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006A ticket pays Rs 1 if at least one six appears when three fair dice are rolled, and nothing otherwise. What is the fair price of the ticket?Market making, betting and sizingWarm upAkuna CapitalChicago · 2026
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Your price, to the nearest paisa band?
Show the worked solution
91/216 of a rupee, about 42 paise. A fair price for a ticket paying Rs 1 is the probability of winning. The fastest route is the complement: the chance of no six on three dice is 5/6 x 5/6 x 5/6 = 125/216, so the chance of at least one six is 1 - 125/216 = 91/216, or 0.421. Adding 1/6 three times gives 50 paise and overcounts.
Why is the price just a probability?
If a raffle pays Rs 100 and you win one time in four, playing many times earns you Rs 25 a ticket on average, so Rs 25 is the break-even price. A ticket paying Rs 1 on some event is worth exactly the probability of that event, because that is its average payout. Trading firms phrase probability questions as prices on purpose: it makes you answer in the units a desk uses, and it sets up the next question, which is where you would quote a bid and an offer.
Of the 216 equally likely rolls of three dice, 125 contain no six, so 91 contain at least one and the ticket's fair price is 91/216 of a rupee, about 42 paise, not the 50 paise that adding 1/6 three times suggests.
Why is at least one a signal to use the complement?
At least one six covers exactly one six, exactly two, or three, and each needs its own count. The opposite event, no six at all, is a single clean case: every die avoids six, and independent dice multiply. So the complement takes one line. Adding 1/6 + 1/6 + 1/6 fails because the three events overlap: a roll of 6, 6, 2 is counted once for the first die and again for the second. With ten dice the same mistake would give a probability above 1.
The relationship
$P(\text{at least one six}) = 1 - \left(\tfrac{5}{6}\right)^3 = 1 - \tfrac{125}{216} = \tfrac{91}{216} \approx 0.421$
(5/6)^3 the chance that each of the three dice avoids a six
91/216 the share of the 216 rolls with at least one six
What it says in wordsThe chance of at least one success is one minus the chance of none.
What does a trader add after the number?
A fair value is the centre of a market, not the market itself. A market maker quotes a bid below 42 paise and an offer above it, and the width depends on how confident they are in the number and how much risk one ticket adds to their book. Here the fair value is exact, so a tight market such as 40 bid, 44 offer is defensible. Saying that sentence turns a probability answer into a trading answer, which is what the question format is inviting.
Where candidates lose it
The fast wrong answer is 50 paise, from adding the chance of a six on each die. It is fast, it feels natural, and it ignores that rolls with two or three sixes get counted more than once.
The second loss is time. In an online assessment where each question has seconds, working exactly one, exactly two and exactly three sixes separately is correct and too slow. The complement is the habit being tested.
What the interviewer asks next
What is the fair price if the ticket pays Rs 1 for each six that appears?
How many dice do you need before at least one six is more likely than not?
Quote me a two-sided market on this ticket and tell me what you do if I lift your offer ten times.
Asked at Akuna Capital, Junior Trader Interview, Chicago, 2026 (Wall Street Oasis): if you win you get 1$. how much money would be a fair bet
018I will draw a card from a shuffled deck. You may pay Rs 6 to play a bet that pays Rs 10 if the card is red. Before deciding, you may pay to be told the card's colour. What is the most you should pay for that information?Market making, betting and sizingCoreOptiverChicago · 2025
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What is the information worth?
Show the worked solution
Rs 2. Without information the bet is worth 0.5 x 10 - 6 = -1, so you decline and your value is 0. With the colour known, you play on red and make 4, and skip black and make 0, which averages 2. Information is worth the improvement in your best decision: 2 - 0 = 2. If it would not change what you do, it is worth nothing.
How do you value a piece of information?
Suppose a weather forecast costs money and you are deciding whether to carry an umbrella. If you would carry it anyway, the forecast is worthless to you; it is valuable only if some answer would change what you do. The value of information is the expected value of your best decision with it, minus the expected value of your best decision without it. Work out both decision trees separately and subtract. Never value information by the size of the payout it relates to.
Blind, the bet has an expected value of minus 1 so you decline and get 0; told the colour first, you play only on red and make 4 half the time, an average of 2, so the information is worth Rs 2.
Why is it not worth Rs 4 or Rs 5?
Rs 4 is what you make when the card is red, but it is red only half the time. Rs 5 is half the payout, which ignores the Rs 6 you pay to play. The information saves you from the losing half of the bet and lets you keep the winning half, and that is worth half of Rs 4, which is Rs 2. Pay more than Rs 2 and you would do better declining the offer of information and declining the bet.
The relationship
$\text{VOI} = E\big[\max(\text{payoff}, 0)\big] - \max\big(E[\text{payoff}], 0\big) = \tfrac12\max(4,0) + \tfrac12\max(-6,0) - \max(-1,0) = 2$
payoff 10 - 6 = 4 on red, -6 on black
E[max(payoff, 0)] your value when you can choose after seeing the colour
max(E[payoff], 0) your value when you must choose blind
What it says in wordsInformation is worth the gap between deciding after you know and deciding before.
When is information worth the most?
Vary the price of the bet. At a price of 5 you are exactly indifferent blind, and the information is worth 2.50, its maximum; at a price of 0 you would always play, and it is worth 0. Information is valuable when you are close to indifferent and the decision could go either way. The formula also has the shape of an option payoff: knowing first lets you exercise only when it pays, which is why traders talk about paying for optionality and paying for information in the same breath.
Where candidates lose it
The trap is answering with the size of the win, Rs 4, or half the payout, Rs 5. Both value the information by the bet it is about, not by the decision it improves.
The second loss is forgetting that without information you would decline. Candidates who compare with playing blind, at -1, get 3. The comparison is always with your best action without the information, which here is to walk away.
What the interviewer asks next
What is the information worth if the bet costs Rs 3?
What if the information is only 80% reliable?
You can pay to see one card of a two-card hand before betting. How do you decide what that is worth?
Asked at Optiver, Quantitative Research, Chicago, 2025 (Wall Street Oasis): Valuing information, taking directional bets when not plus EV.
034Make me a two-way market on the number of heads in 100 flips of a fair coin, and justify the width.Market making, betting and sizingWarm upDRWNew York · 2026
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What is the standard deviation of the number of heads?
Show the worked solution
Centre it at 50 and quote around 46 at 54. The fair value is exactly 50. The standard deviation is √(100 x 0.5 x 0.5) = 5, so settlement lands between 45 and 55 about 73% of the time. A market 4 either side of fair earns 4 per lot on any trade, loses on a single sale at 54 only 18% of the time, and leaves room to move the quote if the other side seems to know something.
Where does the centre come from, and what sets the width?
A shopkeeper selling mangoes by the dozen knows the fair price; the margin he adds depends on how much the price of the next crate can swing and on whether the buyer knows something he does not. The centre of your market is the expected value, and the width is a choice about risk and information, scaled by how much the outcome can move. Here the expected value is 100 x 0.5 = 50, and nobody can know more than you about fresh flips of a fair coin, so the width is about risk alone.
The number of heads in 100 fair flips is centred at 50 with a standard deviation of 5, landing in 45 to 55 72.9% of the time and in 40 to 60 96.5% of the time, so a market of 46 at 54 sits inside one standard deviation and earns 4 per lot on each side.
How do you justify 46 at 54 rather than 49 at 51?
Use the standard deviation as the ruler. The count has variance 100 x 0.5 x 0.5 = 25, so a standard deviation of 5. A quote 4 either side of fair earns 4 on each lot traded, against a settlement that typically moves 5, so every trade has an edge worth a large fraction of its risk. If someone buys at 54, you lose only if the count finishes at 55 or more, about 18% of the time. A tight 49 at 51 earns 1 per lot and a sale at 51 loses whenever the count reaches 52, about 38% of the time. Tighter wins more trades and earns less on each; in an interview game, start around one standard deviation wide and tighten as you learn.
The relationship
$\mu = np = 50 \qquad \sigma = \sqrt{np(1-p)} = \sqrt{25} = 5$
n = 100 number of flips
p = 0.5 chance of heads on each flip
sigma standard deviation of the number of heads
What it says in wordsThe count of heads averages 50 and typically lands within 5 of it.
Then say how you would react to trades, because that is the follow-up. If the interviewer lifts your 54 again and again, either they are testing your nerve or they know something, perhaps that the coin is not fair or that some flips are already done. Repeated one-way trading is information: move your market toward it and cut your size, rather than defending 50. The limitation of the simple answer is exactly that it assumes nobody knows more than you.
Where candidates lose it
The common loss is quoting 50 at 50, or 49.5 at 50.5, and calling it fair. A market maker earns the spread; a zero-width quote gives away every trade at no edge and leaves no room to adjust when the other side knows more.
The second is quoting a width with no reason. Name the standard deviation of 5, then choose a width against it. The number you say matters less than showing that width and risk are linked.
What the interviewer asks next
I buy 10 lots at 54. Where is your new market?
Now 60 flips have already happened and I have seen them. How does your market change?
Make a market on the number of heads squared.
Asked at DRW, Quantitative Trading, New York, 2026 (Wall Street Oasis): Make a market on the number of heads out of 100 coin flips.
077An equity index stands at 20,000 and its implied volatility is 18% a year. Where do you think it closes in four months? Give a central value and a 90% range you would be willing to make a market around.Market making, betting and sizingHardMorgan StanleyTokyo · 2025
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Roughly how wide is a 90% range for the index four months out?
Show the worked solution
Centre on today's level, about 20,000, with a 90% range of roughly 16,800 to 23,600. Four months is a third of a year, so one standard deviation is 18% x sqrt(1/3), about 10.4%. In log terms the 90% band is 1.645 of those either side, which gives 16,767 and 23,601. The median sits a little below 20,000 and the upside tail is longer than the downside.
Why is a single number the wrong answer?
Ask a cab driver how long the airport run takes and a good one says forty minutes, maybe an hour in traffic. The range is the useful part, because you plan your flight around it. A trading interviewer asking where an index closes wants a distribution, because a market maker quotes against the spread of outcomes, not against a guess. Your central value should not be a view on the economy either: with no edge, the best central estimate of a traded index is roughly its forward, which for four months is close to today's 20,000 once financing and dividends roughly offset.
The width comes from the implied volatility the market already quotes. Volatility grows with the square root of time, because independent daily moves add their variances, not their standard deviations. Four months is a third of a year, so one standard deviation is 18% x sqrt(1/3) = 10.4%, about 2,078 index points.
With 18% volatility over four months, the index's middle 90% runs from about 16,767 to 23,601, which is 3,233 points below today's level and 3,601 points above, because a lognormal distribution stretches further up than down.
The relationship
$S_T = S_0\, e^{-\sigma^2 T/2 + \sigma\sqrt{T} Z} \qquad \text{5th, 95th pct} = S_0\, e^{-\sigma^2T/2 \mp 1.645\,\sigma\sqrt{T}}$
S_0 today's level, 20,000
sigma implied volatility, 0.18 a year
T time in years, 1/3
Z a standard normal draw
What it says in wordsLog returns are normal with a standard deviation of sigma times root T, and a small drift correction keeps the mean at today's level.
Why is the range lopsided, and where does the median sit?
A fall of 10% and a rise of 10% are not mirror images in log space. Normal log returns make the upside tail longer: the 90% band stretches 3,601 points up but only 3,233 points down. The same convexity pushes the median below the mean: if the mean is 20,000, the median is 20,000 x exp(-sigma squared T / 2), about 19,892. That gap of about 108 points is small here, but it grows with volatility and time, and a candidate who names it shows they know the difference between the most central outcome and the average one.
What would you add before quoting a market on it?
Two honest caveats. Implied volatility is a price, not a forecast: it tends to sit above the volatility that is later realised, because option sellers charge for bearing crash risk, so the band built from it is usually a little wide. Against that, real index returns have fatter tails than the lognormal, so the 5% tails are more likely to hold a larger move than the curve suggests. Say both, then give your market: a tight two-way price around 20,000 if asked for the level, and the 90% band as the range you would sell outside of.
Where candidates lose it
The first loss is scaling volatility linearly with time: a third of 18% is 6%, which gives a band far too narrow. Volatility scales with the square root of time, so four months is about 10.4%, not 6%.
The second loss is answering with a macro story and a point forecast. The interviewer wants you to use the price the market already gives you, implied volatility, and to say that the honest answer is a distribution with a lopsided shape.
What the interviewer asks next
What 90% range would you give for one week out?
How would the range change if implied volatility jumped to 30%?
If you had to bet on the index finishing above 22,000, what fair probability would you quote?
Asked at Morgan Stanley, Sales and Trading, Tokyo, 2025 (Wall Street Oasis): What do you think this index will close at by the end of the year (4 months from now)
089Make me a market on the number of disposable nappies used in the UK in one day. Build the estimate from stated assumptions and choose a width you would actually trade on.Market making, betting and sizingCoreDRWLondon · 2025
Try it first
If each of four inputs could be about 10 to 25% off in either direction, how uncertain is the product?
Show the worked solution
About 9.4 million a day, and I would open at 8 bid, 11 offered, in millions. Assume about 700,000 births a year, 2.5 years in nappies, six changes a day and 90% disposable: 9.45 million. Multiplying the low and high ends of each input gives 5.5 to 15.0 million, so a quote of 8 at 11 is tight enough to trade and still honest about the uncertainty.
How do you build the estimate so the interviewer can follow it?
Chain it through things you can reason about. Children in nappies are roughly births a year times the years each child spends in them. Assume about 700,000 births a year, a round number worth checking against the latest official statistics, and 2.5 years in nappies: about 1.75 million children. Each child uses about six a day on average, more as a newborn and fewer as a toddler, and assume 90% of families use disposables: 1.75 million x 6 x 0.9 = 9.45 million a day. Say each assumption out loud and give it a range as you go.
Multiplying the four central assumptions gives 9.45 million nappies a day, but multiplying the four lows and the four highs gives 5.5 to 15.0 million, so the honest uncertainty is roughly a factor of 1.6 either side, and a quote of 8 at 11 million sits inside it.
The relationship
$N = B \times Y \times c \times d = 700{,}000 \times 2.5 \times 6 \times 0.9 \approx 9.45\text{ million}$
B births a year, an assumption
Y years a child spends in nappies
c changes a day
d share of families using disposables
What it says in wordsBuild the count from quantities you can defend one at a time, and multiply.
Where should the width of the market come from?
From the ranges, multiplied. A shopkeeper who is unsure of both price and quantity is more unsure of revenue than of either. Put a low and a high on every input and multiply the lows together and the highs together: here 5.5 million to 15.0 million, about a factor of 1.6 either side of the centre. Centre the quote near the middle on a multiplicative scale, the geometric mean of the ends, 9.1 million, which sits close to the central estimate.
Then choose the width you will actually trade. A market as wide as the whole range, 5.5 at 15, is useless: nobody trades against it and it tells the interviewer you have no view. Quote tighter, 8 at 11, and move it as they trade: if they keep buying at 11, raise both sides, because their trades carry information. Say the scope questions too: does the count include adult incontinence products, and a school-age child in night-time pants? Those can move the answer more than any of the four inputs.
Where candidates lose it
The first loss is giving one number, or a market whose width is a round guess such as plus or minus a million, with no link to the assumptions. The interviewer wants to see where the width came from.
The second loss is the opposite: a market so wide it is safe and worthless. Show the full range, then quote a tighter two-way price and explain how you would move it when they trade.
What the interviewer asks next
I buy 5 lots at your offer. Where is your market now?
What single piece of data would you buy to narrow the range most, and why?
How would you size the market if the settlement were a count of nappies sold rather than used?
Asked at DRW, Trading, London, 2025 (Wall Street Oasis): Make me a market on the amount of diapers used in the UK daily
099A stock is worth either 100 or 110, with equal probability. 20% of the traders who arrive know the true value: they buy if it is 110 and sell if it is 100. The other 80% buy or sell at random, half and half. Where should a market maker set its ask so that it breaks even, on average, when someone buys from it?Market making, betting and sizingHardJane StreetNew York · 2025
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Where should the ask be?
Show the worked solution
Set the ask at 106, and by the same logic the bid at 104. If the stock is worth 110, a buy arrives with probability 0.2 + 0.8 x 0.5 = 0.6; if it is worth 100, with probability 0.4. Given a buy, Bayes puts the chance of 110 at 0.6, so the stock is worth 106 to the market maker selling it. The spread of 2 is the price of trading against informed flow.
Why can't the market maker just quote the expected value of 105?
A second-hand car dealer who pays the average price for every car will find that the owners of good cars go elsewhere and the owners of bad ones queue up. Who chooses to trade with you is information. A market maker does not care what the stock is worth on average; it cares what the stock is worth given that someone has just chosen to buy from it. At an ask of 105, noise buyers are harmless, a loss of 5 when the stock is worth 110 and a gain of 5 when it is worth 100. Informed buyers only appear in the 110 world, and they cost 0.5 per arriving trader on average, so 105 is a losing quote.
Tracing who sends a buy order in each world, buys come with probability 0.30 from the 110 world and 0.20 from the 100 world, so a buy lifts the chance of 110 from 0.5 to 0.6 and the break-even ask is 106.
The relationship
$P(110 \mid \text{buy}) = \frac{0.5 \times 0.6}{0.5 \times 0.6 + 0.5 \times 0.4} = 0.6, \qquad \text{ask} = \mathbb{E}[V \mid \text{buy}] = 100 + 0.6 \times 10 = 106$
0.6 the chance of a buy when the stock is worth 110: 0.2 informed plus 0.8 x 0.5 noise
0.4 the chance of a buy when the stock is worth 100: noise only
V the stock's true value
What it says in wordsSet the ask at the value of the stock conditional on being bought from, which Bayes' rule gives directly.
What sets the width of the spread?
The share of informed traders and the size of what they know. With a share alpha informed, a buy is alpha + (1 - alpha)/2 likely in the high world and (1 - alpha)/2 in the low world, and the ask works out to 105 + 5 alpha. The spread is a fee for adverse selection: it is zero when nobody is informed and widens to the full 100 to 110 range when everyone is. The table runs the formula for a few shares. Order processing and inventory costs add to this in real markets, but the information component is what makes spreads jump around earnings and news.
Informed share Ask Bid Spread
0% 105 105 0
10% 105.5 104.5 1
20% 106 104 2
50% 107.5 102.5 5
100% 110 100 10
The break-even spread equals the informed share times the 10-point value gap, so it is 2 at 20% informed and 5 at 50% informed.
What happens after the first trade?
The market maker updates. After one buy, the chance of 110 is 0.6, and if a second buy arrives the same Bayes step lifts it to 0.692, so the next ask is about 106.92. Each order moves the quotes towards the true value, which is how prices come to reflect what the informed traders know. The limitation is that the model has one share size, no inventory risk and no competition between market makers; real desks also skew quotes to manage position, which this puzzle leaves out.
Where candidates lose it
The fast wrong answer is 105, the unconditional expected value. It ignores that the act of buying is evidence: informed traders buy only when the stock is worth 110, so a market maker at 105 loses on every informed buyer and only breaks even on noise traders.
The second loss is overreacting and quoting 110 because some buyers are informed. Most buyers are noise traders, and a quote at 110 drives them away. Bayes gives the exact weight, 0.6 on the high value, and the ask of 106.
What the interviewer asks next
Where should the bid be, and why is the spread symmetric here?
After one buy at 106, where is the next ask?
How does the spread change if half of the traders are informed?
Asked at Jane Street, Generalist, New York, 2025 (Wall Street Oasis): It was a probability theory based quant trading style market making questions which were intense

(5/6)^3	the chance that each of the three dice avoids a six
91/216	the share of the 216 rolls with at least one six

payoff	10 - 6 = 4 on red, -6 on black
E[max(payoff, 0)]	your value when you can choose after seeing the colour
max(E[payoff], 0)	your value when you must choose blind

n = 100	number of flips
p = 0.5	chance of heads on each flip
sigma	standard deviation of the number of heads

S_0	today's level, 20,000
sigma	implied volatility, 0.18 a year
T	time in years, 1/3
Z	a standard normal draw

Try it first

Why is the price just a probability?

Why is at least one a signal to use the complement?

What does a trader add after the number?

Where candidates lose it

What the interviewer asks next

Try it first

How do you value a piece of information?

Why is it not worth Rs 4 or Rs 5?

When is information worth the most?

Where candidates lose it

What the interviewer asks next

Try it first

Where does the centre come from, and what sets the width?

How do you justify 46 at 54 rather than 49 at 51?

Where candidates lose it

What the interviewer asks next

Try it first

Why is a single number the wrong answer?

Why is the range lopsided, and where does the median sit?

What would you add before quoting a market on it?

Where candidates lose it

What the interviewer asks next

Try it first

How do you build the estimate so the interviewer can follow it?

Where should the width of the market come from?

Where candidates lose it

What the interviewer asks next

Try it first

Why can't the market maker just quote the expected value of 105?

What sets the width of the spread?

What happens after the first trade?

Where candidates lose it

What the interviewer asks next