At market close, this market will resolve to a random number I generate between 1 and 6. But with a twist - the number that has the highest % at market close, will have a chance of 35% in the generation, and the remaining numbers a chance of 13% as opposed to the standard 1/6 (16.666666666666667%).

The market close time is 2023/10/24 20:00 (Berlin, Germany time).

Let's say number 1 has the highest % at market close, that means the chances of the market resolving to each of the numbers is:
1 - 35%
2 - 13%

3 - 13%

4 - 13%

5 - 13%

6 - 13%

If several numbers have the highest percentage (approximated based on displayed numbers on the Manifold front-end when viewed through a standard web-browser - which means rounded to the nearest integer) at market close, then all those numbers will have the weight divided between them. The weight being divided will be calculated as 35%+13%*{extra numbers having top %}. So for instance, let's assume 1, 2, and 3 all occupy the top percentage slot at market close. That would result in:
1 - (35%+13%*2)/3=20.333333333333334%
2 - (35%+13%*2)/3=20.333333333333334%

3 - (35%+13%*2)/3=20.333333333333334%
4 - 13%

5 - 13%

6 - 13%

I will not bet on this market myself, since I'm going to be generating the number without using a method that's independently verifiable, so you essentially have to trust me. I'll try to resolve the market as close after the closing time as possible. I will record a video of (1.) asking ChatGPT in a separate chat what the date and time is, and (2.) asking it to run the code below using code interpreter, with the contents of the top_numbers list changed according to the market results:

import random

def weighted_random_choice(top_numbers: list) -> int:

base_weight = 0.13

total_numbers = 6

special_weight = 0.35 + base_weight * (len(top_numbers) - 1)

weights = [special_weight / len(top_numbers) if i in top_numbers else base_weight for i in range(1, total_numbers + 1)]

return random.choices(range(1, total_numbers + 1), weights, k=1)[0]

draw_number = weighted_random_choice([1, 2, 3])

draw_number