Citadel Securities Interview Questions 2026

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Citadel Securities Quant Interview Practice

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A concert ticket costs £80 for admission plus a compulsory £20 booking charge. The box office advertises a 25% student discount, but the discount applies only to admission; the booking charge is never reduced.

What percentage less does a student actually pay for the complete ticket, compared with a customer paying full price? Explain which price belongs in the denominator.

Get hint

Hint 1

Separate the £80 admission price from the fixed £20 charge before applying the concession.

Hint 2

Compute the cash saved, then divide by the original complete bill, not by the admission-only price.

Show solution

The undiscounted total is $\text{£}80+\text{£}20=\text{£}100$. Only the admission component receives the 25% reduction, so discounted admission is $\text{£}80\times(1-0.25)=\text{£}60$, and the booking charge stays £20. The student pays $\text{£}60+\text{£}20=\text{£}80$, saving £20.

The effective reduction on the actual total is $\text{£}20/\text{£}100=0.20=20\%$. Dividing by £80 would give 25%, which correctly describes the admission discount but not the reduction in the final ticket bill. The fixed fee dilutes the advertised percentage. The two totals also cross-check the answer: $\text{£}100\times(1-0.20)=\text{£}80$.

Assumptions

The booking charge is compulsory and identical for students and non-students.
The student discount is exactly 25% of the admission component only.

Final Answer

20% off the complete £100 bill: the student pays £80 rather than £100.

Numerical reasoning and estimation

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An exhibition display can draw power through either of two routes. The main route has one component that works throughout the day with probability 90%. A spare route has a switch and a transformer in series: both must work, with probabilities 80% and 75% respectively.

All three components fail or work independently, and the routes share no hardware. The display stays powered if at least one entire route works. What is the probability it stays powered all day?

Get hint

Hint 1

The spare route works only if its switch and transformer both work.

Hint 2

It is easier to calculate the probability that the main route fails and the spare route also fails, then take the complement.

Show solution

Let $M$ mean the main component works, and $S$ and $T$ mean the spare switch and transformer work. The main route succeeds with probability 0.90. Because the spare components are independent, the spare route succeeds with probability $0.80\times0.75=0.60$, so it fails with probability 0.40.

The main component is independent of both spare components, so the two route-success events are also independent. The display loses power only if both routes fail:

$\Pr(\text{power lost})=(1-0.90)(1-0.60)=0.10\times0.40=0.04.$

Therefore $\Pr(\text{powered})=1-0.04=0.96$.

As a component-state check, both routes fail either when the main and spare switch fail (transformer arbitrary), probability $0.10\times0.20=0.02$, or when the main fails, switch works and transformer fails, probability $0.10\times0.80\times0.25=0.02$. The disjoint cases total 0.04. It would be wrong to treat the spare switch working on its own as sufficient for an alternative power route.

Assumptions

The three component working events are mutually independent.
Each probability refers to the component working for the entire specified day.
Either complete route suffices to power the display, with no shared failure point.

Final Answer

96%.

Probability foundations

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Ava, Ben, Cleo, Dev and Eli are each assigned to exactly one of five numbered workstations, with every possible one-to-one assignment initially equally likely. The coordinator seals the assignment and later checks four statements against it. The reports are guaranteed correct:

True: Ava was assigned station 2.
False: Ben was assigned station 1.
False: Cleo was assigned station 3.
False: Dev was assigned station 1.

No reassignment occurs after the checks. Given these four reports, what is the probability that Eli is at station 1? Explain why multiplying four separate report probabilities would not give the right conditional probability.

Get hint

Hint 1

Fix Ava at station 2 and count permutations of the remaining four people among the remaining stations.

Hint 2

For the denominator, exclude assignments with Ben or Dev at station 1 and those with Cleo at station 3; overlaps in these exclusions require inclusion–exclusion.

Show solution

The initial $5!$ bijections are equally likely. Conditioning on $\text{Ava}=2$ leaves $4!=24$ equally likely assignments of Ben, Cleo, Dev and Eli to stations $\{1,3,4,5\}$.

Within these 24, define the forbidden events $B_1$: $\text{Ben}=1$, $D_1$: $\text{Dev}=1$, and $C_3$: $\text{Cleo}=3$. Each excludes $3!=6$ assignments. $B_1\cap D_1$ is impossible, since one workstation cannot contain two people. The intersections $B_1\cap C_3$ and $D_1\cap C_3$ each have $2!=2$ assignments; all triple intersections are impossible. Hence inclusion–exclusion gives
$N_{\rm survive}=24-(6+6+6)+(2+2)=10.$

For the requested event $\text{Eli}=1$, the forbidden $\text{Ben}=1$ and $\text{Dev}=1$ reports are automatic. Assign Ben, Cleo and Dev to stations 3,4,5: there are $3!=6$ arrangements, of which $2!=2$ place Cleo at the forbidden station 3. Thus $N_{E_1}=6-2=4$, and
$\Pr(Eli=1\mid\text{reports})=\frac4{10}=\boxed{\frac25}.$

The reports are not independent observations of separate random assignments. For example, if Ben occupied station 1 then Dev could not, even before conditioning on the other reports. We condition on the intersection of reports by counting compatible complete bijections, not by multiplying marginal exclusion factors.

Assumptions

The original assignment is a uniformly random bijection, not five independent workstation choices.
The four report outcomes are exact observations of that one fixed assignment, with no reporter error.
Each numbered workstation holds exactly one volunteer and each volunteer occupies exactly one workstation.

Final Answer

$2/5$. Exactly 10 one-to-one assignments survive the four reports; in 4 of them Eli occupies station 1.

Conditional probability

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A portrait artist has three commissions. A small sketch takes 1 hour and pays $120, a charcoal portrait takes 2 hours and pays $180, and a full portrait takes 3 hours and pays $240. The artist starts at noon and attempts the three jobs in a uniformly random order, working continuously until closure.

The venue will unexpectedly close $T$ hours after noon, where $T$ is uniform on $[0,6]$ and independent of the job order. The artist keeps payment for a commission only if that particular portrait is fully finished by closing; an interrupted portrait earns nothing, and later jobs cannot be completed.

What is the expected total amount the artist receives?

Get hint

Hint 1

For any fixed order, a portrait is paid if the shared closing time exceeds its cumulative finish time. Take its payment times that survival probability.

Hint 2

At any finish time $s\in[0,6]$, $P(T\ge s)=1-s/6$. Average the paid receipts over the six equally likely job permutations; do not assign independent closing events to different portraits.

Show solution

For a portrait finishing at time $s$, the probability of payment is $P(T\ge s)=1-s/6$. The single closing time applies to every portrait, so completion indicators are dependent, but linearity of expectation still lets us sum their expected payments.

The six orders and their conditional expected receipts (in dollars) are:

$\begin{array}{c|rrrrrr} \text{Order}&ABC&ACB&BAC&BCA&CAB&CBA\\\hline E[\text{receipts}\mid\text{order}]&190&180&180&160&160&150 \end{array}$

For example, order ABC finishes jobs at hours 1, 3, and 6. Its expectation is $120(5/6)+180(3/6)+240(0)=\$190$. Every order has probability $1/6$, hence

$E[\text{receipts}]=\frac{190+180+180+160+160+150}{6}=\boxed{\$170}.$

Equivalently, averaging finish-time survival probabilities across permutations gives completion probabilities $5/12$ for A, $1/3$ for B, and $1/4$ for C. The independent check is $120(5/12)+180(1/3)+240(1/4)=50+60+60=170$. Neither calculation assumes independent payment events.

Assumptions

All six orders are equally likely, independently of the single uniform closing time.
A portrait pays its full fee only if finished by closure; equality has probability zero.
Durations are deterministic and jobs cannot be resumed after closure.

Final Answer

The expected paid receipts are $\$170$.

Expectation and expected value

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A ferry company pays a forecaster to label its weather instrument standard or premium before deciding whether to run a crossing. The forecaster knows whether the instrument's readings are reliable (type H) or unreliable (type L), but the ferry operator sees only the label. Initially both types are equally likely.

The operator can sail or cancel. Its expected net payoff from sailing is +$4 if the instrument is H and −$6 if it is L; cancellation pays $0. The forecaster receives a $4 fee if the operator sails, regardless of the true type. Sending the standard label costs nothing. Producing a premium-labelled report requires extra checking: it costs the forecaster $1 if H, but $5 if L. Everyone knows these figures; label choice itself is not directly verified.

Treat the interaction as a signaling game. Find a separating perfect Bayesian equilibrium and a pooling perfect Bayesian equilibrium, specifying the operator's beliefs and actions. Can both types pool on the premium label in a pure-strategy equilibrium? Explain the incentive and off-path-belief checks.

Get hint

Hint 1

If the operator believes the probability of a reliable instrument is $\mu$, compute its expected payoff from sailing. Then compare $\mu$ with the break-even posterior.

Hint 2

For separation, test whether an L forecaster would pay $5 to imitate H for a $4 sailing fee. For standard-label pooling, specify what the operator would believe about an unexpected premium label.

Show solution

Let $\mu$ be the operator's posterior probability of type $H$ after observing a label. Sailing yields $4\mu-6(1-\mu)=10\mu-6$ dollars, so the operator sails for $\mu>\frac35$, cancels for $\mu<\frac35$, and is indifferent at $\mu=\frac35$.

Separating equilibrium. Let $H$ choose premium and $L$ choose standard. Since both types have prior probability $\frac12$ and both labels occur, Bayes' rule gives $\mu(H\mid\text{premium})=1$ and $\mu(H\mid\text{standard})=0$. The operator therefore sails after premium and cancels after standard. Forecaster $H$ earns $\$4-\$1=\$3$ using premium, versus $0 from deviating to standard. Forecaster $L$ earns $0 on standard, versus $\$4-\$5=-\$1$ by deviating to premium. Both sender incentive constraints hold. There are no off-path labels in this separating profile.

Pooling equilibrium. Suppose both types choose standard. On that reached path, Bayes' rule requires $\mu(H\mid\text{standard})=\frac12$; the operator cancels since $10(\frac12)-6=-1$. Premium is off path, so its posterior is not fixed by Bayes' rule. Assign $\mu(H\mid\text{premium})=0$ and let the operator cancel there too. Neither sender gains by deviating: premium would produce no fee and cost H $1 or L $5. More generally, any off-path posterior at most $\frac35$ supports cancellation (at equality, choosing cancellation is optimal); if premium instead induced sailing with certainty, H would deviate for a positive $3 payoff.

No premium pooling. If both types choose premium, the operator's on-path belief must remain $\frac12$ and therefore it cancels. Each forecaster would then pay a positive premium-label cost for no sailing fee; switching to free standard earns at least $0, regardless of the operator's off-path action. Thus neither type prefers premium pooling. Reverse separation (L premium, H standard) also fails: the costly L report would lead to cancellation, but L could switch to the standard label, trigger sailing and earn $4. These are pure-strategy pooling/separating conclusions; they do not rule out every possible mixed signaling assessment.

Assumptions

Both types are privately known by the forecaster; no third type or direct verification follows a label.
Payoffs are expected monetary payoffs conditional on quality; agents maximise expected money.
The operator observes only the public label; labels are chosen before the sail/cancel decision.
Beliefs at off-path labels are permissible if they make the operator sequentially rational.

Final Answer

Separating PBE: H sends premium, L sends standard; the operator sails only after premium. Pooling PBE: both send standard, the operator cancels after standard and also cancels after an off-path premium label when its belief $\Pr(H\mid\text{premium})\le\frac35$ (for example, zero). Pooling on premium cannot be a pure-strategy equilibrium.

Game theory and strategic decisions

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A dealer lists identical camera lenses for $40 each and lets a collector order zero, one or two. The collector has privately assessed the resale value V of each lens; from the dealer's perspective V is uniformly distributed between $0 and $100. Buying q lenses gives the collector resale surplus $q(V-40)$, but keeping multiple lenses also incurs a storage and insurance cost of $10$q^{2}$. The collector chooses q to maximise net surplus and, when indifferent, chooses the smaller order.

The dealer sees the order quantity but not V. Which values of $V$ produce each order size, and what should the dealer infer about the expected value of a lens after observing $q=0$, 1 or 2?

Get hint

Hint 1

Write the collector’s net surplus for each permitted quantity. Find where neighbouring choices have equal surplus.

Hint 2

Check that neither non-neighbouring choice dominates the candidate in each interval, then condition a uniform distribution on the resulting value intervals.

Show solution

The three net surpluses in dollars are $U_{0}=0$, $U_{1}=V-40-10=V-50$, and $U_{2}=2(V-40)-40=2V-120$. The first lens becomes worthwhile at $V=50$: $U_{1}-U_{0}=V-50$. Adding the second instead of stopping at one changes surplus by $U_{2}-U_{1}=V-70$. Hence $q=0$ is optimal at $V\le50$, $q=1$ at $50<V\le70$, and $q=2$ above 70. The non-adjacent comparison $U_{2}-U_{0}=2V-120$ cannot overturn this: for $V\le50$ it is negative, and for $50<V\le70$ the one-lens choice exceeds both other options. At $V=50$ or 70 there is a tie and the collector chooses the smaller quantity, as specified.

Conditional on $q=0$, V is uniform on $[0,50]$, giving $\mathbb{E}[V\mid q=0]=25$. After $q=1$ it is uniform on $(50,70]$, giving $\mathbb{E}[V\mid q=1]=60$. After $q=2$ it is uniform on $(70,100]$, giving $\mathbb{E}[V\mid q=2]=85$. The respective order probabilities are $\frac{1}{2}$, $\frac{1}{5}$ and $\frac{3}{10}$; their conditional means reproduce the original mean: $(\frac{1}{2})25+(\frac{1}{5})60+(\frac{3}{10})85=50$. The quantity is therefore an endogenous graded signal of the buyer's private assessment, not an externally assigned demand class.

Assumptions

The buyer knows V and is risk-neutral with the stated deterministic storage cost.
V applies to both identical lenses, and the dealer treats it as continuous Uniform(0,100).
The only allowed quantities are 0, 1 and 2; there are no other taxes, fees or strategic participants.

Final Answer

Order 0 for $V\le\$50$; order 1 for $\$50<V\le\$70$; order 2 for $V>\$70$. Conditional mean values are $25, $60 and $85 respectively (ties at $50 and $70 are probability-zero events).

Market making and microstructure

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Three businesses must settle six invoices on the same day (amounts in dollars): A owes B 90 and C 30; B owes A 50 and C 80; C owes A 40 and B 20. They can use (i) gross settlement, paying every invoice separately; (ii) bilateral netting, cancelling opposite invoices between each pair; or (iii) an enforceable multilateral clearing agreement, replacing all invoices with each business's single net obligation to or from the clearing pool.

If B is solvent, everyone pays in full. If B defaults, a contractual insolvency rule applies after the chosen netting: creditors collect only 50% of each remaining positive payment owed by B. All A and C obligations are paid in full, with no secondary defaults. Payments made to B enter its insolvency estate and cannot be redirected to cure these shortfalls because senior claims have priority. Compare total promised settlement payments and aggregate creditor losses in the solvent and default states. Count cash settled once per obligation; under multilateral clearing, report money paid into the pool, not that same money again as it is distributed out. If B defaults with probability 20%, what is expected creditor loss under each arrangement?

Get hint

Hint 1

For each method, establish the legally enforceable payment obligations before applying the 50% recovery. Gross, pairwise and clearing-wide offsets are not equivalent.

Hint 2

The bilateral obligations reduce to A→B 40, B→C 60, C→A 10. For multilateral netting, compare each business's total payables with its total receivables.

Show solution

Gross invoices. The six amounts total $90+30+50+80+40+20=310$. If B defaults, its outgoing amounts 50 to A and 80 to C suffer 50% shortfalls, so creditor loss is $25+40=65$. The other invoices are still due and paid.

Bilateral netting. Opposite flows cancel pairwise: A owes B $90-50=40$; B owes C $80-20=60$; C owes A $40-30=10$. Promised cash volume is $40+60+10=110$. B's one remaining debt is 60 to C; at 50% recovery, C loses $30$. A's separate 40 obligation to B still settles.

Multilateral clearing. Before default, A owes 120 and is owed 90, so pays net 30. B owes 130 and is owed 110, so pays net 20. C owes 60 and is owed 110, so receives net 50. The clearing pool receives $30+20=50$ and pays 50; nominal cash volume counted as payer-to-pool contributions is 50, not 100 (which would double-count the pool's redistribution). If B defaults, it pays 10 instead of 20, and C receives 40 instead of 50, a creditor loss of 10.

When B is solvent, creditor losses are zero under all three methods. With default probability $0.2$, expected loss is respectively $0.2(65)=13$, $0.2(30)=6$ and $0.2(10)=2$. Netting reduces the legally exposed amount but cannot conjure the missing $10 when B's final net debt is only half recovered. The specified bankruptcy priority is essential: absent an enforceable netting and recovery order, amounts due to and from B could be treated differently.

Assumptions

All figures are nominal dollars and obligations are enforceable under the specified settlement method.
Only B may default; its 50% recovery applies to positive obligations after legal netting.
Receipts paid to B cannot be used to make up the 50% unpaid creditor claims in this stylized priority arrangement.
The multilateral cash volume counts contributions into the pool once, not both sides of clearing.

Final Answer

Solvent gross payment volumes: $310, $110, $50 (gross, bilateral, multilateral). When B defaults, creditor losses: $65, $30, $10; expected losses at 20% default: $13, $6, $2, respectively.

Risk, covariance and diversification

A restaurant orders a 100 kg sack of flour at £4 per kilogram, so the invoice is £400. On arrival the sack contains only 90 kg. Of that delivered flour, a…

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