the first three cuts. But the hybrid monte carlo-determined cut results in only ~20.6% of the sphere remaining uncut versus ~21.1% after your 4th cut. Based on that my solution appears to be "greedier" even though it requires more total cuts to fully mow the sphere.
colab.research.google.com/drive/1ygJjT...
#ThisWeeksFiddler @xaqwg.bsky.social
I also got 12, but used different approach and also not very confident each cut is best... Tried a hybrid monte carlo approach to sample many different combinations of mow cuts and choosing the greediest for each one.
colab.research.google.com/drive/1z6Cly...
colab.research.google.com/drive/16IpyD...
colab.research.google.com/drive/1H9VYz...
#ThisWeeksFiddler
Arrange circles on a triangular lattice, each surrounded by six other circles.
Centers are integer linear combinations of: v1 =(1,0), v2 = (1/2, sqrt(3)/2).
Avg area contributed by each surrounded circle = sqrt(3)/2
For large N, min area of region is ~ sqrt(3)/2 N
length/side ratio = sqrt(1+sqrt(2))
I posted a solution yesterday, but not really very "pretty."
162
th = upper angle
h = height
x = base
Base (left triangle)
x = 4 + 4*tan(th)
Base (middle triangle)
x = sqrt(15)/cos(th) + sqrt(15)*sin(th)
sqrt(15)/cos(th) + sqrt(15)*sin(th) = 4 + 4*tan(th)
th ≈ 0.364864
h = 4 + 4/tan(th) ≈ 14.472
x = h*tan(th) ≈ 5.527
A = 1/2*h*x ≈ 40.000