Millennium Management Interview Questions 2026

A and K are independent uniform variables on the six residues 0,...,5. Fix any displayed pair (a,b). The modular equation \(b\equiv a+K\pmod6\) has exactly one solution \(K\in\{0,\ldots,5\}\), namely \(K=(b-a)\) modulo 6. Therefore
\(P(A=a,B=b)=P(A=a,K=(b-a)\bmod6)=(1/6)(1/6)=\frac1{36}\).

Summing this joint probability over six values of the other variable yields \(P(A=a)=P(B=b)=\frac16\). Because \(\frac1{36}=(1/6)(1/6)\) for every \(a,b\), the outputs A and B are independent.

In particular, after seeing \(A=4\), the second dial shows 2 precisely when \(K=(2-4)\bmod6=4\), which occurs with probability \(\frac16\). Dependence cannot be inferred merely from the fact that one random variable appears in another's construction: the independent uniform offset hides all information about the base modulo 6.

Let H denote poor ventilation, R a red lamp, and A,B condensation in rooms A and B. By Bayes’ rule, \(P(H\mid R)=\frac{P(R\mid H)P(H)}{P(R\mid H)P(H)+P(R\mid H^c)P(H^c)}=\frac{(\frac34\cdot\frac12)}{(\frac34\cdot\frac12+\frac14\cdot\frac12)}=\frac34\); \(P(H^c\mid R)=\frac14\).

Within each true mode, A and B are independent and remain independent of R. Consequently, \(P(A\cap B\mid H,R)=(\frac34)^2=\frac9{16}\), and \(P(A\cap B\mid H^c,R)=(\frac14)^2=\frac1{16}\). Averaging with the posterior weights gives \(P(A\cap B\mid R)=(\frac34)(\frac9{16})+(\frac14)(\frac1{16})=\frac{27+1}{64}=\frac{28}{64}=\frac7{16}\).

Each room's marginal after red is \(P(A\mid R)=P(B\mid R)=(\frac34)(\frac34)+(\frac14)(\frac14)=\frac{9+1}{16}=\frac{10}{16}=\frac58\). If A and B were conditionally independent given red, their joint probability would be \((\frac58)^2=\frac{25}{64}\), but the actual joint probability is \(\frac{28}{64}\). Their conditional covariance as binary indicators is \(\frac{28}{64}-\frac{25}{64}=\frac3{64}>0\). Knowing the actual ventilation mode would remove this shared-source dependence; seeing a noisy proxy does not, because the posterior still mixes two different condensation probabilities.

Let \(T_A\sim\mathrm{Exp}(\lambda_A)\) and \(T_B\sim\mathrm{Exp}(\lambda_B)\) be independent, with rates \(\lambda_A=0.002\) and \(\lambda_B=0.003\) per hour. For lamp A to fail first at time \(t\), A must fail then while B survives beyond \(t\). Independence gives

\[\Pr(T_A<T_B)=\int_0^{\infty}f_A(t)\Pr(T_B>t)\,dt.\]

Using \(f_A(t)=\lambda_Ae^{-\lambda_A t}\) and \(\Pr(T_B>t)=e^{-\lambda_B t}\),

\[\Pr(T_A<T_B)=\int_0^{\infty}\lambda_Ae^{-(\lambda_A+\lambda_B)t}\,dt=\frac{\lambda_A}{\lambda_A+\lambda_B}=\frac{2}{5}.\]

The same calculation with A and B interchanged gives \(\Pr(T_B<T_A)=\lambda_B/(\lambda_A+\lambda_B)=3/5\). The two probabilities sum to one; the equality event has probability zero because these are independent continuous lifetimes.

Lamp B does have a shorter mean lifetime (\(1/0.003\) versus \(1/0.002\) hours), but comparing those means alone does not determine a race probability for arbitrary lifetime distributions. The exact 2/5–3/5 split follows from the exponential densities, their constant hazard rates and independence.

For a repair time \(T\sim\operatorname{Exp}(\lambda)\), a fully observed completion at time t contributes density \(f(t\mid\lambda)=\lambda e^{-\lambda t}\), while a repair known only to last longer than c contributes survival probability \(P(T>c)=e^{-\lambda c}\). Because censoring is independent of completion time, we may multiply these factors (omitting any censoring terms independent of λ).

The three completed times sum to \(1+2+4=7\) days and the two still-open ages add \(3+5=8\) days, giving total observed exposure 15 days. There are \(d=3\) completed repairs, so

\(L(\lambda)=[\lambda e^{-\lambda\cdot1}][\lambda e^{-\lambda\cdot2}][\lambda e^{-\lambda\cdot4}]\cdot e^{-\lambda\cdot3}e^{-\lambda\cdot5}\)
=\(\lambda^{3}e^{-15\lambda}\), \(\lambda>0\).

The log-likelihood is \(\ell(\lambda)=3\operatorname{log}\lambda-15\lambda\). Its score is \(\ell^{\prime}(\lambda)=\frac{3}{\lambda}-15\), which vanishes at \(\hat\lambda=\frac{3}{15}=\)\(\frac{1}{5}\ \text{day}^{-1}\). Because \(\ell^{\prime\prime}(\lambda)=-\frac{3}{\lambda^{2}}<0\), this is the unique global maximum for \(\lambda>0\). The exponential mean is \(\frac{1}{\lambda}\), giving an estimated mean 5 days.

More generally, for d completed jobs and total exposure \(T>0\), the likelihood (up to λ-free constants) is \(\lambda^{d}e^{-\lambda T}\), so \(\hat\lambda=\frac{d}{T}\) when \(d>0\). If \(d=0\), \(\ell(\lambda)=-\lambda T\) decreases strictly for every \(\lambda>0\). Its supremum occurs as \(\lambda\to0^{+}\); over the strictly positive exponential-rate parameter space there is no attained interior MLE, and the associated estimated mean tends to infinity. Treating unfinished repairs as zero-duration completions would both invent events and lose the exposure information.

The familywise error rate (FWER) is the probability of one or more false rejections across the fixed family, not the error probability for one isolated test.

When all eight null hypotheses are true, independence gives
\[ P(\text{no false rejections})=(1-\alpha)^8, \quad \mathrm{FWER}=1-(1-\alpha)^8. \]
Requiring this to be at most 0.05 yields
\[ (1-\alpha)^8\ge0.95 \quad\Longleftrightarrow\quad \alpha\le\boxed{1-0.95^{1/8}\approx0.00639115}. \]
This is the exact Šidák threshold under independence. If only \(m\le8\) of the hypotheses are true, their rejection events are still independent under the stated model and \(\mathrm{FWER}=1-(1-\alpha)^m\le1-(1-\alpha)^8\); thus the all-true case is worst.

Bonferroni instead applies the union bound:
\[ \mathrm{FWER}=P(\text{at least one true-null rejection}) \le \sum_{i\in\text{true nulls}}P(\text{reject }i) \le8\alpha. \]
It therefore sets \(\alpha_{B}=0.05/8=\boxed{0.00625}\), slightly more conservative but valid even if tests are dependent. At this value, independent all-null tests yield \(1-(1-0.00625)^8\approx0.04892<0.05\).

At a naive 5% level on each of eight independent true-null tests, the overall chance of a false claim is
\[ 1-0.95^8\approx\boxed{0.3366}, \]
far above the director's 5% limit. Picking the family after looking at observed results would require a different analysis; the specified family is fixed in advance.

The four observations are \((x_i,y_i)=(1,23),(2,26),(3,29),(4,32)\). An unrestricted affine fit \(y=a+bx\) can reproduce each exactly with \((a,b)=(20,3)\); because distances vary and zero residual is minimal, this is the unique least-squares solution. Equivalently, the normal equations \(\sum(y_i-a-bx_i)=0\) and \(\sum x_i(y_i-a-bx_i)=0\) give the same coefficients.

With the intercept constrained to zero, minimize \(\sum_i(y_i-b_0x_i)^2\). Differentiating yields
\[ 0=-2\sum_i x_i(y_i-b_0x_i),\qquad b_0=\frac{\sum_i x_iy_i}{\sum_i x_i^2}. \]
Here \(\sum x_i=10\), \(\sum x_i^2=30\), and the actual charges satisfy \(y_i=20+3x_i\), so
\[ \sum x_iy_i=20(10)+3(30)=290, \qquad b_0=\frac{290}{30}=\boxed{\frac{29}{3}}. \]
The forced-origin slope is inflated by
\[ b_0-3=20\frac{\sum x_i}{\sum x_i^2}=\frac{20}{3}\ \text{dollars/km}, \]
because the missing $20 fixed fee is pushed into the per-kilometre coefficient. The prediction error (prediction minus actual) at distance \(x\) is
\[ \frac{29}{3}x-(20+3x)=\frac{20}{3}(x-3). \]
It is negative at 1 and 2 km, zero at 3 km, and positive at 4 km. Fitting through the origin asserts that a zero-kilometre dispatch would cost $0, contradicting the known fixed fee; that is a substantive model restriction, not a display choice.

Write \(S(v)=\sum_{j=1}^{4}1\{v_j=y_j\}\) for the exact returned score of prediction vector \(v\). First query \(v=(0,0,0,0)\), giving \(s_0=S(0,0,0,0)\). This baseline is the number of zero labels.

Recover labels sequentially. Suppose \(y_1,\ldots,y_{i-1}\) are already known. Submit the adaptive vector \(v^{(i)}=(y_1,\ldots,y_{i-1},1,0,\ldots,0)\), where position \(i\) is 1 and all later positions are 0, and observe score \(s_i\). The hypothetical prediction vector with the same recovered first i−1 labels but with position i still zero would score exactly \(s_0+\sum_{j<i}y_j\): changing a previously recovered position from the baseline 0 to its correct label adds one point only when that label is 1.

Thus the new score difference is

\(d_i=s_i-[s_0+\sum_{j<i}y_j]=+1\text{ if }y_i=1,\text{ and }-1\text{ if }y_i=0.\)

So \(y_i=(d_i+1)/2\). Apply this for \(i=1,2,3,4\). The method uses one baseline and four further queries, each informed by the earlier answers; it identifies every one of the \(2^{4}=16\) possible label vectors. Since the four labels are independent fair bits, their entropy is \(H(Y)=4\) bits. Because the transcript is a deterministic function of Y and uniquely determines Y, the mutual information is \(I(Y;\mathrm{transcript})=H(Y)=4\) bits. The exact scores have revealed the entire hidden target vector despite never displaying individual labels.

Now submit the classifier that remembers the four identifier–label pairs. Its validation error on the reused holdout is \(0/4=0\). For a genuinely fresh identifier, its label remains an independent \(\operatorname{Bernoulli}(1/2)\) random variable, so any prediction based only on previous labels and identifiers errs with probability \(1/2\). The expected fresh-sample error is 50%, not zero. A memorizing model can therefore have perfect apparent validation performance with no generalizable signal. Reusing a supposedly secret holdout with informative score feedback is not an independent final test.