For #thisweeksfiddler by @xaqwg.bsky.social , we find the probability that you win a best-of-N series, given that you win vs. lose the very first game.

For #thisweeksfiddler, we race among randomly selected loops of length 1, 3, 3.5, 4.5 miles at a 10 min/mi pace. If unfinished loops don't count and we have 65 min left to race, how can we maximize the avg total distance completed?

For #thisweeksfiddler by @xaqwg.bsky.social , we start with a set of vouchers ($10, $10, $10, $25), and can bet any of them on either side of an even odds game.

For #thisweeksfiddler, we're blocking our friend from placing their square on a board of side length B using 3 of our own.

#thisweeksfiddler @xaqwg.bsky.social
Other bikers will sprint if their strength >0.5, while we sprint if our leg strength >t. What is our chance of winning?

For #thisweeksfiddler, what's the maximum score we can reach in bowling if we knock down a given number of pins? @xaqwg.bsky.social

For the small cases, you can have F(0)=1 and F(<0)=0

Google Colab

I modeled the sphere using voxels, and tried to come up with the greediest steps, getting 11.
colab.research.google.com/drive/185hNX...

For @xaqwg.bsky.social 's #thisweeksfiddler, we're in a race where we increase our speed continuously to the end. (1+b)v(2x) = v(x), where x is the distance remaining in the race and b is the factor we increase our speed.

I reasoned that for two random points to be collinear to a given position, they must be on the same line oriented at angle theta away. The probability this happens is proportional to the length of the line segment contained within the square oriented at that theta. Then avg over all theta.

Updated plot for the simulation

My attempts for #thisweeksfiddler by @xaqwg.bsky.social : how often can each point in a square be covered by a randomly placed, long line? Simulated (Left) and analytical.

My findings for #thisweeksfiddler by @xaqwg.bsky.social : How long is a river of spaces in a text? thefiddler.substack.com/p/how-long-i...

For the EC, there are multiple points A from which we can reach point B in a given layer, and we must sum over the product of ways from A to B and the number of ways to reach point A. This number blows up quickly to over 1 billion.

For the regular fiddler, there is only one way to travel from points A to B within one layer, therefore the number of ways to reach each point in a given layer is identical and equal to the sum of ways to reach all points A in the current layer.

Can You Permeate the Pyramid? How many paths can you find down a pyramid? (Okay, it’s really a rhombus.) What about a bipyramid? (Yes, it’s really a bipyramid.)

For #thisweeksfiddler by @xaqwg.bsky.social thefiddler.substack.com/p/can-you-pe..., my approach: We consider separately the number of ways to descend to a point in the next layer and the ways to travel between two points in a given layer.

For #thisweeksfiddler by @xaqwg.bsky.social , we evaluate win probabilities for a best of 7 series. My graphical approach:

Google Colab

colab.research.google.com/drive/1woSjU...

My findings for #thisweeksfiddler:
thefiddler.substack.com/p/how-many-r...

My findings for #Thisweeksfiddler, looking through the coprime forest. thefiddler.substack.com/p/can-you-se...
@xaqwg.bsky.social

For #ThisWeeksFiddler: Another graph!
@xaqwg.bsky.social

For #thisweeksfiddler by @xaqwg.bsky.social , how often will a 1-seed win a single elimination tournament, if in any matchup between teams with seeds M and N, the M-seed wins with probability N/(M+N)?

My findings for #Thisweeksfiddler by @xaqwg.bsky.social , where we try to escape pie, only to find pi.

colab.research.google.com/drive/1ZTfQc...

Google Colab

colab.research.google.com/drive/1PfBxj...

For #thisweeksfiddler by @xaqwg.bsky.social : Setting up a chain of dominoes, using the geometric distribution:

"including the domino that causes the chain reaction)? More precisely, if this median number is M, then you would expect to have placed fewer than M dominoes at most half the time", so 69?

Google Colab

colab.research.google.com/drive/1PfBxj...

Varying the total number of hats:

For #ThisWeeksFiddler, picking rabbits out of a hat. A near linear trend.