Finite games are all definite, either player 1 as a winning strategy or player 2 has, all other "outcomes" are just mental illnesses. Get over it, math doesn't care about your feelings.
whoever needs to use the bathroom first loses, or if you die of thirst or hunger, that could also disqualify one from such a theoretically limitless dilemma.
Alternately, if you have a mathematical way to measure boredom, and also introduce a rule that the person to truly become bored with the game first (or more bored, measured quantitatively somehow) would actually win automatically after a draw… or the game could just be determined by some arbitrary momentary measure of chaotic systems outside the game that the players can’t see or affect, and then giving the win to one of the players based on some hidden scoring matrix of outside variables probing purely environmental or coincidental variables, to generate an arbitrary-enough-seeming-to-the-players (though not likely enough for the mathematicians) winner, in the event of a repeated draws which outnumber the lower of the two single largest numbers that each player could think of, probed proper to match before each challenge, unbeknownst to the players, that way they cannot strategize to give a dishonest answer to affect play somehow towards their advantage (by saying a lower number than the highest possible [countable-to-by-this-person] number that they had ever actually been taught to count up to)
Ok, i dont understand this level of math, but cant you force a win in a 2 player game of non-infinite moves? Why wouldnt you be able to? Genuinely asking
You could design a different game that does though. E.g. A Tic-tac-toe variant but the player who starts looses if they don't get 3 in a row by the end of the game.
You can force a win in a specific 2 player game of non-infinite moves(if it has no draw condition). But yeah , it doesn't apply generally to any 2 player game of non-infinite moves. And the converse also doesn't apply generally.
Hey now, just because someone isn't pro-choice doesn't mean they're pro-AD. Honestly, people nowadays think everyone who disagrees with them on one thing must have every unhinged belief under the sun.
The axiom of determinacy, which implies some of (or all?) of the statements in op, and is more or less stated at the end. AD implies ~AC but they're not equivalent.
If I open up a pack of biscuits, and we each take turns eating a biscuit, AD says that there's a dominant strategy that can ensure that I eat the last biscuit. (e.g. there's only 1 biscuit; I win, or there's an odd number of biscuits; I win)
i.e. AD says you can rig games like this from the start
Bertrand Russell coined an analogy: for any (even infinite) collection of pairs of shoes, one can pick out the left shoe from each pair to obtain an appropriate collection (i.e. set) of shoes; this makes it possible to define a choice function directly.
For an infinite collection of pairs of socks (assumed to have no distinguishing features such as being a left sock rather than a right sock), there is no obvious way to make a function that forms a set out of selecting one sock from each pair without invoking the axiom of choice
So mathematicians always make the assumption that they can make a set from an infinite list of other non-empty sets based on this hunch, rather than any concrete choice function. And then they build mansions on top of this foundation, and use it to score chicks and ferraris, smh
36 Comments
Tilgare@lemmy.world · 140 pts · 233d
The bait and switch on this one really caught me off guard and gave me a great laugh. Good post.
edinbruh@feddit.it · 68 pts · 233d
Finite games are all definite, either player 1 as a winning strategy or player 2 has, all other "outcomes" are just mental illnesses. Get over it, math doesn't care about your feelings.
Dupelet@piefed.social · 44 pts · 233d
That would be true, except draws exist
SeptugenarianSenate@leminal.space · 11 pts · 233d
Tie goes to winner of the next game, easy fix
OhNoMoreLemmy@lemmy.ml · 29 pts · 233d
Congrats, the game is now non-finite (you can just keep drawing forever).
josephc@lemmy.ml · 5 pts · 233d
Draw goes to the player who moved second?
SeptugenarianSenate@leminal.space · 4 pts · 233d
whoever needs to use the bathroom first loses, or if you die of thirst or hunger, that could also disqualify one from such a theoretically limitless dilemma.
Alternately, if you have a mathematical way to measure boredom, and also introduce a rule that the person to truly become bored with the game first (or more bored, measured quantitatively somehow) would actually win automatically after a draw… or the game could just be determined by some arbitrary momentary measure of chaotic systems outside the game that the players can’t see or affect, and then giving the win to one of the players based on some hidden scoring matrix of outside variables probing purely environmental or coincidental variables, to generate an arbitrary-enough-seeming-to-the-players (though not likely enough for the mathematicians) winner, in the event of a repeated draws which outnumber the lower of the two single largest numbers that each player could think of, probed proper to match before each challenge, unbeknownst to the players, that way they cannot strategize to give a dishonest answer to affect play somehow towards their advantage (by saying a lower number than the highest possible [countable-to-by-this-person] number that they had ever actually been taught to count up to)
okwhateverdude@lemmy.world · 8 pts · 233d
For games that allow that, yep
Buddahriffic@lemmy.world · 6 pts · 233d
You can't draw in pictionairy.
okwhateverdude@lemmy.world · 4 pts · 233d
🥁🐍
ViatorOmnium@piefed.social · 8 pts · 233d
Let's play tic-tac-toe?
eestileib@lemmy.blahaj.zone · 2 pts · 232d
All other outcomes are a collaborative aesthetic exploration of a game tree subject to a variety of constraints.
The joy of the game, and indeed the value of the game, does not consist simply of winning. Even in go.
TomMasz@piefed.social · 40 pts · 233d
I thought I understood sets until I saw a show on PBS where a guy showed how there were different infinities using them and I realized I knew nothing.
AFKBRBChocolate@lemmy.ca · 22 pts · 233d
I have a friend who had the license plate "ALEPH NUL" which I thought was good nerd humor.
Rusty@lemmy.ca · 6 pts · 232d
The movie theater in Futurama is called Lowe's Aleph-Null-Plex.
AFKBRBChocolate@lemmy.ca · 3 pts · 232d
Nice, I missed that one
davidgro@lemmy.world · 24 pts · 233d
On the other hand, he Doesn't think you can double a sphere by cutting it into 5 pieces and reassembling them, so there's that.
Zizzy@lemmy.blahaj.zone · 20 pts · 233d
Ok, i dont understand this level of math, but cant you force a win in a 2 player game of non-infinite moves? Why wouldnt you be able to? Genuinely asking
Klear@quokk.au · 42 pts · 233d
Tic-tac-toe always ends in a draw with perfect play.
Zizzy@lemmy.blahaj.zone · 13 pts · 233d
Mmm, i did overlook the, very obvious in hindsight, draw outcome. Thanks
Venator@lemmy.nz · 3 pts · 232d
You could design a different game that does though. E.g. A Tic-tac-toe variant but the player who starts looses if they don't get 3 in a row by the end of the game.
Klear@quokk.au · 1 pts · 232d
Your point being?
Venator@lemmy.nz · 3 pts · 232d
You can force a win in a specific 2 player game of non-infinite moves(if it has no draw condition). But yeah , it doesn't apply generally to any 2 player game of non-infinite moves. And the converse also doesn't apply generally.
trevdog@lemmy.world · 17 pts · 233d
assuming a draw condition is impossible maybe
BB84@mander.xyz · 4 pts · 233d
For a finite game with no draws you are indeed able to.
FishFace@piefed.social · 17 pts · 233d
Hey now, just because someone isn't pro-choice doesn't mean they're pro-AD. Honestly, people nowadays think everyone who disagrees with them on one thing must have every unhinged belief under the sun.
SuperEars@lemmy.world · 7 pts · 233d
I am behind the times on some abbreviations. And dense.
What is AD in this context?
FishFace@piefed.social · 12 pts · 233d
The axiom of determinacy, which implies some of (or all?) of the statements in op, and is more or less stated at the end. AD implies ~AC but they're not equivalent.
tetris11@feddit.uk · 2 pts · 232d
If I open up a pack of biscuits, and we each take turns eating a biscuit, AD says that there's a dominant strategy that can ensure that I eat the last biscuit. (e.g. there's only 1 biscuit; I win, or there's an odd number of biscuits; I win)
i.e. AD says you can rig games like this from the start
Venator@lemmy.nz · 5 pts · 232d
It's been a while since I've done products of sets, but what if one of the sets in the product is a set of empty sets?
davidagain@lemmy.world · 4 pts · 232d
Then it's not empty. If it were a union of empty sets, that would be empty.
serra@slrpnk.net · 5 pts · 232d
I hope that at least he believes in the Axiom of Choice.
SlurpingPus@lemmy.world · 8 pts · 232d
Another comment in the thread says that “isn't pro-choice” is exactly about the rejection of the axiom.
tetris11@feddit.uk · 5 pts · 232d
For anyone wondering what this is
So mathematicians always make the assumption that they can make a set from an infinite list of other non-empty sets based on this hunch, rather than any concrete choice function. And then they build mansions on top of this foundation, and use it to score chicks and ferraris, smh
davidgro@lemmy.world · 2 pts · 231d
... That's the joke. (That he doesn't)
mEEGal@lemmy.world · 5 pts · 233d
Fucking relatable !
I'm that guy