Equity Research puzzles, solved step by step

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.

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.

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.

The reroll right is worth Rs 75. A single roll is worth Rs 350 on average. With the reroll, keep 4, 5 and 6, which average Rs 500, and reroll 1, 2 and 3 for an expected Rs 350. Half the time you get Rs 500 and half the time Rs 350, so the game is worth Rs 425, Rs 75 more than without the right.

What is the rule for keeping or rerolling?

Think of a job offer in hand while you wait on another interview. You take the offer if it beats what you expect the other process to give you, not if it beats your dream job. Keep any result worth more than the expected value of the alternative; here the alternative is a fresh roll worth Rs 350. So 4, 5 and 6 are kept, and 1, 2 and 3 are rerolled.

How do you value the right itself?

Value the game with the right, then subtract the game without it. With the right: half the time you hold 4, 5 or 6, averaging Rs 500; half the time you reroll and expect Rs 350. That is Rs 425. Without it the game is worth Rs 350. The right is worth the difference, Rs 75, and all of it comes from the three bad faces being replaced.

This is an option, and a research interviewer asks it to see whether you value flexibility correctly. The right has value only because you can refuse it when the first roll is good. Its value is the gain in the bad states, Rs 250, 150 and 50 on faces 1, 2 and 3, averaged over six faces: Rs 75. The same logic prices an option to expand a project or to delay an investment.

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.

What do the numbers look like at each stake?

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.

The expected price is Rs 512, 2.4% above today. Weight each outcome by its chance: 0.6 x 560 plus 0.4 x 440 gives 336 plus 176. Today's Rs 500 sits exactly halfway between Rs 440 and Rs 560, so the market is pricing roughly a 50% chance of good results. Your 60% view is what separates Rs 512 from Rs 500, and that probability is what you would have to defend.

Why is the expected price not the likely price?

A cricket fan who thinks her team wins 60% of the time does not expect the team to win exactly 0.6 of a match; she expects the win most often and a loss sometimes. Expected value is the probability-weighted average of every outcome, and it can be a price that never actually trades. Here the stock will be at Rs 560 or Rs 440 tomorrow, never at Rs 512, yet Rs 512 is the right number to compare with today's price.

What does today's price tell you about the market's view?

Run the calculation backwards. If Rs 500 is the market's expected price, the chance q of the good outcome solves 560q + 440(1 - q) = 500, so q = 60 / 120 = 50%. The interesting number is not Rs 512 but the gap between your 60% and the market's 50%: that gap is the whole of your view. An analyst would next ask what evidence justifies seeing more upside than the market does.

Say the limitations. Two outcomes are a simplification of a whole spread of possible moves, and a 2.4% expected gain on one event is small next to the Rs 60 swing either way. The expected value is a way to state a view precisely, not a reason on its own to act on it.

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.

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.

The market is pricing about a 23.5% chance of approval; at 30% the stock is worth about Rs 201. Set the price equal to the probability-weighted outcomes: 190 = p x 320 + (1 - p) x 150, so p = 40/170 = 23.5%. At 30%, the value is 150 + 0.3 x 170 = Rs 201, about 5.8% above the price, ignoring the time value of money and any premium for bearing a binary risk.

How does a price turn into a probability?

Think of a ticket that pays Rs 320 if it rains tomorrow and Rs 150 if it does not. Nobody would sell it below Rs 150 or buy it above Rs 320, and where it trades in between tells you how likely the crowd thinks rain is. Where the price sits between the two outcomes is the market's probability: Rs 190 is 40 of the 170 rupees from the downside to the upside, 23.5% of the way.

What does the gap between 23.5% and 30% actually buy you?

The expected edge is the difference in probability times the spread: 6.5 points x Rs 170, about Rs 11 a share, or 5.8% of the price. But the outcome is binary: the stock either falls 21% to Rs 150 or rises 68% to Rs 320, and on a 30% view it falls seven times in ten. That is why desks size binary positions small and spend their effort asking how confident the 30% itself is, not just whether it beats 23.5%.

What does the simple formula leave out?

Three things worth naming. The decision may be months away, so the outcomes should be discounted. Investors want paying for binary risk, so part of the gap between your view and the price is a premium, not a mistake. And the implied probability is only as good as the two end values: if the rejection value is Rs 160 instead of Rs 150, the implied probability drops from 23.5% to 18.8%. Most of the real work is in the end values, not the division.

Randomise: pay Rs 3.15 with probability about 0.159 and Rs 3.14 otherwise. Pi is 3.14159, which is 15.9% of the way from 3.14 to 3.15. Weighting the two payable amounts by those odds gives an expected payment of exactly pi. Neither side is favoured on average, which is the only sense in which an amount you cannot pay can be paid fairly.

What does fair mean when the exact amount cannot be paid?

Five children and four mangoes: nobody can get four fifths of a mango each without a knife, so a fair parent draws lots and one child misses out, with every child facing the same chance. When the exact amount is impossible, fair means fair on average: the expected payment equals what you owe. Rounding to 3.14 short-changes the restaurant every time; rounding to 3.15 overpays every time. Only a random choice between the two neighbours can hit pi exactly.

How do you find the right odds?

Call the chance of paying 3.15 by the letter p. The expected payment is 3.14 plus p times one paisa, and you want that to equal 3.14159. So p is the fraction of the gap you still owe: 0.00159 over 0.01, which is 0.1593. Say it as a balance point: the further pi sits towards 3.15, the more often you pay 3.15.

How do you actually produce odds of 0.159 with a coin?

This is the follow-up that separates people. Write p in binary: 0.0010100011 and so on. Flip a fair coin to generate your own random binary digits, one at a time, and compare each with p's digit in the same place. At the first flip that differs, you know whether your random number is below p or above it, and you pay 3.15 or 3.14 accordingly. Each flip has a one in two chance of settling it, so on average two flips decide, even though p itself is irrational. The same idea, called randomised roundingRounding a fractional quantity up or down at random, with odds set so that the expected result equals the exact fraction., is used to split odd lots of shares fairly across client accounts.

Pitch B, at 1.1x against 0.5x for Pitch A. Weight each outcome by its probability. A gives 10% of 5x plus 90% of nothing, which is 0.5x: on average it loses half the money. B gives 60% of 1.5x plus 40% of 0.5x, which is 0.9 plus 0.2, 1.1x. A would need better than a one in five chance of the 5x just to break even.

Why does the big number not win?

A lottery ticket that costs Rs 100 and pays Rs 500 one time in ten is a bad ticket, however good Rs 500 sounds. An expected value multiplies each outcome by its probability, so a large payoff is shrunk by a small chance before it counts. For A, 5x shrinks to 0.5x. For B, a modest win that happens more often than not, plus a partial loss that still returns half the money, adds up to 1.1x.

What would make Pitch A worth taking?

Work backwards from breakeven. A returns your money on average only if the chance of the 5x is one in five, 20%. So the real question about A is not how big the upside is but whether you can defend a probability above 20%, double what the pitch claims. That is how a buy-side analyst would push back on a story built around one dramatic outcome: ask for the odds, then check them.

Is the higher expected value the whole answer?

No, and saying so earns the extra point. B still loses half the money 40% of the time, so the expected value tells you which pitch to prefer, not how much to put in. Position size depends on how bad the bad outcome is and how often it comes. The limitation of the question is that it hands you the probabilities. In practice those are the hardest number to estimate, and a pitch with a vivid upside tends to come with an optimistic probability attached.