Jump Trading Interview Questions 2026

A width labelled \(k\) cm can be any value in the half-open interval \([k-0.5,k+0.5)\) cm: for instance, a width of exactly 18.5 cm would be labelled 19, not 18. The four true widths therefore lie respectively in
\[ [17.5,18.5),\quad [20.5,21.5),\quad [23.5,24.5),\quad [14.5,15.5). \]
Their total \(T\) lies in \([76,80)\) cm. All four rounding errors can point in the same direction, so adding only the displayed widths would not give a guarantee. Nevertheless, the upper endpoint 80 is excluded by the stated tie rule. Thus \(T<80\) for every admissible set of widths, and the 80 cm shelf is guaranteed to work.

The shorter 79 cm shelf is not guaranteed: widths 18.4, 21.4, 24.4 and 15.4 cm all round to their respective labels, yet total \(79.6\) cm. Conversely, the labels themselves could be exact true widths, totalling 78 cm, which would fit. Hence the 79 cm fit is genuinely undetermined by the measurements. The tie rule determines whether the total can equal 80 cm; either way, the 80 cm shelf fits the frames.

Fix Noor's location; rotational symmetry means this does not change the probability. The unordered set of her two neighbours is uniformly distributed among \(\binom{11}{2}=55\) possible pairs. Of the other eleven guests, seven are not trained. Both neighbours are untrained in \(\binom{7}{2}=21\) pairs. Hence
\[ \Pr(\text{at least one trained neighbour}) =1-\frac{\binom72}{\binom{11}2} =1-\frac{21}{55}=\boxed{\frac{34}{55}}. \]
An ordered cross-check gives \(1-(7/11)(6/10)=34/55\). Multiplying \((7/11)^2\) would be wrong: after one untrained neighbour is seated, only six untrained people remain among ten available guests for the second adjacent seat.

Write \(X\) for A's arrival time and \(Y\) for B's. Their joint density is constant, \(\frac1{2\times3}=\frac16\), on \(0\le x\le2\) and \(0\le y\le3\). Let \(C\) be \(\lvert X-Y\rvert\le1\). Conditional probabilities are ratios of the corresponding areas, since the density cancels.

For A to be first, we need \(x<y\le x+1\). For every \(x\in[0,2]\), the allowed Y-interval \((x,x+1]\) has width 1 and stays inside [0,3]. Its area is therefore \(\int_{0}^{2}1\,dx=2\).

For B to be first, the allowed Y-interval is \([\max(0,x-1),x)\), with width \(x\) for \(0\le x\le1\) and width 1 for \(1\le x\le2\). Its area is \(\int_0^1x\,dx+\int_1^2 1\,dx=\frac12+1=\frac32\).

These disjoint ordered regions partition \(C\) apart from a zero-area diagonal. Hence \(\operatorname{area}(C)=2+\frac32=\frac72\) and \(\Pr(X<Y\mid C)=\frac{2}{7/2}=\frac47\). As a cross-check, \(\Pr(C)=\frac{7/2}{6}=\frac7{12}\) and \(\Pr(X<Y\text{ and }C)=\frac26=\frac13\); \(\frac{1/3}{7/12}=\frac47\).

Let \(C_n\) be the number of cycles, including singletons, in a uniformly random permutation of \(n\) labels. Starting from a uniform permutation on \(n-1\) labels, insert label \(n\) at one of \(n\) equally likely cycle slots. One slot forms a new one-element cycle; each of the other \(n-1\) inserts \(n\) into an existing cycle. This construction yields a uniform permutation on \(n\) labels, and therefore

\(\mathbb E[C_n]=\mathbb E[C_{n-1}]+1/n\), with \(\mathbb E[C_1]=1\).

Induction gives \(\mathbb E[C_6]=H_6=1+1/2+1/3+1/4+1/5+1/6=49/20\). But a fixed point needs no tray. For each parcel \(i\), the probability that its tag remains correct is exactly \(1/6\); summing the six indicators yields an expected one fixed point, irrespective of dependence between them. Therefore the number of trays \(T\) equals \(C_6\) minus the number of fixed points, and

\(\mathbb E[T]=49/20-1=29/20\).

Since a tray costs \(12\), the expected bill is \(12(29/20)=\boxed{\$17.40}\). For a check, the same construction for \(n=3\) gives one identity permutation with zero trays, three transpositions with one tray, and two 3-cycles with one tray. The average is \(5/6=H_3-1\), as predicted. A long cycle is still only one tray, while a fixed point costs nothing: counting misplaced parcels would charge for the wrong object.

All values below are measured in vouchers at the round in question. Only Ben discounts, by \(\delta_B=0.8\); Ada's factor is \(\delta_A=1\).

Round 2 (Ada proposes). If Ben rejects, his round-3 fallback is five vouchers, worth \(0.8(5)=4\) in round 2. The least acceptable whole-voucher offer gives Ben four and Ada six. Ben accepts on equality. Ada prefers six now to her fallback of five, so this offer is made.

Round 1 (Ben proposes). If Ada rejects, she receives six vouchers in round 2 and, with no discounting, values that at six in round 1. Ben must therefore give Ada at least six. He keeps four, strictly better than waiting to receive four in round 2, which is worth \(0.8(4)=3.2\) in round 1. Thus Ben offers \((\text{Ada},\text{Ben})=(6,4)\), accepted by Ada on indifference.

Round 0 (Ada proposes). Ben can get four in round 1, worth \(0.8(4)=3.2\) now. Because transfers are indivisible, the smallest acceptable current transfer is \(\lceil3.2\rceil=4\) vouchers. That leaves Ada six. Passing also gives Ada six in round 1, with no discount, so she is indifferent. Under the specified proposer tie rule she passes, and Ben's round-1 offer of \((6,4)\) is accepted. An accepted opening offer was possible; delay is selected by the tie rule, not forced by rejection.

With dollar-level rather than $10-voucher indivisibility, the round-2 split is $60/$40, and the round-1 split is again $60/$40. At round 0, Ben needs only \(0.8(\$40)=\$32\); Ada can keep $68, which strictly beats the $60 she gets from passing. She offers $68/$32 and Ben accepts. The coarse transfer unit removes Ada's strict gain from an immediate agreement. In general, for a continuation split of \(a+b=N\) indivisible units, the minimum acceptable current offer to a responder discounted by \(\delta_B\) is \(\lceil\delta_B b\rceil\); compare the proposer's \(N-\lceil\delta_B b\rceil\) with their own discounted continuation \(\delta_A a\). When \(\delta_A=1\) and \(\lceil\delta_B b\rceil=b\), those payoffs tie, permitting delay under the stated tie rule.

Let \(C(s)\) be the cash cost of covering \(s\) missing radios. By the rules, \(C(0)=0\), \(C(1)=110+30=140\), and \(C(2)=220+30+70=320\). With one owned radio, doing nothing incurs $140.

If the kiosk buys one at price \(b\), it has two radios and incurs no shortage bill. Relative to no trade, the improvement is \(140-b\). Buying is therefore non-worsening exactly for \(b\le140\).

If it sells its radio at price \(a\), it has none, receives \(a\), but now owes a $320 shortage bill. The relative improvement is \(a-(320-140)=a-180\). Selling is non-worsening exactly for \(a\ge180\). At equality the kiosk is indifferent; strict inequalities make the trade profitable.

If instead it begins fully stocked, no trade has zero shortage cost. Buying one extra radio creates a surplus whose terminal value is $90, so the maximum bid is $90. Selling one creates a $140 shortage, so the minimum ask is $140. Thus the marginal inventory value changes with holdings. In particular, the second missing radio is more costly than the first; using the $140 first-shortage charge for every marginal trade would misprice the sale.

Expected loss before insurance is \(0.3\times40+0.2\times100\) = £32. The deductible pays £0, £10, £70 in the three states, so its expected reimbursement is \(0.3\times10+0.2\times70\) = £17. The proportional policy pays \(\frac{17}{32}L\); its expected reimbursement is \(\frac{17}{32}\times32\) = £17. Equal payouts and equal £20 premiums imply the same expected final wealth, but not the same distribution.

After the premium, wealth before theft is £180. With the deductible the owner bears \(\min(L,30)\), giving final wealth \((180,150,150)\) at losses \((0,40,100)\). Under proportional coverage the owner bears \(\frac{15}{32}L\), giving \((180,161.25,133.125)\). Both expected wealth values are £165: for the deductible, \(0.5\times180+0.5\times150=165\); for proportional, \(0.5\times180+0.3\times161.25+0.2\times133.125=165\).

Expected square-root utilities are therefore

\[\begin{aligned}U_d&=0.5\sqrt{180}+0.3\sqrt{150}+0.2\sqrt{150}\approx12.8319283,\\U_p&=0.5\sqrt{180}+0.3\sqrt{161.25}+0.2\sqrt{133.125}\approx12.8253276\end{aligned}\].

The deductible is preferred by approximately 0.0066007 utility units. There is also a structural proof. Conditional on a positive loss (probability 0.5), deductible wealth is £150 for certain. Proportional wealth is £161.25 with conditional probability \(\frac{0.3}{0.5}=0.6\) and £133.125 with probability 0.4, with the same conditional mean £150. Strict concavity of \(\sqrt{w}\) makes the certain £150 strictly preferable to that spread. The deductible allocates more protection to the severe, low-wealth state. Equal insurer payouts alone miss this risk-distribution effect.