The expected maximum of two dice, and other classic quant probability questions

“Roll two dice. What's the expected value of the larger one?” This question, or a close cousin, comes up a lot in quant trading and research interviews. It's popular because it's easy to state, has a clean answer and leads naturally into harder follow-ups. The interviewer isn't really testing whether you know the answer. They're watching how you get there and how quickly you adapt when they change it.

Here's how a typical sequence of questions goes, from warm-up to stretch, with the maths for each.

Warm-up: at least one six in two rolls

Use the complement. The chance of no six on one roll is 5/6, so on two rolls it's 25/36. The chance of at least one six is 1 − 25/36 = 11/36, about 30.6%.

Say “I'll use the complement” before you start. It tells the interviewer you're picking a method, not stumbling into one.

Core: the expected maximum of two dice

The direct approach is to find the probability that the maximum equals each value. The maximum is at most k when both dice are at most k, which has probability (k/6)². So the probability the maximum is exactly k is the difference between “at most k” and “at most k − 1”.

Sanity check before you say it: one die averages 3.5, so the larger of two should be noticeably higher but well short of 6. 4.47 passes.

The faster way: add up the tail probabilities

For any random variable that takes values 1, 2, 3 and so on, the expected value equals the sum of P(X ≥ k). This trick is worth knowing cold because it makes the follow-ups much quicker.

A nice extra: the minimum and maximum of two dice add up to the sum of the dice, which averages 7. So E[min] = 7 − 161/36 = 91/36 ≈ 2.53. Mentioning that unprompted is the kind of thing interviewers remember.

Harder: three dice, then n dice

With the tail method, three dice is the same calculation with cubes.

If they ask what happens as n grows, the sum shrinks towards zero and the expected maximum approaches 6. That's obvious once said, but say it, along with the fact that it gets there fairly fast: by three dice you're already just under 5.

Hardest: the reroll game

“Roll a die and I pay you the face value in pounds. After the first roll, you may roll again once, but then you must take the second result. What's the game worth?”

Work backwards. If you reroll, you get a fresh roll worth 3.5 on average. So keep any first roll above 3.5 (a 4, 5 or 6) and reroll a 1, 2 or 3.

Stretch: two rerolls

Now you get up to three rolls. With two rolls left, the game is worth 4.25 (from above). So on the first roll, keep only results above 4.25, which means a 5 or 6.

The interviewer may then ask what you'd pay to play, or whether you'd take a bet at a given price. Answer with the expected value, then mention variance or risk if the stakes are large relative to your bankroll. They want to see you treat it as a trading decision, not just a sum.

What to do while you're working

1. Say your method before the maths. “I'll use complements” or “I'll work backwards from the last roll.”
2. Write as you go. Most interviews give you paper or a whiteboard. Small tables like the ones above stop you dropping a term.
3. Sanity check out loud. “That's between 3.5 and 6, which makes sense.”
4. Give exact fractions, then decimals. 161/36 first, then “about 4.47”.
5. If you're stuck, simplify. Try one die, or a coin, then build up. Interviewers often hint when they see you do this.

Practising these properly

The questions themselves are well known. What varies is the follow-up, and how calm you are when it arrives. Practise in ladders: one idea, then a harder version straight after, so you get used to reusing your last answer. That's how RoundReady's quant track is set up: five questions that go Warm-up, Core, Harder, Hardest, Stretch, with a scratchpad beside you that evaluates the lines of working you type.

If you want a check on your own: try the expected maximum of two uniform random numbers between 0 and 1. Using the same idea with integrals, the answer is 2/3. For n of them, it's n/(n + 1).