Astronomical denialism

Sirley L. Quimby’s four non-transitive dice , which we discussed last week, have faces numbered 1 2 16 17 18 19, 3 4 5 20 21 22, 6 7 8 9 23 24, and 10 11 12 13 14 15, and they can be played in the following two-player game : one player chooses one of the four dice, and the other chooses one of the three; then each player rolls their own die, and the one with the highest score wins. The question is: If the first player chooses the first die, which die should they choose as the second die to maximize their probability of winning?

Each of the 6 sides of one die can be combined with each of the 6 sides of the other, so for each pair of dice, there are 36 possible outcomes when both players roll their respective dice. If the first player chooses the first die and the second player chooses the second, the possibilities are:

1-3 1-4 1-5 1-20 1-21 1-22

2-3 2-4 2-5 2-20 2-21 2-22

16-3 16-4 16-5 16-20 16-21 16-22

17-3 17-4 17-5 17-20 17-21 17-22

18-3 18-4 18-5 18-20 18-21 18-22

19-20 19-21 19-22

The first player wins 12 out of 36 possible outcomes (in bold), meaning the probability of winning for the second player if they choose the second die is 2/3. Is this the best option, or can the probability of winning be increased by choosing a different die?

A summer lapse led me to say that I hadn't received any proof of the uniqueness of Sicherman's dice, but that's not true: there wasn't one, but two, although a hasty reading led me to consider them incomplete. However, my mistake proved fruitful (something that often happens in mathematics and science in general), as it sparked an interesting discussion on the topic (see comments from one and three weeks ago). I highlight a simplified proof by Salva Fuster:

“I have the impression that perhaps it could be simplified a little if we start from the solution of two conventional dice {1, 2, 3, 4, 5, 6} and see what happens when we modify the two values ​​6 by 5 and 7, by 4 and 8 and by 3 and 9. The case with 3 and 9 would have a die {1, 2, 2, 2, 2, 3}, which would give four sums 11. The case with 4 and 8 would have five possibilities for the die whose highest value is 4, but four of them are instantly discarded, leaving only one valid: {1, 2, 2, 3, 3, 4}. The cases with 5 and 7 or the one of keeping the two values ​​6 probably require a little more work, but I think that an idea to apply would be that if one of the intermediate values ​​of one die goes up by one unit, one of the intermediate values ​​of the other die must go down by one unit.”

The CVA (Old Outpost Account) once again proves its effectiveness.

True, false… or debatable?

Changing the subject, and even the subject (astronomy is no longer a branch of mathematics, as it was when they were called “ exact sciences ”):

What do you think of these three statements (or rather, denials) that seem like those of a drunken flat-earther?

The Earth does not revolve around the Sun.

Jupiter is not a gas planet.

The speed of light is not insurmountable.