Derivatives Foundation puzzles, solved step by step

You lose everything about 2.0% of the time, 0.52 to the sixth power, and the expected result is about minus Rs 265. Rs 63,000 covers exactly six bets: 1, 2, 4, 8, 16 and 32 thousand. A win at any of them recovers every earlier loss and nets Rs 1,000, which happens 98.0% of the time. Six losses in a row cost all Rs 63,000. Weighted, 980 of expected winnings against 1,246 of expected loss leaves minus Rs 265.

Why does a plan that wins 98 times in 100 still lose money?

Picture a friend who sells phone insurance to classmates for Rs 50 a month. Month after month nobody drops a phone, and the Rs 50 notes pile up; it feels like free money until the month three phones go into a pond. A win rate tells you how often you are paid, not how much you are paid against how much you can lose, and the expected value needs both. Doubling after every loss builds exactly that shape: Rs 1,000 collected almost every time, and Rs 63,000 handed back rarely. The rare branch is 63 times the size of the common one, so a 2% chance of it more than cancels a 98% chance of the small win.

How do you lay out the six bets in the room?

Write the ladder down before computing anything. The stakes are 1, 2, 4, 8, 16 and 32 thousand, which add to 63 thousand exactly, so the seventh bet of 64 thousand can never be placed. If the first win comes at bet k, it pays 2 to the power k minus 1 thousand, and the losses before it add to one thousand less than that, so every winning session nets exactly plus Rs 1,000. There are only two outcomes, and the table shows how quickly the chance of reaching each rung falls: by the sixth bet you are staking Rs 32,000 to recover Rs 31,000 of losses and win one more thousand.

Is there a faster way to see the sign without the ladder?

Yes, and it is the one a trader reaches for first. Every rupee placed on red loses 4 paise on average, whatever happened on the previous spin, because the wheel has no memory. The expected result of any staking plan is the edge per rupee times the expected total amount staked: here minus 4% of Rs 6,633, which is minus Rs 265, the same figure as the ladder. Doubling raises the amount you put down when you are losing; it cannot change the sign of the edge. On a fair 50/50 wheel the same plan has an expected value of exactly zero, with the same lopsided shape.

Why does a desk interviewer care about a roulette plan?

Because the shape is the shape of selling far out-of-the-money options, or of adding to a losing position to get back to flat. Both produce a long run of small gains and a rare large loss, and a good-looking track record says almost nothing about the tail. Repetition makes the rare branch common: play 50 sessions and the chance of at least one ruin is 1 minus 0.98 to the 50th, about 63%. The limitation to state is that the plan assumes no table limit; a casino maximum bet cuts the ladder short and makes ruin more likely, not less.

Yes: buy at 22, because the ticket is worth 23.08. A standard deck has 12 picture cards out of 52, a 23.1% chance, so the fair value of a Rs 100 payout is 12/52 x 100 = 23.08. You buy at the offer, 22, which is below fair, and expect to make about Rs 1.08 a ticket. Selling at 20 would give away about Rs 3.08. The edge is small next to the outcome, so it pays only over many tickets.

How do you value the ticket?

If a friend offers you Rs 100 when a coin lands heads, you would pay up to Rs 50 and no more; the coin's odds set the price. A ticket that pays a fixed amount on an event is worth the payout times the probability of the event, before any talk of the quote. A standard deck has four suits, each with a jack, a queen and a king, so 12 picture cards in 52. The ticket is worth 12/52 x 100 = 3/13 x 100 = 23.08. Notice that the ace is not a picture card; counting it is the quickest way to get 16/52 and the wrong side.

Which side of the quote do you compare with?

A quote has two prices, and you can only use one of each: you buy at the offer and sell at the bid. Trade only when fair value sits outside the quote: buy when it is above the offer, sell when it is below the bid, and pass when it falls between them. Here fair is 23.08 against a 22 offer, so you buy and expect 1.08. The figure runs two other quotes for contrast: at 22 bid, 24 offered, fair is inside and there is no trade; at 24 bid, 26 offered, you sell at 24 and expect 0.92.

How much would you buy, and what could change your mind?

Each ticket pays 0 or 100, so its standard deviation is 100 x the square root of 0.231 x 0.769, about Rs 42, against an edge of Rs 1.08. The edge is real but small next to the noise, so it is worth taking in size over many independent draws and worth very little on a single ticket. Before trading, ask the two questions a desk would: is the deck standard and well shuffled, and does the person quoting know something you do not, such as a card already removed? The limitation of the clean answer is that it trusts the setup; in a game where the quoter controls the deck, a quote this generous is itself a warning.

Bet 20% of your capital each time, for a growth rate of about 2.0% per bet. The Kelly fraction at even money is the edge, p minus q, which is 0.6 minus 0.4. The expected log growth is 0.6 x ln(1.2) + 0.4 x ln(0.8) = 0.0201, so capital typically grows by a factor of about 7.5 over 100 bets. Bet twice that, 40%, and the growth turns slightly negative, -0.24% a bet: anything beyond about 39% loses money despite the edge.

Why not bet everything on a coin that favours you?

A street vendor with a stall that makes money six days in seven does not spend the whole float on stock each morning, because the seventh day would end the business. Expected value is the right guide for a single bet with money you can replace, but when you must survive to play again, what matters is the growth of your capital over many bets, and that is governed by the average of the logarithm, not the average of the rupees. Betting all of it has the highest expected value and a certainty of ruin: one loss and there is nothing left to compound. The logarithm punishes that loss infinitely, which is the mathematical way of saying you cannot come back from zero.

Where does the 20% come from, and what if the odds are not even?

Differentiate g with respect to f and set it to zero: p over (1 + f) equals q over (1 - f), which gives f = p - q. At even money that is the edge, 0.2. For a bet that pays b to 1 the formula becomes (bp - q) over b, edge over odds. Say the general form, because the interviewer will often move the payout: at 2 to 1 on the same coin the stake becomes (1.2 - 0.4) / 2 = 40%, and at odds where bp is below q the right stake is zero. The growth at 20% is 0.0201 per bet, which compounds to a factor of e to the 2.01, about 7.5 times, over 100 bets on the typical path. That is the median outcome, not the average: a few lucky paths do far better and pull the mean up.

Why do desks bet less than Kelly?

The curve is flat on the left and steep on the right. Half Kelly, 10%, still delivers 1.50% a bet, three quarters of the peak, with far smaller swings. Double Kelly, 40%, delivers -0.24%, already below zero, with wild swings: the crossing point is about 39%, and anything beyond it shrinks capital in the long run even though every single bet has positive expected value. Over-betting is punished much harder than under-betting, and in real trading the 60% is an estimate rather than a fact, so sizing at a fraction of Kelly is the standard discipline. The limitation to state: Kelly assumes you know p, that bets are independent and repeated, and that you can resize freely; on a desk, position limits and the uncertainty in your edge usually bind before the formula does.

The typical player ends with about 3% of the starting capital, while the average outcome is about 13,781 times the start. With 60 wins and 40 losses the stack is multiplied by 1.5 to the 60 and 0.5 to the 40, which is 0.033. Betting half the stack is 2.5 times the Kelly fraction of 20%, and anything beyond twice Kelly turns a positive-edge game into a losing one for the person actually playing it.

Why can every bet be favourable and the typical result still be a loss?

Think of a shopkeeper who doubles the stock every time a line sells and halves it every time it does not. A good line sells six times in ten, yet after a hundred cycles the shelf is nearly bare, because growth is a chain of multiplications and the order of operations does not rescue you. Expected value is an average across many imaginary players, but your capital follows one path, and a path is a product of factors, not a sum. Betting half the stack makes each win a factor of 1.5 and each loss a factor of 0.5. Sixty wins and forty losses, the most likely split, multiply to 0.033. The arithmetic mean of 13,781 is carried by the rare paths with 80 or 90 wins, which almost nobody lives on.

What is the right quantity to look at instead of the mean?

Look at the expected growth rate of the logarithm of capital, because the log turns products into sums and the law of large numbers then works. Each bet changes log capital by 0.6 x log(1.5) + 0.4 x log(0.5), which is -0.034, a negative number, so the typical player loses about 3.4% of capital per bet on average. Over 100 bets that is a factor of exp(-3.4), the 0.033 you saw above. The chance of finishing below the start is 76%, so the mean is a number you will almost never meet.

Where does the sizing go wrong, and what would fix it?

The Kelly fraction for an even-money bet is 2p minus 1, here 20% of capital, and it maximises the growth rate at +0.0201 per bet, which turns 100 bets into a typical multiple of about 7.5. Growth is zero at exactly twice Kelly, 40% here, and negative beyond it, so a half-stack bettor is on the wrong side of the hill even though the coin is in their favour. The limitation is that the Kelly rule assumes you know p exactly and will play many times; with an uncertain edge, desks size below Kelly on purpose, often at half of it.

At most Rs 100/11, about Rs 9.09. A total of 10 comes from (4, 6), (5, 5) and (6, 4), three of the 36 equally likely outcomes, so the chance is 1/12 and the odds are 11 to 1 against. A fair bet matches the odds: risk 1 to win 11, so to win Rs 100 the fair stake is 100/11. Risk Rs 10 and you lose about Rs 0.83 per game on average.

Why is the answer odds, not a probability?

A friend offers to pay you Rs 100 if your cricket team wins a match it wins one time in twelve. Paying Rs 20 to enter is a bad deal, and the way you know is by comparing what you risk to what you get: you lose Rs 20 eleven times and win Rs 100 once. A bet is fair when the stake and the prize are in the same ratio as the losing and winning chances, which is what odds express directly. One in twelve means eleven losses per win, so a fair stake is one eleventh of the prize: Rs 100/11, which is Rs 9.09. Anything above that is paying for the privilege of playing.

How do you check the stake with expected value?

Set the expected profit to zero. With stake s, you win Rs 100 with chance 1/12 and lose s with chance 11/12, so the expectation is 100/12 - 11s/12, which is zero when s = 100/11. The expected value route and the odds route must agree, and saying both in the room shows the interviewer you can move between them. Then say the direction: a lower stake is a profitable bet for you, a higher one is profitable for the house, and at Rs 10 the house edge is 8.3% of your stake, which is roughly the edge a casino takes on its better games.

What would a trader add after the number?

That the fair stake is a ceiling, not an offer. A trader asked to risk money quotes below fair, say Rs 8, and explains the gap as the edge for taking the other side, because a bet at exactly fair value makes nothing over many plays. The limitation is that 1/12 is the chance only if the dice are fair and the counting is right; if the interviewer changes the target to 7, the chance rises to 6/36 and the fair stake jumps to Rs 20, so the first thing to re-check whenever the rules change is the count of winning cells.