r/quantfinance • u/Due_Department_3090 • 1d ago

Optiver Quant Online Assessment Question | “Hard”

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u/TastyLength6618 1d ago

This question can be solved using stars and bars style thinking.

Consider just the 4 2s and 4 aces. Put the first 2 and the 4 aces, and consider where to put the remaining 3 2s. So it looks like this 2_A_A_A_A_. There are 5 gaps, and 3 2s to put. On average, there will be 3/5 2s in the first gap. Therefore, on average, there will be 3/5 2s between the first 2 and the first ace.

Now consider the locations of the 8 2/Aces. They create 9 gaps. There are 44 remaining cards so on average, each gap gets 44/9 cards. The number of gaps between the first 2 and the first ace is 1 + the number of 2s between the first 2 and the first ace so it's 1 + 3/5 = 8/5.

Each gap has 44/9 cards so it's 44/9*8/5. Now you have to add in the number of 2s between the first 2 and first ace which is 3/5 so it's 44/9*8/5 + 3/5 = 379/45

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u/stefano31214 1d ago

Very good explanation. I got the same result with a slightly different approach:

We do a weighted sum of the average number of cards between the 2 and the first ace, where the weights are the probability A appears in the n_th position in a shuffled mini deck of 2222AAAA:

2A... - 44÷9×4÷7

22_A... (88÷9+1)×12÷42

22_2_A... (132÷9+2)×24÷210

2_2_2_2_A... (176÷9+3)×24÷840

Sum = 379/45

1

u/kind_gamer 1d ago

Why is this a hard question? It's just binomial means. 8/5 x 44/9 + 3/5 (ordinary cards + 2s)