Suppose You Have A Fair Coin That Can Be Flipped Independent

Suppose you have a fair coin that can be flipped indepe

Suppose you have a fair coin that can be flipped independently for an unbounded number of times. Is it possible to design a procedure that samples a number uniformly random over {1, 2, 3, 4, 5, 6}, i.e., to simulate a fair die? Consider analyzing different types of algorithms: (1) an algorithm that never fails and outputs the correct distribution perfectly but might not terminate; (2) an algorithm that might fail but produces the correct distribution conditioned on not failing; and (3) an algorithm that does not fail and outputs approximately correct results. Argue that if you can simulate a fair die using a fair coin (either perfectly or almost correctly), this implies you can generate random outputs over {1, 2, 3, 4, 5} as well, either perfectly or almost correctly, respectively. The argument should be simple, using only one sentence.

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Simulating a fair die from a fair coin involves transformations that preserve uniform randomness; since any outcome over {1, 2, 3, 4, 5, 6} can be mapped or reduced to outcomes over {1, 2, 3, 4, 5} through such transformations, the ability to generate one implies the ability to generate the other under the same conditions (perfect or near-perfect fairness).

The challenge of simulating a fair die from a biased coin (p ≠ 1/2) hinges on creating a fair coin flip from the biased coin, which can be achieved via established probabilistic techniques, such as von Neumann’s method. For the three case analyses:

• Perfect, non-terminating algorithm: Using von Neumann’s method iteratively ensures a perfectly fair coin flip but does not guarantee termination, as some sequences may be infinitely long—hence, it may never produce an output in finite time.
• Failing algorithm with correct output conditioned on success: Such an algorithm can run until a fair flip is generated or fail; conditional correctness is maintained if it succeeds, but the expected running time depends heavily on p, especially when p is close to 0 or 1, since the probability of success per attempt is 2p(1-p).
• Almost correct, non-failing algorithm: An approximation can be achieved through methods like the iterative bias correction or finite-precision representation, resulting in a bias arbitrarily close to 0.5, with running time depending on the desired accuracy and possibly on p.

The running time for the bias correction algorithms generally depends on the bias p, with worst-case expected times increasing as p approaches 0 or 1, due to the lower probability of success in each iteration.

Prohibition of Finite-Termination Algorithms for Uniform {1,2,3,4,5}

Assuming a fair coin, any algorithm that always terminates in finite steps and outputs an element uniformly at random over {1, 2, 3, 4, 5} would require the total number of outcomes from the coin tosses to match the size of the set (i.e., 5). However, the outcome space of k coin flips has size 2^k; since 2^k can never be exactly divisible by 5 for any finite k, there’s no way to partition the sample space into 5 equally likely outcomes, thus proving such an algorithm cannot always terminate finitely while producing a uniform output over {1, 2, 3, 4, 5}.

More formally, using a counting argument: For each positive integer k, the total number of outcomes is 2^k; for the algorithm to produce a uniform outcome in {1, 2, 3, 4, 5}, 2^k must be divisible by 5, which is impossible due to the properties of powers of 2 modulo 5. Therefore, no finite procedure can always terminate and produce a perfect uniform sample over {1, 2, 3, 4, 5}.

Generating a Uniform Random Subset and Probabilities

Part (a): To generate a uniformly random subset of [n], flip an independent fair coin for each element: include the element in the subset if it lands heads, exclude if tails. As each element has a 50% chance independently, all 2^n subsets are equally likely, making the distribution uniform over all subsets.

Part (b): For independent uniform subsets X and Y:

• Probability that X is a subset of Y (X ⊆ Y): For each element, the probability that it is in X is 1/2; that it is in Y is 1/2; and for X ⊆ Y, if x ∈ X, then x must be in Y. Since the choices are independent, the probability that for each element, either it is not in X or it is in Y, sums to a total probability of (3/4)^n, but more precisely, for each element, the probability that x ∈ X implies x ∈ Y is 1/3, so overall probability of X ⊆ Y is (3/4)^n (this specific probability calculation uses the fact that for each element, the probabilities of different inclusion scenarios are symmetric).Alternatively, a more straightforward calculation is that each element is included in both X and Y with probability 1/4, in X only with 1/4, in Y only with 1/4, or in neither with 1/4, leading to the probability that X ⊆ Y as (1/2)^n.
• Probability that X ∪ Y = [n]: This occurs only if for every element, at least one of X or Y contains it—that is, no element is outside both sets. For each element, the probability that it's in at least one of X or Y is 1 - probability it is in neither, which is 1 - (1/2)*(1/2) = 3/4. Since each element's inclusion is independent, the probability is (3/4)^n.

Learning about a Noisy Function and Query Algorithm

Given a linear function F and a damaged table with up to 1/5 of entries corrupted, the key challenge is to estimate F(z) with probability at least 1/2 using minimal queries. Since some entries may be corrupted, querying a single index z may yield incorrect results. To ensure a high probability of correctness, a simple randomized algorithm randomly samples a small subset of the table entries, queries these the minimal number of times possible, and then uses a statistical inference method—such as majority voting or median—to decide on the estimate for F(z). Repeating this process three times and taking the majority vote increases the probability that the estimated value matches F(z) at least 50%, regardless of which entries are corrupted, due to the assumption that less than one-fifth of entries are damaged, and the randomness in sampling aids in averaging out the errors.

This approach leverages the structure of the linear function and the assumption of bounded corruption, applying probabilistic error reduction techniques, like Chernoff bounds, to control the likelihood that the estimate deviates significantly from the true F(z). The minimal number of queries is determined by the desired confidence level—using as few as three sample queries and a majority rule ensures the probability exceeds 1/2, satisfying the problem's criteria.

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