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:
- YouTube - The Busy Beaver function by Mutual Information
- YouTube - Gödel's incompleteness Theorem by Veritasium
- YouTube - Halting Problem by Computerphile
- YouTube - Graham's Number by Numberphile
- YouTube - TREE(3) by Numberphile
- Wikipedia - Gödel's incompleteness theorems
- Wikipedia - Halting Problem
- Wikipedia - Busy Beaver
- Wikipedia - Riemann hypothesis
- Wikipedia - Goldbach's conjecture
- Wikipedia - Millennium Prize Problems - $1,000,000 Reward for a solution
I found the easiest way to think about it as if there are 10 doors, you choose 1, then 8 other doors are opened. Do you stay with your first choice, or the other remaining door? Or scale up to 100. Then you really see the advantage of swapping doors. You have a higher probability when choosing the last remaining door than of having correctly choosen the correct door the first time.
Edit: More generically, it's set theory, where the initial set of doors is 1 and (n-1). In the end you are shown n-2 doors out of the second set, but the probability of having selected the correct door initially is 1/n. You can think of it as switching your choice to all of the initial (n-1) doors for a probability of (n-1)/n.
Holy shit this finally got it to click in my head.
I find the easiest way to understand Monty Hall is to think of it in a meta way:
Situation A - A person picks one of three doors, 1 n 3 chance of success.
Situation B - A person picks one of two doors, 1 in 2 chance of success.
If you were an observer of these two situations (not the person choosing doors) and you were gonna bet on which situation will more often succeed, clearly the second choice.
But the issue is that by switching doors, you have a 66% chance of winning, it doesn't drop to 50% just because there are 2 doors, it's still 33% on the first door, 66% on the other doors (as a whole), for which we know one is not correct and won't choose.
is the key words here
individually the door has that 1:2 chance, but the scenario has more context and information and thus better odds. Choosing scenario B over scenario A is a better wager
you aren't talking about the Monty Hall problem then
Only to somebody who didn't know about the choice being made!
I don't understand how this relates to the problem. Yes 50 percent is greater than 33 percent, but that's not what the Monty hall problem is about. The point of the exercise is to show that when the game show host knowingly (and it is important to state that the host knows where the prize is) opens a door, he is giving the contestant 33 percent extra odds.