You are offered two dice games and asked to decide which one gives you a better chance of winning. - **Game 1:** Roll one fair six-sided die 4 times. You win if at least one roll shows a 6. - **Game 2:** Roll a pair of fair six-sided dice 24 times. You win if at least one of the 24 rolls shows double sixes — both dice landing on 6 on the same roll. Which game has the higher probability of winning? Justify your answer with a calculation. This is a phone-screen question, so be prepared to reason through the comparison out loud, ideally without relying on a calculator. ```hint Complement rule "At least one success" is hard to compute directly. Use $P(\text{at least one}) = 1 - P(\text{zero successes})$: find the per-roll failure probability, then raise it to the number of rolls (rolls are independent). ``` ```hint Comparing without a calculator You need to compare $(5/6)^4$ with $(35/36)^{24}$. Notice $(35/36)^{24} = \left((35/36)^6\right)^4$, so it suffices to compare $(35/36)^6$ with $5/6$. Bernoulli's inequality $(1-x)^n \ge 1 - nx$ settles that comparison cleanly. ``` ```hint Why intuition misleads The expected number of successes is $4 \cdot \tfrac{1}{6} = \tfrac{2}{3}$ in Game 1 and $24 \cdot \tfrac{1}{36} = \tfrac{2}{3}$ in Game 2 — identical. Equal expected counts do **not** imply equal probabilities of "at least one." Think about what happens to $1-(1-p)^{c/p}$ as $p$ shrinks with $c$ fixed. ``` ### Constraints & Assumptions - All dice are fair: each face of each die has probability $1/6$. - All rolls are independent, and the two dice within a roll of Game 2 are independent. - Game 1 uses exactly 4 rolls; Game 2 uses exactly 24 rolls. - Winning means at least one success — extra successes beyond the first don't matter. - In Game 2, a success is both dice showing 6 on the *same* roll (probability $1/36$ per roll). ### Clarifying Questions to Ask - Are the dice fair, and are all rolls independent of one another? - In Game 2, does "(6,6)" mean both dice show a 6 on the same roll, i.e., one event with probability $1/36$ per roll? - Is the win condition purely "at least one success," with no bonus for multiple successes? - Are we comparing only the probabilities of winning, or is there also a payout/stake difference between the games? - Am I expected to produce exact numbers, or is a rigorous inequality argument for which probability is larger sufficient? ### What a Strong Answer Covers - **Correct setup via the complement:** translating "at least one" into $1 - (\text{failure prob})^{n}$ with the right per-roll failure probabilities ($5/6$ and $35/36$). - **A correct comparison:** either accurate numeric evaluation of both probabilities, or a clean analytic argument (e.g., Bernoulli's inequality) proving which is larger without a calculator. - **Intuition for the result:** articulating why the naive "equal expected number of successes" argument fails — the at-least-one probability is not linear in the number of trials. - **Communication under phone-screen conditions:** stating assumptions, walking through arithmetic clearly, and sanity-checking the final numbers (both should be near $1/2$). ### Follow-up Questions - What is the minimum number of rolls of a single die needed for the probability of seeing at least one 6 to exceed $1/2$? What about the minimum number of rolls of a pair of dice for at least one double six? - Fix the expected number of successes at $c = np$ and let $p \to 0$. What does the probability of at least one success converge to, and why? - Which is more likely: at least one 6 in 6 rolls of a die, or at least two 6s in 12 rolls? How would you set up that comparison? - How would your answer change if the win condition were "exactly one success" instead of "at least one"?