Da Vinci Trading Interview Questions 2026

Drinking needs \(20\times3=60\) L. Cooking needs 30 L. Dishwashing itself needs 50 L of initially fresh water. Those uses consume \(60+30+50=140\) L of new supply before floor cleaning.

Recovery provides \(0.60\times50=30\) L of dishwater that may be reused only for floors. Since floors need 40 L, they still require \(40-30=10\) L of fresh water. Therefore
\[ W_{\mathrm{new}}=60+30+50+\max(40-30,0)=\boxed{150\text{ L}}. \]
The same answer follows by balancing total uses: \(60+30+50+40=180\) L of water demand, of which 30 L is met by water already supplied for dishwashing, leaving 150 L of external supply. This recovered 30 L must not be counted again as new water. Nor can it offset drinking or cooking, which require fresh water.

Containers cannot be divided, so the minimum number is \(\lceil150/20\rceil=\boxed{8}\). The site receives 160 L and has 10 L spare if forecasts are exact.

The organiser has only one decision variable, \(q=P(\text{use machine A})\), shared by both sponsor outcomes. By conditioning on the selected machine,
\[ P(G)=0.8q+0.2(1-q)=0.2+0.6q, \qquad P(B)=0.2q+0.6(1-q)=0.6-0.4q. \]
To promise a 50% Gold rate, \(0.2+0.6q=0.5\) requires \(q=0.3/0.6=\boxed{1/2}\). But the 50% Blue rate requires \(0.6-0.4q=0.5\), or \(q=0.1/0.4=\boxed{1/4}\).

Both weights are individually feasible, but they differ. A machine-selection policy cannot simultaneously use machine A with probability 1/2 and 1/4. If the organiser honours Gold, Blue's rate would be \(0.6-0.4(1/2)=0.4\), only 40%. If they honour Blue, Gold's rate would be \(0.2+0.6(1/4)=0.35\), only 35%.

At the permitted endpoints \(q=0\) and \(q=1\), the output pairs are \((\text{Gold }20\%,\text{ Blue }60\%)\) and \((\text{Gold }80\%,\text{ Blue }20\%)\). Every achievable pair lies on the line segment connecting those points; \((50\%,50\%)\) is not on that segment. Thus the promises are incompatible under the one-parameter policy, even though each separate target is attainable.

Let \(E\) and \(D\) denote easy and difficult parcels, and \(S\) mean on-time completion. Under the current policy,

\(\Pr(E\cap U)=0.7(0.2)=0.14\), \(\Pr(D\cap U)=0.3(0.8)=0.24\), so \(\Pr(U)=0.38\).

The joint probability of urgent and on time is \(0.14(0.95)+0.24(0.60)=0.133+0.144=0.277\). Conditional on observing an urgent dispatch, the success probability is consequently

\(\Pr(S\mid U)=0.277/0.38=\boxed{277/380\approx0.72895}\).

Under a policy that forces urgent delivery for every parcel, the type mix stays at 70% easy and 30% difficult. By the problem's explicitly stated policy-invariance rule, the success probability is

\(\Pr(S\mid \operatorname{force}U)=0.7(0.95)+0.3(0.60)=0.665+0.180=\boxed{0.845}\).

These expressions use different denominators and weights. Observed urgent parcels are only \(0.14/0.38=7/19\) easy, whereas the entire population is \(7/10\) easy. The currently urgent group is selected towards difficult cases; forcing the action changes that composition. For an additional consistency check, routine parcels have current success rate \([0.7(0.8)(0.85)+0.3(0.2)(0.30)]/0.62=0.494/0.62\approx0.7968\), even though urgent delivery improves the success rate within both types. This reversal is a selection effect, not evidence that urgent service is harmful. No unmeasured causal premise is needed because the problem supplies the policy-invariant type-specific rates explicitly.

Let \(T\in\{1,2,\ldots\}\) be the first post-due-date day on which the borrower returns the book. Conditional daily return probability \(p=1/4\) makes T geometric: \(P(T=t)=(3/4)^{t-1}(1/4)\). Because the return day is excluded, there are \(T-1\) chargeable complete days, but the fee stops at four days' worth. Define \(F=3\cdot\operatorname{min}(T-1,4)\) dollars.

For a nonnegative integer N, \(\operatorname{min}(N,4)=\sum_{k=1}^{4}1\{N\ge k\}\). Taking expectations gives the tail-sum identity
\[\begin{aligned}E[F]&=3\sum_{k=1}^{4}P(T-1\ge k)=3\sum_{k=1}^{4}(3/4)^{k}\\&=3[(3/4)(1-(3/4)^{4})/(1-3/4)]\\&=9(1-81/256)=1575/256\end{aligned}\] dollars ≈ $6.15.

For an independent direct distribution check, the possible fees are $0 for \(T=1\), $3 for \(T=2\), $6 for \(T=3\), $9 for \(T=4\) and $12 for \(T\ge5\). Thus
\[E[F]=3(3/16)+6(9/64)+9(27/256)+12(81/256)=1575/256\].

The cap limits charges, not the stochastic time until return. Counting the return day would shift the whole distribution and produce the wrong fee.

Work in thousands of dollars. Let \(e\) be the probability of a challenge and \(d_L,d_H\) the low- and high-cost incumbent mobilization probabilities. If mobilization occurs, the incumbent retains its profit of 8 and pays cost \(c\), whether or not entry occurs. If it does not mobilize, it retains 8 only when there is no entry. Thus the expected incremental value of mobilization for type \(c\) is
\[ (8-c)-8(1-e)=8e-c. \]
The low type \(c_L=2\) defends whenever \(e>1/4\); the high type \(c_H=4\) defends whenever \(e>1/2\), and each is indifferent at its threshold.

The challenger's expected payoff from filing (before mixing) is
\[ 10(1-q)-3,\qquad q=0.4d_L+0.6d_H, \]
where \(q\) is the unconditional chance that the incumbent mobilizes. Not filing pays zero. Indifference requires \(q=0.7\).

At \(e=1/2\), the low-cost incumbent strictly mobilizes, so \(d_L=1\); the high type is indifferent and may mix. Setting \(0.4+0.6d_H=0.7\) yields \(d_H=1/2\). The challenger then receives \(10(0.3)-3=0\) if it files and 0 otherwise, validating its \(e=1/2\) mix. The low-cost type's defense advantage is \(8(1/2)-2=2>0\), so it never prefers to deviate. The high-cost type's advantage is \(8(1/2)-4=0\), allowing its 50% mixture.

These are the only equilibrium probabilities: for \(e<1/4\) both types prefer not to defend, making filing strictly attractive; for \(1/4<e<1/2\) only the low type defends, giving \(q=0.4\) and still making filing attractive; for \(e>1/2\) both defend, giving \(q=1\) and making filing unattractive. At \(e=1/4\), even maximal low-type defense gives \(q\le0.4<0.7\). Consequently only the high-type indifference point \(e=1/2\) can support the entrant's mixing. The positive mobilization cost paid without entry is what makes the defense types' thresholds depend on the filing probability.

For each type, apply the stipulated decision order: buy if \(a\) is within the type's buying limit; otherwise sell if \(b\) meets its selling threshold; otherwise leave. The dealer's profit from a purchase is \(100-b\) and from a sale is \(a-100\). Since there is exactly one customer, these actions are mutually exclusive.

Enumerating the three bids and three asks gives the following expected profits (pounds per customer), with rows \(b=90,94,98\) and columns \(a=102,106,110\):

For example, at \(b=\text{£90}\) and \(a=\text{£106}\), type I sells to the dealer for £90, yielding £10; types II and III buy for £106, yielding £6 each. Their probabilities are \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{1}{6}\), so expected profit is \(\left(\frac{1}{2}\right)\times10+\left(\frac{1}{3}\right)\times6+\left(\frac{1}{6}\right)\times6=5+2+1=\text{£8}\).

At ask £102 every type buys, earning £2 regardless of the bid. At ask £110, type III buys for a £10 margin; types I and II may sell at the bid if their thresholds are met. Applying those possibilities gives the last column. Every other feasible pair earns at most £\(\frac{20}{3}\) or £6, so £8 is a strict global maximum and \((90,106)\) is the unique optimal quote pair.

A model that independently added probabilities of selling at the bid and buying at the ask could double count types for which both quotes are appealing (for instance, type II at bid £94 and ask £106). The customer's buy-first choice, not two independent order streams, determines the actual trade.

Let W be the bankroll immediately before a round. A loss gives \(W(1-f)\). A win gives \(W[1+\operatorname{min}(f,0.20)]\), because the cash bonus is capped at \(0.20W\) and the stake itself is returned. For \(0\le f<1\), expected log growth relative to log W is

\[g_{p}(f)=p\operatorname{log}[1+\operatorname{min}(f,0.20)]+(1-p)\operatorname{log}(1-f).\]

For \(0\le f\le0.20\) this is the ordinary even-money Kelly objective, with derivative \(g^{\prime}_{p}(f)=\frac{p}{1+f}-\frac{1-p}{1-f}=\frac{(2p-1)-f}{1-f^{2}}\), and strictly negative second derivative wherever \(0<p<1\). Its unconstrained maximiser is \(2p-1\). For \(f>0.20\), the win term is constant but the loss penalty continues to increase; indeed \(g^{\prime}_{p}(f)=-\frac{1-p}{1-f}<0\). Therefore staking more than 20% can never be optimal. On \([0,0.20]\) the concave objective is maximised by clipping the unconstrained value to that interval: \(f^{*}=\operatorname{max}(0,\operatorname{min}(2p-1,0.20))\). This is a global, not merely local, optimum because the function decreases strictly above the cap.

With \(p=0.70\), the uncapped Kelly fraction would be 0.40, so the cap binds: \(f^{*}=0.20\), at the kink where the left derivative is positive and the right derivative negative. If \(p\le\frac{1}{2}\), \(f^{*}=0\); if \(\frac{1}{2}<p<\frac{3}{5}\), \(f^{*}=2p-1\); if \(p\ge\frac{3}{5}\) (and \(p<1\)), \(f^{*}=0.20\). A fraction \(f=1\) has a strictly positive chance \(1-p\) of zero final wealth, hence expected log wealth \(-\infty\) and cannot be optimal. No borrowing or shorting is allowed, so negative f and \(f>1\) are excluded. This repeated-round solution depends on the cap scaling with current bankroll; a fixed dollar cap independent of wealth would produce a different dynamic problem.