Solution: Let $ H $ be the number of heads in 6 flips of a fair coin, so $ H \sim \text{Binomial}(6, \frac{1}{2}) $. Let $ D $ be the outcome of a fair 6-sided die, uniformly distributed over $ \{1,2,3,4,5,6\} $. We seek $ P(H = D) $.

["Title: How to Calculate the Probability That Heads Equal Die Roll in a 6-Flip Experiment", "Understanding the probability of matching outcomes between coin flips and a die roll might seem simple at first, but it reveals elegant principles of probability theory. In this article, we explore the probability $ P(H = D) $, where $ H $ counts the number of heads in 6 fair coin flips, and $ D $ is an outcome of a fair 6-sided die. We show how to compute this probability using the binomial and discrete uniform distributions.", "---", "### Understanding the Variables", "Let:\n- $ H \sim \ ext{Binomial}(6, 0.5) $: The random variable representing the number of heads in 6 independent flips of a fair coin.\n- $ D \in {1,2,3,4,5,6} $, uniformly distributed: $ P(D = d) = \frac{1}{6} $ for all $ d $.", "We want to compute $ P(H = D) $, the probability that the number of heads matches the value rolled on the die.", "Because $ H $ is discrete and $ D $ is discrete, we compute this joint probability using the law of total probability:", "[\nP(H = D) = \sum_{d=1}^{6} P(H = d) \cdot P(D = d) = \frac{1}{6} \sum_{d=1}^{6} P(H = d)\n]", "Note: $ H $ cannot equal 0 because the minimum $ H = 0 $ corresponds to 6 tails, and the die only shows values from 1 to 6, so $ P(H = 0) = 0 $. Hence, the sum starts at $ d = 1 $.", "---", "### Computing $ P(H = d) $ for $ d = 1, 2, ..., 6 $", "Since $ H \sim \ ext{Binomial}(6, 0.5) $, we use the binomial probability mass function:", "[\nP(H = d) = \binom{6}{d} \left( \frac{1}{2} \right)^d \left( \frac{1}{2} \right)^{6-d} = \binom{6}{d} \left( \frac{1}{2} \right)^6 = \frac{\binom{6}{d}}{64}\n]", "Now compute each:", "- $ P(H = 1) = \frac{\binom{6}{1}}{64} = \frac{6}{64} $\n- $ P(H = 2) = \frac{\binom{6}{2}}{64} = \frac{15}{64} $\n- $ P(H = 3) = \frac{\binom{6}{3}}{64} = \frac{20}{64} $\n- $ P(H = 4) = \frac{\binom{6}{4}}{64} = \frac{15}{64} $\n- $ P(H = 5) = \frac{\binom{6}{5}}{64} = \frac{6}{64} $\n- $ P(H = 6) = \frac{\binom{6}{6}}{64} = \frac{1}{64} $", "Now sum:", "[\n\sum_{d=1}^{6} P(H = d) = \frac{6 + 15 + 20 + 15 + 6 + 1}{64} = \frac{63}{64}\n]", "---", "### Final Computation", "Now plug into the total probability formula:", "[\nP(H = D) = \frac{1}{6} \cdot \frac{63}{64} = \frac{63}{384} = \frac{21}{128}\n]", "---", "### Final Answer", "[\n\boxed{P(H = D) = \frac{21}{128} \approx 0.1641}\n]", "This means there is roughly a 16.41% chance that the number of heads in 6 fair coin flips equals the integer rolled on a fair 6-sided die.", "---", "### Why This Is Interesting", "This problem beautifully combines two core concepts:\n- The binomial distribution modeling discrete outcomes in repeated trials (coin flips),\n- The discrete uniform distribution capturing randomness in a fair die.", "By conditioning on the die outcome, we illustrate conditional probability and demonstrate a practical real-world application of probability theory. Whether for games, experiments, or statistical modeling, this kind of calculation is foundational in understanding random events.", "If you're curious about other outcomes—like $ P(H < D) $ or $ P(H = 2D) $—the same method applies: use the PMF of $ H $, multiply by the single-point distribution of $ D $, and sum over valid $ d $.", "---", "Keywords: probability of heads matching die roll, binomial distribution, fair coin flips, fair die, $ P(H=D) $, discrete probability, combinatorics, statistical theory."]