The Squid Game framing made it feel approachable at first, which was a trap in itself because I started thinking about it narratively instead of mathematically.
Model the process as a Markov chain where the state is the number of revealed safe steps. Derive the survival probability for the i-th player by conditioning on the number of steps revealed before their turn, then use dynamic programming or combinatorial arguments to find the distribution of revealed steps. Finally, compute the probability that at least one player survives using the complement and the independence of player outcomes given the revealed steps.
Pro tip: Clearly state your assumptions (e.g., players choose randomly among unrevealed steps, and once a step is revealed, all subsequent players know the safe panel). This shows you can handle ambiguity and communicate effectively, which is crucial for a data science role at Citadel.
Let X_k be the number of steps revealed after k players have crossed. Initially X_0 = 0. When a player crosses, they may reveal new steps if they encounter unrevealed ones.
For the i-th player, given X_{i-1} = r, they survive if they correctly guess all unrevealed steps they encounter. The probability is (1/2)^{B - r} if they must guess all remaining steps, but they only guess until they hit a revealed step? Actually, they cross sequentially: they start at step 1, if revealed they proceed; if unrevealed, they guess and if correct, that step becomes revealed and they continue. So survival probability given r revealed steps is (1/2)^{B - r} only if the revealed steps are the first r? Not necessarily; revealed steps can be anywhere. Need to consider the positions of revealed steps.
The process of revealing steps is equivalent to each player, when encountering an unrevealed step, flips a fair coin. If correct, the step is revealed and they continue; if wrong, they die. The number of steps revealed after a player survives is the number of consecutive unrevealed steps they correctly guess starting from the first unrevealed step they encounter.
Use dynamic programming: let P(r, i) be the probability that after i players, r steps are revealed. Derive recurrence relations based on the probability that the next player reveals k new steps (which depends on the positions of revealed steps). Alternatively, note that the set of revealed steps after i players is simply the set of steps that have been guessed correctly by any player so far. Since players only guess at unrevealed steps, the process is equivalent to each step being independently 'solved' by the first player who encounters it, with probability 1/2 each time? Not independent because once solved, it's known.
The probability that at least one player survives is 1 - P(no player survives). No player survives if the first player dies, or the first survives but the second dies, etc. This can be computed using the distribution of revealed steps and the survival probabilities.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use the convolution formula for the sum of independent continuous random variables, integrating the product of their PDFs. Since X and Y are uniform on (0,1), the integrand is 1 over the overlapping region, so the integral reduces to the length of the intersection of intervals. Compute the piecewise result for 0 < s < 1 and 1 ≤ s < 2, and state the PDF is zero elsewhere.
Pro tip: Sketch the unit square in the (x,y)-plane and note that the line x+y=s cuts it; the PDF value is the length of that slice. This geometric view makes the triangular PDF intuitive and shows you can derive it without heavy algebra.
Write f_S(s) = ∫ f_X(x) f_Y(s-x) dx, using the fact that X and Y are independent. Since f_X(x)=1 for 0<x<1 and f_Y(y)=1 for 0<y<1, the integrand is 1 when both conditions hold.
For a fixed s, the integrand is nonzero only when 0 < x < 1 and 0 < s-x < 1, i.e., s-1 < x < s. The integration interval is the intersection of (0,1) and (s-1, s).
For 0 < s < 1, the intersection is (0, s), length s. For 1 ≤ s < 2, the intersection is (s-1, 1), length 2-s. Outside [0,2], the length is 0.
Combine the pieces: f_S(s) = s for 0 < s < 1, f_S(s) = 2-s for 1 ≤ s < 2, and 0 otherwise. This is the triangular distribution on [0,2] with peak at s=1.
Verify that the PDF integrates to 1 over [0,2] and that it is symmetric about s=1, as expected for the sum of two i.i.d. uniforms.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Coupon collector, basically a warm-up at this point.
Recognize this as the coupon collector's problem and use linearity of expectation by decomposing the total rolls into phases based on the number of distinct faces seen. Compute the expected rolls for each phase and sum them to get the final answer.
Pro tip: Mention that this is a classic coupon collector problem and that the expected value is n * H_n, where H_n is the nth harmonic number. For n=6, this is approximately 14.7, but the exact answer is 147/10 or 14.7.
State that this is the coupon collector's problem: we need to collect all 6 faces of a fair die. The expected number of rolls is the sum of expected rolls to get each new face.
Let X_i be the number of rolls needed to go from i-1 distinct faces to i distinct faces. The total rolls T = X_1 + X_2 + ... + X_6.
When i-1 faces have been seen, the probability of getting a new face on a roll is p_i = (6 - (i-1))/6 = (7-i)/6. Thus, X_i follows a geometric distribution with success probability p_i, so E[X_i] = 1/p_i = 6/(7-i).
By linearity of expectation, E[T] = sum_{i=1}^6 E[X_i] = sum_{i=1}^6 6/(7-i) = 6 * (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) = 6 * H_6.
Compute H_6 = 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 = 49/20 = 2.45. Therefore, E[T] = 6 * 2.45 = 14.7. So the expected number of rolls is 14.7.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
I knew by symmetry that E[X] = 0 and that Var(X) + Var(Y) + Var(Z) = Var(X+Y+Z) = E[X²+Y²+Z²] = 1 since x²+y²+z²=1 everywhere on the sphere.
Recognize that the x-coordinate of a uniformly random point on the unit sphere is uniformly distributed on [-1, 1]. Then compute the variance of a uniform distribution on [-1, 1] using the formula Var(X) = E[X^2] - (E[X])^2, yielding 1/3. Alternatively, derive it using spherical coordinates and integration.
Pro tip: Mention that the uniform distribution on the sphere is not the same as uniform in spherical coordinates; the density includes a sin(θ) factor. This shows deeper understanding and avoids a common pitfall.
Explain that by symmetry, the x-coordinate has a uniform distribution on [-1, 1]. Justify this by noting that the surface area of a spherical cap is proportional to its height.
Since the distribution is symmetric around 0, the expected value E[X] is 0.
For a uniform distribution on [-1, 1], E[X^2] = ∫_{-1}^{1} x^2 * (1/2) dx = 1/3. Alternatively, use the formula for the variance of a uniform distribution on [a, b]: (b-a)^2/12 = (2)^2/12 = 1/3.
Var(X) = E[X^2] - (E[X])^2 = 1/3 - 0 = 1/3.
If time permits, derive using spherical coordinates: X = sinθ cosφ, with θ uniform on [0, π] with density sinθ/2 and φ uniform on [0, 2π]. Compute E[X^2] = 1/3.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Frame the problem as a dynamic programming problem where the state is the number of draws remaining and the number of items still to be kept. Derive the optimal threshold policy: at each step, accept the current draw if it exceeds a threshold that depends on the remaining draws and slots. Then compute the expected total value using backward induction or by solving the recurrence.
Pro tip: Mention that the optimal policy is a threshold policy and that the thresholds can be precomputed; also note that the expected value can be computed via DP in O(nk) time, and for large n,k, approximations or continuous-time analogs may be used.
Let V(i, j) be the maximum expected total value when there are i draws remaining and j items still to be selected. Base cases: V(i, 0) = 0, V(0, j) = -infinity (or 0 if j=0).
For i > 0 and j > 0, if we accept the current draw x, we get x + V(i-1, j-1); if we reject, we get V(i-1, j). The optimal decision is to accept if x >= V(i-1, j) - V(i-1, j-1). Thus, V(i, j) = E[max(x, V(i-1, j) - V(i-1, j-1))] + V(i-1, j-1).
Compute V(i, j) for all i from 1 to n and j from 1 to k using the recurrence. The threshold for state (i, j) is t(i, j) = V(i-1, j) - V(i-1, j-1). The expected value is V(n, k).
The optimal strategy is: when i draws remain and j items are needed, accept the current draw if it exceeds t(i, j); otherwise reject. This is a time-dependent threshold policy.
The DP takes O(nk) time and O(k) space if optimized. For large n, k, one can use continuous approximations or note that the thresholds converge to a limiting curve.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.