Alice and Bob want to agree on a bit $0$ or $1$. Both know it would be fair to pick that at random, but there's no way they could meet to throw a dice and no third party they could trust. Are there known protocols that allow either side to prove that they did not manipulate the outcome?  My idea is: Both randomly (or not - but that would be in their best interest) pick $0$ or $1$, write it down, put it in an envelope and send it to the other person. Then the XOR of both bits could not have been manipulated unfairly by either party.  The problem I see with that: How can Alice prove to Bob that she did not open his envolope before making her choice?  Maybe one could instead choose an initial string $n$ and use the first $k$ with $f^k(n) &lt; a$ as the chosen value. Now if $f$ is a cryptographic hashing function and $a$ is small Alice would have taken a long time if she were to predict $k$ before sending her own choice.
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Several people are playing a game with random events and require a way to produce a random number. (Such as dice rolls or a lottery.)  Can this be done such that each player has the power to be reasonably sure that the random number was fairly selected, without having to trust anyone else?  This question stems from a discussion in comments to my answer on the . I came up with a protocol but someone else spotted a flaw. This site seems more appropriate. Please feel free to make use of my answer to that question.
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I think that the question is contained in the title. Suppose we begin from something that is false for example $1=0$. Is it possible using only $\Rightarrow$ (and of course $\lnot ,\wedge,\lor$)  to prove any possible statement? It has been ages since i studied logic at the university so i would prefer a simple explanation (if possible)
68164
Let $H:\{0,1\}^*\to\{0,1\}^n$ be a cryptographic hash function as a black-box, and suppose we have unlimited space.  As I understand, finding $x$ such that $H(x)=0$ (if such exists) would require a preimage attack, and avg. time $O(2^n)$ (linear in the size of the output). On the other hand, finding $x\neq y$ such that $H(x)\oplus H(y)=0$ could use the birthday attack, and therefore avg. time $O(2^{n/2})$.  My question is, if something better can be said for finding 3 (or more) distinct values $x,y,z$ such that $H(x)\oplus H(y)\oplus H(z)=0$. It seems clear that this could be done using less samples of $H$, but it's not clear to me if the time complexity could be improved.
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Is there a known pair of distinct bit strings (A,B) such that SHA-1(A) == SHA-1(B)?  If the answer is no, then how can SHA-1 be considered broken?
51661
Consider two fair $k$-sided dice with the numbers 1 through $k$ on their faces, obtaining values $X_1$ and $X_2$. What is $\mathbb{E}[\max(X_1, X_2)]$ and what is $\mathbb{E}[\min(X_1, X_2)]$.
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Given that $X_1, X_2...$ are mutually independent random variables. For each $i$ with $1\le i \le n$   the variable $X_i$ is equal to either $0$ or $n+1$ $E(X_i)$ = $1$   also.. if $X_i$ is equal to either $0$ or $n+1$, doesn't that mean that all $X_i$ need to be $0$?  Thanks
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The assumptions are:    Alice and Bob have perfectly synchronized clocks Alice and Bob have successfully exchanged a pair of entangled photons   The idea is simply to have Alice and Bob perform the Quantum Eraser Experiment (doesn't need to be the delayed choice).   Alice and Bob agree on a specific time when Bob's photon will be between the "path marker" (which is usually just after the slits) and the detector.  If Alice acts collpasing the wave-function on her photon, the interference pattern will disappear. If not it won't.  Alice and Bob can be spatially separated...  What am I misunderstanding?   The only meaningful difference from this spatially separated quantum eraser experiment to one done on tabletop is that you won't be able to use a coincidence detector, but that is not impeditive to identifying the interference pattern, just will make errors more probable. Which we should be able to deal with a appropriate protocol...  There is a experimental paper with a small amount of citations pointing out to the breaking of complementarity in a very similar setup:
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I came across the following problem in a book:  Give a combinatorial proof of  $$ {n \choose 0} + {n \choose 2} + {n \choose 4} + \, \, ... \, = {n \choose 1} + {n \choose 3} + {n \choose 5} + \, \, ...$$ using the "weirdo" method (i.e., where one of the elements is chosen as special and included-excluded -- I'm sure you get the idea).  After days of repeated effort, the proof has failed to strike me. Because every time one of the elements is excluded, the term would be ${n-1 \choose k}$ and not $ {n \choose k}$, which is not the case in either of the sides of the equation.
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Alice and Bob play the following game with an $n \times n$ matrix, where $n$ is odd. Alice fills in one of the entries of the matrix with a real number, then Bob, then Alice and so forth until the entire matrix is filled. At the end, the determinant of the matrix is taken. If it is nonzero, Alice wins; if it is zero, Bob wins. Determine who wins playing perfect strategy each time.    When $n$ is even it's easy to see why Bob wins every time. and for $n$ equal to $3$ I have brute-forced it. Bob wins. But for $n = 5$ and above I can't see who will win on perfect strategy each time. Any clever approaches to solving this problem?
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Would someone please advance/discuss some real-life situations falsities $1, 2$? I'd like to intuit why these are false. As a neophyte, since I still need to compute the probabilities for the examples in the two answers to apprehend them, I haven't naturalised these concepts yet. Thus, I beg leave to ask about other real-life examples which require less or no computations.  I tried and would appreciate less numerical examples than  and , and  Examples 1.22 and 1.23 on P39 of Introduction to Pr by Bertsekas.  $1.$ Pairwise Independence $\require{cancel} \cancel{\implies}$ (Mutual) Independence.      $2.$ Pairwise Independence $\require{cancel} \cancel{\Longleftarrow}$ (Mutual) Independence.         P38 defines (Mutual) Independence :    For all $S \subseteq \{1, ..., n - 1, n\} $ and events $A_i$, $Pr(\cap_{i \in S} A_i) = \Pi_{i \in S} A_i.$
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A question in probability by a non-mathematician:  A fair coin is tossed $2N$ times. Is it unlikely that we get exactly $N$ heads and $N$ tails?  From one side, this must be the most likely result! But intuitively, if someone reports to me that they threw 2,000,000 coins and got exactly 1,000,000 heads and 1,000,000 tails I would be a little suspicious.  If this really is suspicious, how does it compare to biased results? Obviously, 2,000,000 heads is a much more suspicious result. But, for what value of $K$ is $N+K$ heads and $N-K$ tails as suspicious as $N$ heads and $N$ tails?  Equations are very welcome, but please try to provide some aid in understanding them.  ADDED: Henning Makholm's answer helped me understand better what I want to ask: I always assume a fair coin and would like to compare the probability of the split N/N with the probability of getting either more than more than N+K or less than N-K heads. By ncmathsadist's answer , the probability for the N/N split goes like 1 over the square root of N.  What about the probability to get more than either more than N+K or less than N-K heads? For what value of K (approximately) does it equal to 1/sqrt(n pi)?  CLARIFICATION: If f(N) is the probability to get an N/N split in 2N tosses of a fair coin, and g(N,K) is the probability to get more that N+K heads or less than N-K heads in 2N tosses of a fair coin, then I am asking for a formula which, given N, approximates the values of K for each f(N) is closest to g(N,K). This is a formula that takes N and returns K so we can call it h(N)=K. Which kind of function is h? My guess from a little experementing with the online software in Brian M. Scott's answer is that the function h is apporximately the SQRT function, multiplied by some constant, or something like that.
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Consider the  of Bennett et. al. How does one prove that this protocol would never violate the conservation of energy? At the face of it, it doesn't seem to be something obvious, as the various measurements, communications, rotations etc. don't seem to be able to account for the differences in the Hamiltonians of Alice's and Bob's labs. The state $\psi$ might have an expected energy $E_1$ in Alice's lab and a completely different expected energy $E_2$ in Bob's. Where does the difference come from?
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$\newcommand{\P}{\mathbb{P}}$Is it valid to interpret $\P(B|A)$ as "infinite probability" when $\P(A) = 0$?
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this is a rewrite of another recent question of mine [1] that wasnt stated well (it had a semi obvious simplification, mea culpa) but I think theres still a nontrivial question at the heart of it. have seen similar problems in the literature but not this one in particular.  will write it in terms of bit-vectors because thats easiest for me.      let there be a set of bit-vectors of size $n$, $v_1, v_2, v_3, ... , v_n$. consider the bitwise XOR operation. given a target vector $v_0$. find a subset of vectors such that the bitwise XOR of the set equals the target vector. what is an efficient (or ideally, optimal) algorithm to find a subset?   the brute force algorithm enumerates the powerset of size $2^n$ and lists the 1st subset found. (slightly?) more efficient would look at 1-positions in the target and exclude subsets that do not have at least 1 vector with a 1 in a 1-position of the target.  the subset may or may not exist. it may or may not be unique.  closely related questions: (1) find the smallest subset, (2) output T/F depending on whether such a subset exists.  have a suspicion one of these problems is NP complete.  looking for references, insight etc. it would be interesting to know if there are "hard" vs "easy" inputs etc  as I wrote on the other question this seems closely related to the subset sum problem (see eg garey &amp; johnson ref) which is known to be NP complete but this seems to have "slightly" less complexity because its simpler to compute a vector bitwise XOR than a binary sum (the sum can have more binary digits).  the problem also seems closely connected to bin fu's recent question [2]  [1]   [2]
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Let's suppose there is a chessboard (with 64 squares), and there is a penny on each square. Each of the pennies has a side (head or tails) facing up, chosen at random.  I look at the board and choose my favorite penny. I then tell my two mathemetician friends that I will reveal which penny is my favorite to one of them, who then must flip over exactly one of the pennies on the chessboard, while the other is in another room not looking. Once the coin is flipped over, the uninformed mathematician is summoned into the room and must deduce which penny is my favorite only by examining the coins on the chessboard.  What strategy can the two mathematicians devise to ensure that my favorite penny can always be deduced?
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I was asked this question today in an interview and am still not sure of the answer. The question is as follows.   Part one:  Say I have flipped 100 coins already and they are hidden to you. Now I reveal these coins one by one and you are to guess if it is heads or tails before I reveal the coin. If you get it correct, you get $\$1$. If you get it incorrect, you get $\$0$. I will allow you to ask me one yes or no question about the sequence for free. What will it be to maximize your profit?  My approach for this part of the problem was to ask if there were more heads than tails. If they say yes, I will just guess all heads otherwise I just guess all tails. I know the expected value for this should be greater than 50 but is it possible to calculate the exact value for this? If so, how would you do it?   Part two: Same scenario as before but now I will charge for a question. I will allow you to ask me any amount of yes or no questions as I go through this process for $\$1$. What is your strategy to maximize your profit?  I was not sure about the answer to this part of the question. Would the best option be to guess randomly? I think the expected value of this should be 50. I am not sure about the expected value of part one but if it is greater than 51, I think I could also use that approach. Anyone have a good idea for this part?
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There are several proofs I have seen of this, but they all seem to use choice subtely at some point. Is there any way to prove this without choice, or is it possibly unproveable?