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.

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.