Tom Keith @tomkeith.bsky.social · 07/10/2025#ThisWeeksFiddler @xaqwg.bsky.social: Anita the Ant To figure out what was going on, I made an animation. I found that Anita the Ant's path ended up inscribing a 2-3-2 isosceles triangle. 021
Tom Keith @tomkeith.bsky.social · 23/09/2025For a large-enough number of n participants, the average longest increasing subsequence is roughly proportional to the square root of n. 010
Tom Keith @tomkeith.bsky.social · 23/09/2025#ThisWeeksFiddler @xaqwg.bsky.social Here are the average lengths of the longest increasing subsequences of all the permutations of n elements. (The numerators in the table come from oeis.org/A003316.) 121
Tom Keith @tomkeith.bsky.social · 27/08/2025#ThisWeeksFiddler @xaqwg.bsky.social This chart shows how the number of mulligans you have affects your expected score. 031
Tom Keith @tomkeith.bsky.social · 19/08/2025#ThisWeeksFiddler @xaqwg.bsky.social: This chart shows the probability of being able to achieve various dollar amounts of cash if you start with three $10 vouchers and one $25 voucher and play optimally for each amount. 010
Tom Keith @tomkeith.bsky.social · 15/07/2025🤦 I can't subtract! 299 - 119 is 180, the same as you got. 100
Tom Keith @tomkeith.bsky.social · 15/07/2025#ThisWeeksFiddler @xaqwg.bsky.social I wrote code to emulate playing the game of bowling. I found 420,571 possible states in the game including 25,646 terminal states. With this list in hand, it was a matter of finding the terminal states that best qualified in each case. 221
Tom Keith @tomkeith.bsky.social · 08/07/2025Extra credit: How many different parallelograms in the field of stars on the US flag? I found 324 different shapes of parallelogram. The smallest appeared at 124 different locations and orientations. 000
Tom Keith @tomkeith.bsky.social · 08/07/2025#ThisWeeksFiddler @xaqwg.bsky.social How many different equilateral triangles in the game of Dozo? I found 11 different sizes of triangle. The smallest appeared at 36 different locations and orientations on the Dozo board. The largest appeared just once. 121
Tom Keith @tomkeith.bsky.social · 02/07/2025I was pleased to see there were more than zero but few enough to be listed. 000
Tom Keith @tomkeith.bsky.social · 01/07/2025#ThisWeeksFiddler @xaqwg.bsky.social EEC: There is a faint message scrawled onto the second scroll which you can barely make out: "Nullus numerus repetitur", i.e., "No number is repeated". How many combinations are possible if none of the eight numerical inputs is used more than once? 120
Tom Keith @tomkeith.bsky.social · 25/06/2025I think I have a greedier cut for the second borehole. It removes 22% of the original sphere. That's going to affect which are the greediest boreholes after that. Here are my first three cuts. 110
Tom Keith @tomkeith.bsky.social · 24/06/2025#ThisWeeksFiddler @xaqwg.bsky.social Here is "a" solution to the extra credit. Very likely not "the" solution because I could only approximately calculate the volume remaining and couldn't figure out a way to find the greediest cylinder at each step. I obliterated the sphere in 12 steps. YMMV. 221
Tom Keith @tomkeith.bsky.social · 27/05/2025#ThisWeeksFiddler @xaqwg.bsky.social A 10-long river like this one happens only 405 times out of every 1,073,741,824. 041
Tom Keith @tomkeith.bsky.social · 21/05/2025After some code optimization and more than 8 hours of computer time, I was able to calculate the 7th bipyramid. It took 463 times as long as the time to calculate the 6th bipyramid. 000
Tom Keith @tomkeith.bsky.social · 19/05/2025Regular: How many distinct paths are there from the top of the rhombus to the bottom if (1) you never visit the same point twice and (2) you move only downward or sideways? Let's count them up ... 020
Tom Keith @tomkeith.bsky.social · 19/05/2025#ThisWeeksFiddler @xaqwg.bsky.social Extra credit: How many distinct paths are there from the top of a triangular bipyramid to the bottom, if (1) you never visit the same point twice and (2) you move only downward or sideways? Short answer: A lot, especially with larger bipyramids. 261
Tom Keith @tomkeith.bsky.social · 12/05/2025This is the answer I got. Somehow I was expecting something simpler. 010
Tom Keith @tomkeith.bsky.social · 12/05/2025Extra credit: Within the interval (a, b), how likely is it that p4 > p7? Let c = value of p where p7 = p4. Then the answer is (b - c) / (b - a). 120
Tom Keith @tomkeith.bsky.social · 12/05/2025#ThisWeeksFiddler @xaqwg.bsky.social This chart shows how the likelihood of each series result depends on the probability p of the Celtics winning an individual game. The likelihood of "Celtics in 5" (blue line) is greater than the likelihood of any other result when p is between 0.6 and 0.75. 141
Tom Keith @tomkeith.bsky.social · 29/04/2025But what if 2 trees are equally far away, such as (7,4) and (8,1)? Now the relative size of the gaps depends on the exact visibility. (The tree at (8,1)'s gap is larger 60% of the time.) It doesn't settle one way or the other no matter how large your visibility gets. 010
Tom Keith @tomkeith.bsky.social · 29/04/2025As your visibility grows, the gaps get smaller but their size relative to each other stays the same. 110
Tom Keith @tomkeith.bsky.social · 29/04/2025#ThisWeeksFidder @xaqwg.bsky.social The largest gaps seem to lie on either side of the nearest trees. The nearest tree is at (2,1) and it separates the 2 largest gaps. The next nearest trees are at (3,1), (3,2), (4,1), and (4,3). The illustration shows gaps when visibility is 100. 131
Tom Keith @tomkeith.bsky.social · 31/03/2025#ThisWeeksFiddler 28 Mar 2025 EC There seems to be a pattern for the number of teams that can never win a tournament. If the number of teams in the tournament is 2^k, then the number of teams that can never win is 2^(k-1) - k + 1 (in the general case). 010
Tom Keith @tomkeith.bsky.social · 31/03/2025#ThisWeeksFiddler 28 Mar 2025 I tried running the tournament with different boost values, namely 0.5, 1.5, 2.5, and 3.5. Every team won except for the team with power index 3. 130
Tom Keith @tomkeith.bsky.social · 25/03/2025I got similar results to yours, with first seed's probability of winning bottoming out at k = 12. 210
