# D K

less info
reputation
110
bio website location age member for 1 year, 10 months seen 4 hours ago profile views 176

# 106 Actions

 4h comment Probabilities in a riddle involving axiom of choice yes the order would be: 1)describe the probabilistic strategy 2)opponent choses a sequence 3)probabilistic variable i is instanciated 8h accepted Probabilities in a riddle involving axiom of choice 8h comment Probabilities in a riddle involving axiom of choice How about describing the riddle as this game, where we have to first explicit our strategy, then an opponent can choose any sequence. then it is obvious than our strategy cannot depend on the sequence. The riddle is "find how to win this game with proba (n-1)/n, for any n." 16h comment Probabilities in a riddle involving axiom of choice ah ok I see where the misunderstanding comes from, it's true that "independently" is ambiguous, because only one random variable is involved here. But I think it still has a mathematical meaning in the sense "it does not depend on the opponent's choice", namely we have $\exists x \forall y$ where $x$ is our strategy and $y$ is our opponent's strategy (i.e. the sequence), and we still win this game because we can choose devise a (probabilistic) strategy that works on all sequences. 2d comment Probabilities in a riddle involving axiom of choice Our choice of index $i$ is made randomly, but for this we only need the uniform distribution on $\{0,\dots,n\}$. It is made independently of the opponent's choice. Dec12 comment Probabilities in a riddle involving axiom of choice yes but the point is that we can win again any strategy of the opponent, even if he chooses the sequence after we chose our (probabilistic) strategy. This way we avoid talking about probabilities on sequences. Dec12 comment Probabilities in a riddle involving axiom of choice ok this helps, but when you say "suppose $\vec u$ is chosen randomly",this is where the problem arises. Isn't there a way to ask the riddle so that the notion of probability of victory makes sense, without encountering this problem (for instance agains an opponent who chooses the sequence). Dec12 awarded Nice Question Dec10 comment Probabilities in a riddle involving axiom of choice @JoelDavidHamkins That is why I prefer the adversary formulation: you fix a strategy, and then the opponent chooses a sequence. Dec10 comment Probabilities in a riddle involving axiom of choice @JoelDavidHamkins You're right, thanks for correcting. Dec9 comment Probabilities in a riddle involving axiom of choice @AlexanderPruss Thank you for explaining the Brown-Freiling argument, I understand your point better. Would this change if we change the riddle to "an opponent selects a sequence, and then you play with a (possibly probablistic) strategy"? Or in the other order (opponent plays after) ? Dec9 comment Probabilities in a riddle involving axiom of choice I don't get why we need a probability measure on the sequences. Why can't we say that "winning with probability at least $\frac{N-1}{N}$" means that no matter the sequence chosen by an adversary opponent, we will win with probability at least $\frac{N-1}{N}$? Dec9 comment Probabilities in a riddle involving axiom of choice That is why there is +1 in the definition of $M$. If the max occurs twice, no one will be silent and they will all make a correct guess. Dec9 comment Probabilities in a riddle involving axiom of choice I'm not sure I agree. I think we can make sense of "fails with probability at most $1/N$", by saying that for all fixed sequence, the probability (which comes from the strategy) of failing is at most $1/N$. Moreover I don't understand your counter-example, because no matter how you choose the sequence, the strategy still has $\frac{N-1}{N}$ chance of guessing correctly. Dec9 comment Proof that finite prefixes do not alter the set of factors of infinite sequences if considering long enough factors if $m\geq n$, in particular $\eta$ have same factors of length $n$ as $\xi$, which is enough for the above proof to work. Dec9 comment Probabilities in a riddle involving axiom of choice Indeed, we could modify the question to say "one is allowed to remain silent, and all the others have to guess the content of a closed box". Dec9 comment Probabilities in a riddle involving axiom of choice @JoelDavidHamkins Yes in all cases "wrong" means "can be wrong", sorry for the shortcut, they can always be right if they are "lucky". Dec9 comment Probabilities in a riddle involving axiom of choice with infinitely mathematicians numbered by $\mathbb N$, the guy number $i$ can just start looking at the sequence from index $i$. This way, finitely many will be wrong (the ones with $i\leq M$). Dec9 comment Proof that finite prefixes do not alter the set of factors of infinite sequences if considering long enough factors The only factors of length $n$ that are not common between $\eta$ and $w\cdot\xi$ have to end at position $i<|w|+n$, by definition of $\eta$. This is why $m=|w|+n$ is sufficient to ensure the result. Dec9 comment Probabilities in a riddle involving axiom of choice Yes I find it's one of the most surprising use of the axiom of choice :)