Old Mission Capital Interview Questions 2026

For \(n\) paid loaves, the number of qualifying groups of four is \(\lfloor n/4\rfloor\). Since the baker never treats free bread as additional paid bread, the delivered total is exactly

\[D(n)=n+\left\lfloor\frac n4\right\rfloor.\]

The demand requires \(D(n)\ge24\). Paying for 19 gives \(D(19)=19+4=23\), which is insufficient. Paying for 20 gives \(D(20)=20+5=25\), which meets demand with one extra loaf. Because \(D(n)\) is nondecreasing in the nonnegative integer \(n\), failure at 19 proves that no smaller order works. The minimum is therefore \(n=20\), costing \(20(\$6)=\boxed{\$120}\), and it delivers 25 loaves, leaving one spare.

The fifth free loaf appears only at paid quantity 20. Counting previous free loaves towards another four-loaf group would change the promotion's rules and incorrectly lower the paid count.

Let \(S\) denote one of the 24 complete one-to-one assignments, initially uniform, and let \(G\) be the independent random reshuffle. Each possible value \(g\) of \(G\) is a bijection on the set of assignments. Therefore, for any specified final assignment \(s\), the equation \(g(S)=s\) has exactly one solution \(S=g^{-1}(s)\). Conditional on any fixed reshuffle,
\[ P(g(S)=s\mid G=g)=P(S=g^{-1}(s))=\frac1{24}. \]
Independence lets us use the same original uniform law in every condition. Averaging over the uneven reshuffle lottery gives
\[ P(G(S)=s)=0.7\frac1{24}+0.2\frac1{24}+0.1\frac1{24} =\boxed{\frac1{24}}. \]
Thus the probability distribution over final assignments is uniform; the lottery over permutations does not need to be uniform. The relevant property is that each allowed mapping is invertible and chosen without looking at the initial assignment.

If the starting assignment is fixed, this lottery can produce at most three final assignments, not all 24, so the final distribution is not uniform. Independence also matters: if the reshuffle were selected because of a particular initial assignment, conditioning on that reshuffle could distort the original uniform law. The preservation result applies to an independent mixture of bijections on an initially uniform state space.

Let \(F_{1}\) be first-attempt failure and \(S_{2}\) be second-attempt success. Initially 3 of 5 keys fail and 2 succeed, giving \(P(F_{1})=3/5\) and \(P(F_{1}^{c})=2/5\).

If the first key fails, it is one of the three nonworking keys and is removed. Four keys remain, of which two work: \(P(S_{2}\mid F_{1})=2/4=1/2\). If the first key succeeds, it is returned, so all five keys remain with two working ones: \(P(S_{2}\mid F_{1}^{c})=2/5\). The full probability of second success is

\[P(S_{2})=(3/5)(1/2)+(2/5)(2/5)=3/10+4/25=15/50+8/50=23/50\].

The joint probability \(P(F_{1}\cap S_{2})=(3/5)(1/2)=3/10=15/50\). Bayes' rule gives \(P(F_{1}\mid S_{2})=(15/50)/(23/50)=15/23\approx0.65217\). This posterior exceeds the prior failure probability 3/5 because a failed first attempt removes a dud and raises the second attempt's chance of success. Assuming the second attempt is always a fresh one-of-five draw would miss the result-dependent change in sample space.

Let \(N\) be the number of optional riders. The four responses are independent \(\operatorname{Bernoulli}(1/2)\), so \(N\sim\mathrm{Binomial}(4,1/2)\). Because you are guaranteed to ride, the actual occupancy is \(1+N\), never zero, and your fare is \(80/(1+N)\).

By the binomial distribution,
\[ E\!\left[\frac{80}{1+N}\right] =\frac{80}{16}\left(\frac{\binom40}{1}+\frac{\binom41}{2} +\frac{\binom42}{3}+\frac{\binom43}{4}+\frac{\binom44}{5}\right). \]
The bracket is \(1+2+2+1+1/5=31/5\), and \(80/16=5\). Thus the expected payment is \(5\times31/5=\boxed{\$31}\).

For an independent analytic check, use \(1/(k+1)=\int_0^1x^k\,dx\):
\[ E\!\left[\frac{1}{1+N}\right] =\int_0^1\left(\frac{1+x}{2}\right)^4\!dx =\frac{1-(1/2)^5}{5(1/2)}=\frac{31}{80}. \]
Multiplying by $80 again gives $31.

The average passenger count is \(E[1+N]=1+4(1/2)=3\). Dividing by that mean gives \(\$80/3\approx\$26.67\), not the average of the actual fare shares. In general, \(E[1/K]\ne1/E[K]\) when occupancy \(K\) varies. Low-occupancy rides produce disproportionately large shares, which raises your expected bill.

1. Optimal race-2 decisions. Whether the opponent uses a specialist or not, adding your own specialist raises the probability that you win race 2 by exactly 1/2: the win probability changes either from 1/2 to 1 (if the opponent uses no specialist) or from 0 to 1/2 (if they do). This is true even when the opponent's first-race lineup is hidden.

If you won race 1, winning race 2 changes your prize from $30 (a split) to $100 (a sweep). The expected marginal prize from fielding a saved specialist is \(\frac{1}{2}(\$100-\$30)\)=$35, more than its $28 cost. You use it. If you lost race 1, winning race 2 changes your prize from $0 to $30. The marginal expected prize is \(\frac{1}{2}(\$30-\$0)\)=$15, less than $28. You bench it. These are strict choices, independent of posterior roster uncertainty.

2. Race-1 strategic payoff matrix. Now use those optimal continuation decisions. Entries below are A's expected net dollars, with A choosing the row and B the column:

To verify each cell:

• Both use: race 1 is a coin toss; neither has a specialist left for race 2, also a coin toss. Each club has a 1/4 chance of a $100 sweep and a 1/2 chance of a $30 split prize. Gross expectation = \(\$25+\$15=\$40\); subtract the $28 first-race cost to obtain $12.
• A uses, B saves: A wins race 1 for certain. B, having lost, benches in race 2; A has no specialist, so race 2 is a coin toss. A's gross prize is \(\frac{1}{2}\cdot\$100+\frac{1}{2}\cdot\$30=\$65\), minus $28 = $37. B receives $30 on half the paths, so B's net is $15.
• A saves, B uses: symmetrically, A loses race 1 and benches in race 2, which is a coin toss. A gets a $30 split prize with probability 1/2, giving $15; B receives $37 net.
• Both save: race 1 is a fair coin toss. Its winner fields their saved specialist in race 2 and sweeps with certainty, paying $28 and collecting $100; the loser benches and collects nothing. Each club's expectation is \(\frac{1}{2}(\$100-\$28)\)=$36.

Suppose the opponent uses their specialist in race 1 with probability p. A earns \(12p+37(1-p)=37-25p\) by using now, versus \(15p+36(1-p)=36-21p\) by saving. Both choices are played with positive probability in a symmetric mixed equilibrium only if

\(37-25p=36-21p\) ⇒ \(p=\frac{1}{4}\).

The common expected payoff is \(37-\frac{25}{4}\)=\(\frac{\$123}{4}=\$30.75\). The off-diagonal first-race choices also form two asymmetric pure equilibria, but \(p=\frac{1}{4}\) is the symmetric mixed equilibrium requested.

3. Correct posterior beliefs after race 1. Suppose A saved their specialist. The equilibrium prior that B used theirs in race 1 is 1/4. A first-race win can occur only when B also saved and a fair coin favoured A: probability \(\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{8}\). Hence

\(\Pr(B\text{ used in race 1}\mid A\text{ saved},\ A\text{ won})\)=0.

A loss occurs either because B used theirs (probability 1/4) or because B saved and won the coin toss (probability \(\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{8}\)). Thus \(\Pr(A\text{ lost}\mid A\text{ saved})=\frac{5}{8}\), and

\(\Pr(B\text{ used in race 1}\mid A\text{ saved},\ A\text{ lost})=\frac{1/4}{5/8}\)=\(\frac{2}{5}\).

These posterior-weighted branches also verify A's expected payoff from saving. If A saved and won race 1 (probability 3/8), they field their specialist, beat the known specialist-free opponent and earn \(\$100-\$28=\$72\). If A saved and lost (probability 5/8), they bench. Conditional on that loss, B already used their specialist with probability 2/5, in which case race 2 is a fair coin and A earns $30 with probability 1/2; otherwise B saved, won race 1 and uses their specialist, so A loses race 2. The loss-branch expectation is \(\frac{2}{5}\cdot\frac{1}{2}\cdot\$30\)=$6. Total saved-strategy expectation = \(\frac{3}{8}\cdot\$72+\frac{5}{8}\cdot\$6=27+3.75\)=$30.75, matching the matrix. No coach has been allowed to condition on an unseen opposing lineup.

If C of the three earlier buy orders cancel, 3−C orders remain ahead of yours. With \(N\) incoming sellable cards and price-time priority, you receive a card exactly when \(N\ge(3-C)+1\), equivalently \(N+C\ge4\).

The count \(C\) is \(\operatorname{Binomial}(3,1/2)\), independent of \(N\). Conditional on \(N=1\), filling requires \(C=3\), with probability 1/8. If \(N=2\), filling requires \(C\ge2\), with probability \(\frac{3+1}{8}=\frac12\). If \(N=3\), filling requires \(C\ge1\), with probability \(1-\frac18=\frac78\). Averaging over the specified N distribution,
\[\begin{aligned}P(\text{fill})&=(1/4)(1/8)+(1/2)(1/2)+(1/4)(7/8)\\&=\frac1{32}+\frac14+\frac7{32}=\frac12\end{aligned}\].

The expected cost of queuing plus the guaranteed backup is \(46\cdot P(\text{fill})+56\cdot P(\text{no fill})=46(1/2)+56(1/2)=\$51\). Immediate execution costs $50 with certainty, so crossing the ask is $1 cheaper in expectation. Cancellations behind your order only reduce the number of later orders; they never remove any order standing between yours and incoming supply.

As an independent state calculation, after processing each of the three orders ahead, let a be the number surviving; each survives with probability 1/2, so \(P(a=0,1,2,3)=\frac{(1,3,3,1)}8\). The arriving-volume tail \(P(N\ge a+1)\) equals 1, 3/4, 1/4, 0 respectively. Their weighted sum is \(\frac18\cdot1+(3/8)(3/4)+(3/8)(1/4)+\frac18\cdot0=\frac12\), confirming the fill calculation.

Let initial wealth be \(W\) and the random bonus be \(X\in\{0,200\}\). The expected utility of retaining the bonus is

\[E[u(W+X)]=-e^{-W/200}\left(\tfrac14+\tfrac34e^{-1}\right).\]

The certainty equivalent \(C\) of \(X\) is defined so that a sure $C produces this same utility: \(-e^{-(W+C)/200}=E[u(W+X)]\). Cancelling \(-e^{-W/200}\) and taking logarithms gives

\[e^{-C/200}=\tfrac14+\tfrac34e^{-1},\qquad C=-200\ln\!\left(\tfrac14+\tfrac{3}{4e}\right)\approx\$128.5252.\]

Replacing \(X\) with a sure $150 net of fee \(f\) is preferred precisely when \(150-f\ge C\), because exponential utility is strictly increasing in wealth. Hence the largest acceptable fee is

\[\boxed{f_{\max}=150-C=150+200\ln\!\left(\tfrac14+\tfrac{3}{4e}\right)\approx\$21.4748}.\]

At equality, the technician is indifferent; a larger fee makes the risky bonus preferable. The result is independent of \(W\): exponential utility with constant absolute risk aversion allows the \(e^{-W/200}\) factor to cancel. For a risk-aversion coefficient \(a>0\), more generally \(C(a)=-a^{-1}\ln E[e^{-aX}]\). As \(a\to0\), the logarithm's first-order expansion gives \(C(a)\to E[X]=0.25(0)+0.75(200)=\$150\). The willingness to pay for a sure $150 therefore tends to zero in the risk-neutral limit. The positive fee here is a genuine risk premium, not an expected-value improvement.