Equity Research puzzles, solved step by step

Puzzles: 100
Traced to a firm: 19
Topics: 11
Hard: 30

Topic

All topics Probability and brainteasers12 Expected value and decisions8 Market sizing and estimation12 Returns and compounding9 Valuation riddles12 Three statement riddles10 EPS and share count9 Cost of capital and rates8 Growth, mix and unit economics8 Mental maths6 Data and reasoning traps6

Level

Any Warm up Core Hard

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Any Reported at a firm Standard

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002A private company is worth somewhere between Rs 0 and Rs 100 crore to its owner, every value equally likely, and the owner knows the exact figure. Under your management it would be worth 1.5 times whatever it is worth to the owner. The owner accepts any bid at or above the company's value to them. How much should you bid?Expected value and decisionsHardBuy-side equity researchHedge fund long/short
Try it first
Pick your bid before you work it.
Show the worked solution
Bid nothing: every positive bid loses money on average. If you bid b, the owner accepts only when the value is below b, so accepted deals average b/2. Worth 1.5 times that to you, they return 0.75b for a price of b, a loss of a quarter of the bid each time. At Rs 60 crore you would win 60% of the time and lose Rs 15 crore on every win.
Why does the average value of Rs 50 crore mislead?
Imagine buying a used car from a private seller who has driven it for five years. If the seller happily accepts your first offer, the most useful thing you have learnt is what the seller knows about the car. The same logic runs here. The owner says yes only when your bid is above what the company is worth to them, so acceptance itself is bad news about the value. Across all companies the average is Rs 50 crore, but you never get to buy the average company; you buy the ones worth less than your bid.
A bid of Rs 60 crore is accepted only when the owner's value is below Rs 60 crore, so the accepted cases average Rs 30 crore, worth Rs 45 crore to you, and every accepted deal loses Rs 15 crore, a quarter of the bid.
How do you show that no bid works?
Take a general bid b. The owner accepts with chance b/100, and given acceptance the value is spread evenly from 0 to b, averaging b/2. To you that is worth 1.5 x b/2 = 0.75b. You pay b and receive 0.75b, so each accepted deal loses 0.25b, whatever b is. Multiply the loss by the chance of acceptance and the expected result is minus b squared over 400, which is zero only at b = 0. At a bid of Rs 60 crore that is minus Rs 9 crore.
The relationship
$E[\text{profit}] = \frac{b}{100}\left(1.5\cdot\frac{b}{2} - b\right) = -\frac{b^2}{400}$
b your bid, Rs crore
b/100 the chance the owner accepts
b/2 the average value of a company whose owner accepts
What it says in wordsThe chance of a deal times what each deal makes, and each deal loses a quarter of the bid.
This is the winner's curseThe tendency of the winning bid in an auction or negotiation to be the one that most overestimated the value, because the other side or other bidders knew better., and it is why a buy-side analyst asks who is on the other side of a trade. When the seller knows more than you do, the times you get filled are skewed towards the times you were wrong. The answer changes only if your edge is large enough: with a multiple of 2 instead of 1.5, accepted deals exactly break even.
Where candidates lose it
The fast answer takes the Rs 50 crore average, multiplies by 1.5 and bids anything under Rs 75 crore. It ignores that the owner chooses whether to sell, so the companies you actually buy are not a random sample.
The second loss is getting the zero answer and not generalising it. Say that the loss is a fixed quarter of any bid, so no bid escapes it, and name the multiple at which the answer flips.
What the interviewer asks next
What multiple of the owner's value would you need before any positive bid breaks even?
How does the answer change if the owner does not know the value either?
Where do you see the winner's curse in IPO allotments or block trades?
027You can repeat a bet that wins 60% of the time and pays even money, as often as you like. What fraction of your capital should you stake each time to grow your money fastest, and what happens if you stake double that fraction?Expected value and decisionsHardHedge fund long/shortMulti-manager pod
Try it first
Stake double the growth-maximising fraction. What happens to your money over many bets?
Show the worked solution
Stake 20% of capital each time; at 40% your money slowly shrinks. The Kelly fraction for an even-money bet is the win chance minus the loss chance, 0.6 minus 0.4. At 20% the typical path grows about 2.0% a bet. At 40% each bet still has a positive expected gain, but growth is about -0.24% a bet, below zero, because big losses compound harder than big wins.
Why is the bet with the highest expected value not the one that grows fastest?
Imagine a shopkeeper who puts half the till into stock every morning. A good day lifts the till by half; a bad day cuts it in half. One of each leaves 1.5 x 0.5 = 0.75 of where she started, even though the good and bad days were the same size. Wealth compounds, so what matters over many bets is the average of the logarithm of each outcome, not the average outcome. Expected value per bet rises in a straight line with the stake; growth rises, peaks and then falls.
The relationship
$g(f) = 0.6\,\ln(1+f) + 0.4\,\ln(1-f) \qquad f^{*} = p - q = 0.2$
f the fraction of capital staked on each bet
g(f) growth of capital per bet on the typical path, in log terms
p - q the edge: win chance less loss chance
What it says in wordsGrowth per bet is the chance-weighted log of what each outcome does to your capital, and it peaks when you stake your edge.
Growth per bet peaks at 2.01% when 20% of capital is staked, falls to zero at about 38.9%, and turns negative at 40%, so overbetting a real edge can make you poorer over many bets.
What do the numbers look like at each stake?
Stake Expected gain per bet Growth per bet, typical path
10% +2.0% +1.50%
20% +4.0% +2.01%
30% +6.0% +1.47%
40% +8.0% -0.24%
50% +10.0% -3.40%
Expected gain keeps rising with the stake while growth peaks at 20% and turns negative near 40%.
Read the two columns against each other. Every row has a positive expected gain, yet the 40% and 50% rows lose money on the path you will actually live through. Half Kelly at 10% keeps about 75% of the maximum growth with far smaller swings, which is why many desks size below full Kelly. Say the limitation too: the formula assumes you know the 60% exactly. Real edges are estimates, and overestimating one pushes you towards the overbetting side of the curve.
Where candidates lose it
Candidates maximise expected value and conclude you should stake everything, because every bet is favourable. That answer goes broke on the first loss. The interviewer is testing whether you know that repeated bets compound, so the log of wealth is what you should maximise.
The second loss is saying 60%, the win probability, as the stake. The Kelly fraction for even money is the edge, 0.6 minus 0.4, not the win chance.
What the interviewer asks next
The bet now pays 2 to 1 with a 40% win chance. What is the Kelly fraction?
Why might a portfolio manager size positions at half Kelly?
How does position sizing on a stock idea resemble this bet, and where does the analogy break?
052A game multiplies your stake by 1.5 when a fair coin lands heads and by 0.6 when it lands tails, and you must stake everything you have on every flip. The expected return per flip is plus 5%. After 100 flips, what does a typical player hold?Expected value and decisionsHardHedge fund long/shortLong-only asset management
Try it first
Before working it: after 100 flips the typical player holds...
Show the worked solution
About 0.5% of the starting stake. One head and one tail together multiply wealth by 1.5 x 0.6 = 0.90, so the typical growth factor per flip is the square root of 0.90, or 0.949, a loss of about 5.1% a flip. The median path has 50 heads and 50 tails and ends at 0.9 to the power 50, about 0.005. The average ends near 131.5x, carried by rare lucky paths.
Why does a plus 5% game shrink the typical player?
Think of a shop that raises a price 50% one month and cuts it 40% the next. The two changes average plus 5%, yet an item tagged Rs 100 ends at Rs 90. Wealth compounds by multiplying, not by adding. So the rate that decides where one player ends up is the geometric mean of the multipliers, not their arithmetic average. Here the geometric mean is the square root of 1.5 x 0.6, which is 0.949 a flip, and 0.949 applied a hundred times is a very small number.
The relationship
$g = \sqrt{1.5 \times 0.6} = 0.949 \qquad g^{100} = 0.9^{50} \approx 0.005$
g the typical growth factor per flip, the geometric mean of the two multipliers
1.5, 0.6 the multipliers on heads and on tails
0.9^50 fifty head and tail pairs, the median outcome after 100 flips
What it says in wordsThe typical player's wealth grows at the geometric mean of the multipliers, and here that mean is below one.
On a log scale the average of all players climbs in a straight line to about 131.5 times the stake, while the typical player falls to about 0.5% of it; and only 13.6% of all players finish above where they started.
If the typical player loses, where does the plus 5% average come from?
From a very small number of paths with far more heads than tails. You need at least 56 heads in 100 flips just to finish ahead, and only about 13.6% of players get there. The average is pulled up by the few players who land 70 or more heads and finish hundreds of thousands of times richer, while most players finish near zero. A player with exactly 70 heads ends at about 468,733 times the stake. The mean is a true number, but almost no individual player experiences it.
What does this have to do with running money?
A portfolio compounds exactly like the game. Volatility pulls the growth rate below the average return by roughly half the variance, so a strategy with a positive expected return can still shrink a typical account if it is run at too much size. Here the average return is 5% with a swing of 45% either way; half of 45% squared is about 10%, which is why the typical path loses about 5% a flip. The fix is sizing, not the odds: staking a quarter of wealth each flip, the Kelly fractionThe share of wealth to stake on each bet that maximises the long-run growth rate of wealth. here, lifts the typical player to about 1.86x after 100 flips.
Where candidates lose it
The trap is answering with the expected value, 1.05 to the power 100, about 131.5 times the stake. That is the average across every possible player and the right answer to a different question. The interviewer asked what a typical player holds, which is the median.
The second loss is calling the game bad. The odds are good; the sizing is bad. Say that staking a fraction of wealth each flip turns the same odds into a growing account, and you have shown why the question is asked on an investing desk.
What the interviewer asks next
What fraction of your wealth should you stake each flip to maximise long-run growth?
How many heads out of 100 do you need to finish ahead?
Would you play this game once for your whole savings? Would you play it 100 times with a quarter each time?

b	your bid, Rs crore
b/100	the chance the owner accepts
b/2	the average value of a company whose owner accepts

f	the fraction of capital staked on each bet
g(f)	growth of capital per bet on the typical path, in log terms
p - q	the edge: win chance less loss chance

Stake	Expected gain per bet	Growth per bet, typical path
10%	+2.0%	+1.50%
20%	+4.0%	+2.01%
30%	+6.0%	+1.47%
40%	+8.0%	-0.24%
50%	+10.0%	-3.40%

g	the typical growth factor per flip, the geometric mean of the two multipliers
1.5, 0.6	the multipliers on heads and on tails
0.9^50	fifty head and tail pairs, the median outcome after 100 flips

Try it first

Why does the average value of Rs 50 crore mislead?

How do you show that no bid works?

Where candidates lose it

What the interviewer asks next

Try it first

Why is the bet with the highest expected value not the one that grows fastest?

What do the numbers look like at each stake?

Where candidates lose it

What the interviewer asks next

Try it first

Why does a plus 5% game shrink the typical player?

If the typical player loses, where does the plus 5% average come from?

What does this have to do with running money?

Where candidates lose it

What the interviewer asks next