Scratch-Off (Timed, 13 Questions)

How it works

You start with a $100 bankroll and pick a ticket priced at $1, $5, $10, $20, $50, or $100. The ticket flips over to reveal its true prize table before you commit to anything - three outcomes (a $0 tier, a mid tier, and a jackpot tier) with exact fractional odds out of 8 or 16. Real lotteries hide this table; the game shows it to you because reading it is the whole skill.

Before you can scratch, you answer a set of probability questions about the ticket you are holding: expected value, Bayes / conditional probability, no-replacement batch odds, combinatorics, the geometric distribution (expected tickets to first win), and an occasional classic puzzle (Monty Hall, a birthday-problem variant, coupon collector). Then comes the decision stage: buy anywhere from 0 to 10 tickets of this design at this price. Choosing 0 walks away and refunds the ticket price in full - the game still shows what the ticket would have paid, for the record.

This variant gives each ticket 13 questions on its own countdown of 45 seconds per question - 9:45 per ticket, resetting with every new ticket. Hitting zero marks all unanswered questions incorrect and jumps you to the buy decision; the timer never costs money directly and never ends the session. The session ends only when you cash out or your bankroll hits $0.

How scoring works

A ticket that shows multiplier m pays exactly m times the ticket price - the three tiers are $0, a mid prize of 1x or 2x, and a jackpot of 4x, 5x, 6x, or 8x on a normal ticket, or 8x, 10x, or 12x on a rare mispriced one. The $0 tier always covers between 50% and 80% of the probability mass.

70% of generated tickets are fair-ish: expected value at or below 1x the price, a genuine house edge. 30% are hot: EV strictly above 1x, mispriced in the player's favor. The split is constant and is never adjusted based on whether you are winning or losing.

Walking away at the quantity step (choosing 0) refunds your ticket price - a guaranteed EV of exactly 1.0x, better than most tickets in the game.

Compute EV before the buy decision, every time

The flip reveal hands you the full distribution, so EV is one line of arithmetic: multiply each prize by its probability and add. On a $10 ticket showing P($0) = 5/8, P(2x = $20) = 2/8, P(4x = $40) = 1/8: EV = (2/8)($20) + (1/8)($40) = $5.00 + $5.00 = $10.00 - exactly 1.0x, the best a fair-side ticket ever gets. A harsher fair shape like P($0) = 6/8, P(1x = $10) = 1/8, P(5x = $50) = 1/8 gives EV = (1/8)($10) + (1/8)($50) = $7.50, a 25% house edge.

The end-of-session screen states the game's own summary: most tickets pay out about 7/8 of the price - a 12.5% house edge. Any ticket whose EV multiple comes out at or below 1 is a losing or at best break-even bet, and the refund option (buy 0) strictly dominates buying it.

The no-replacement questions are the interview trap

The batch question tells you a ticket came from a batch of N with exactly W winners, and the previous ticket from the batch won nothing. The answer is W/(N-1), not W/N: removing a known non-winner shrinks the pool but leaves all W winners in it. Treating the draws as independent (answering W/N) is the gambler's-fallacy mistake in reverse - here the odds really do move, in your favor.

The geometric question is the other classic: if P(win something) = p, the expected number of tickets to the first win is 1/p. A ticket with p = 6/16 takes 16/6, about 2.67 tickets on average - and every one of those tickets still loses money in expectation if EV is below the price.

Thirteen questions is nearly the whole bank - drill for coverage

The bank holds 18 questions: six topics (EV, Bayes, batch conditional, combinatorics, geometric, classic) in three phrasings each, where the third phrasing uses formal notation - P(B|A), E[X], C(n,r), X ~ Geometric(p). Every set guarantees all six topics and exactly 1 to 3 notation questions; at 13 questions you see most topics twice, so a shaky topic will be found out. Pace at the built-in 45 seconds per question.

The recurring computations to have cold: EV as a probability-weighted sum; P(top prize | win) = P(top)/P(win); batch conditional W/(N-1); C(n,r) up to 220 (pools 6-10, picks 2-4); E[tickets to first win] = 1/p; and P(lose then win) = (1-p) x p for the two-ticket geometric variant.

A worked example

You buy a $10 ticket. The flip reveal shows: $0 with probability 5/8, $10 (1x) with probability 2/8, $40 (4x) with probability 1/8.

EV = (5/8)(0) + (2/8)($10) + (1/8)($40) = $2.50 + $5.00 = $7.50. Expected profit = $7.50 - $10.00 = -$2.50, a 25% house edge. This is a fair-side shape - decline it.

Sample questions on this ticket: P(win nothing) = 5/8. P(top prize | win something) = (1/8) / (3/8) = 1/3. P(win something) = 3/8, so expected tickets to first win = 8/3, about 2.67.

At the decision stage you set quantity to 0. Your $10 is refunded and your bankroll is unchanged - an EV of exactly $10.00 versus the $7.50 of playing. Walking away outperformed the ticket by $2.50 of expectation, which is the lab's title acted out.

Common mistakes

• Buying before doing the EV multiplication. The table is face-up - a ticket that pays 7/8 of its price loses you 12.5 cents per dollar in expectation, every time.

• Answering W/N on the batch question. After a non-winner is removed, W winners remain among N-1 tickets: the answer is W/(N-1).

• Sizing up on a fair ticket to chase the jackpot. Quantity multiplies expected loss just as faithfully as expected gain.

• Ignoring the walk-away option. Buying 0 refunds the full price - a guaranteed 1.0x that beats roughly 70% of tickets generated.

• Sprinting the long set. 9:45 is the same 45 seconds per question as the short mode - unanswered questions at 0:00 are all marked wrong, so distribute the clock rather than hoarding it.

Why interviews test this

Expected value over an explicit discrete distribution is the single most common opener in quant interviews - it tests whether you instinctively price a bet instead of reacting to its best case. The follow-ups here mirror real follow-ups: conditional probability without replacement (does the candidate update on removed outcomes?), the geometric distribution (time to first success), and Bayes on nested events.

The deeper lesson interviewers probe is bet selection and sizing: a trader who can compute EV but still takes -EV bets, or who sizes the same regardless of edge, fails the job even if the arithmetic is right. Declining the bet - quantity 0 - is frequently the correct answer, and saying so out loud is a strong interview signal.