Jason Weisman
@jlw12.bsky.social
Expected score with no mulligan: 19933/4096 miles
Expected score with mulligan: 21921/4096 miles
Expected score with mulligan: 21921/4096 miles
August 26, 2025 at 10:23 PM
Expected score with no mulligan: 19933/4096 miles
Expected score with mulligan: 21921/4096 miles
Expected score with mulligan: 21921/4096 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)."
August 21, 2025 at 5:30 AM
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.
August 20, 2025 at 6:27 PM
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.
August 5, 2025 at 11:46 AM
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...
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...
June 26, 2025 at 6:57 PM
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...
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.
June 24, 2025 at 10:34 PM
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.
#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...
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...
Google Colab
colab.research.google.com
June 24, 2025 at 3:08 PM
#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...
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...
length/side ratio = sqrt(1+sqrt(2))
November 24, 2024 at 11:20 PM
length/side ratio = sqrt(1+sqrt(2))
I posted a solution yesterday, but not really very "pretty."
November 17, 2024 at 2:39 PM
I posted a solution yesterday, but not really very "pretty."
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
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
November 16, 2024 at 11:05 PM
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
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