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submitted 1 year ago* (last edited 1 year ago) by cll7793@lemmy.world to c/nostupidquestions@lemmy.world

What concepts or facts do you know from math that is mind blowing, awesome, or simply fascinating?

Here are some I would like to share:

  • Gödel's incompleteness theorems: There are some problems in math so difficult that it can never be solved no matter how much time you put into it.
  • Halting problem: It is impossible to write a program that can figure out whether or not any input program loops forever or finishes running. (Undecidablity)

The Busy Beaver function

Now this is the mind blowing one. What is the largest non-infinite number you know? Graham's Number? TREE(3)? TREE(TREE(3))? This one will beat it easily.

  • The Busy Beaver function produces the fastest growing number that is theoretically possible. These numbers are so large we don't even know if you can compute the function to get the value even with an infinitely powerful PC.
  • In fact, just the mere act of being able to compute the value would mean solving the hardest problems in mathematics.
  • Σ(1) = 1
  • Σ(4) = 13
  • Σ(6) > 10^10^10^10^10^10^10^10^10^10^10^10^10^10^10 (10s are stacked on each other)
  • Σ(17) > Graham's Number
  • Σ(27) If you can compute this function the Goldbach conjecture is false.
  • Σ(744) If you can compute this function the Riemann hypothesis is false.

Sources:

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[-] kogasa@programming.dev 5 points 1 year ago* (last edited 1 year ago)

Let's name the goats Alice and Bob. You pick at random between Alice, Bob, and the Car, each with 1/3 chance. Let's examine each case.

  • Case 1: You picked Alice. Monty eliminates Bob. Switching wins. (1/3)

  • Case 2: You picked Bob. Monty eliminates Alice. Switching wins. (1/3)

  • Case 3: You picked the Car. Monty eliminates either Alice or Bob. You don't know which, but it doesn't matter-- switching loses. (1/3)

It comes down to the fact that Monty always eliminates a goat, which is why there is only one possibility in each of these (equally probable) cases.

From another point of view: Monty revealing a goat does not provide us any new information, because we know in advance that he must always do so. Hence our original odds of picking correctly (p=1/3) cannot change.


In the variant "Monty Fall" problem, where Monty opens a random door, we perform the same analysis:

  • Case 1: You picked Alice. (1/3)
    • Case 1a: Monty eliminates Bob. Switching wins. (1/2 of case 1, 1/6 overall)
    • Case 1b: Monty eliminates the Car. Game over. (1/2 of case 1, 1/6 overall)
  • Case 2: You picked Bob. (1/3)
    • Case 2a: Monty eliminates Alice. Switching wins. (1/2 of case 2, 1/6 overall)
    • Case 2b: Monty eliminates the Car. Game over. (1/2 of case 2, 1/6 overall)
  • Case 3: You picked the Car. (1/3)
    • Case 3a: Monty eliminates Alice. Switching loses. (1/2 of case 3, 1/6 overall)
    • Case 3b: Monty eliminates Bob. Switching loses. (1/2 of case 3, 1/6 overall)

As you can see, there is now a chance that Monty reveals the car resulting in an instant game over-- a 1/3 chance, to be exact. If Monty just so happens to reveal a goat, we instantly know that cases 1b and 2b are impossible. (In this variant, Monty revealing a goat reveals new information!) Of the remaining (still equally probable!) cases, switching wins half the time.

this post was submitted on 03 Aug 2023
622 points (97.4% liked)

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