Fresh Grade 月入17萬的Intern | 2024/25 畢業生最高人工的行業

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近二十年，一些高科技交易公司和 原本的莊家 Market Maker 透過Quan Trading、Algo Trading等策略大行其道賺大錢，Quantitative Trader 交易員的薪酬也水漲船高，現時這些公司已經給出年薪25萬美元(沒有打錯)，大約港幣16萬以上的底薪去請 Intern 或者Fresh Graduate。以下是幾間在香港有辦公室，在2024和2025年都有Opening請Trader和Engineer的交易公司：

Quan Trader Intern Grad 薪酬

Intern/Grad Job	年薪(美元)	月薪(港幣)
D.E. Shaw Group	264K	171,600
Tower Research Capital	260.6K	169,447
Citadel	260K	169,002
Jane Street	250K	162,497
Hudong River Trading	249.6K	162,887

他們對量化交易員的要求是要天才，是要心算很強的人，不只是普通計算兩位數連點數互相相乘可以在兩秒內作答的人，還需要是可以短時間迅速可以計算機會率、統計難題的人。通常這些天才玩Poker都很厲害，讀書背景也很強，所以不要問這些公司在學歷方面要求什麼了，無論學士/Master/博士畢業都會請。

另外還有幾間交易公司其實提供更高的薪酬，包括 Radix Trading、Five Rings、Susquehanna (SIG)等，開出2.5萬至2.9萬美元的薪酬，不過他們在香港就沒有辦公室或請人。

以下是一些面試題目的參考

Question

Suppose we play a game with a die where we roll and sum our rolls. We can stop at anytime and our score is the sum. However, if our sum is a multiple of 10, our score will be zero and our game is over. What strategy will yield the greatest expected score? What about the same game played with values other than 10?

Answer

1. Let’s define E(s) as the expected value of the game when our current sum is s.
2. If s is a multiple of 10, E(s) = 0.
3. For any other s, we have two choices:
   a) Stop and take s as our score
   b) Roll again
4. If we roll again, there’s a 1/6 chance for each possible outcome. So: E(s) = max(s, (E(s+1) + E(s+2) + E(s+3) + E(s+4) + E(s+5) + E(s+6)) / 6 or 0)
5. We can solve this recursively, starting from s = 9 and working our way down.
6. For s = 9, E(9) = max (9, (0+11+12+13+14+15)/6 = 10.9) > 9
7. For s = 8, E(8) = max(8, (9 + 0 + 11 + 12 + 13 + 14) / 6) = 9.8 > 8
8. We can continue this process for lower values of s.
9. The optimal strategy will be to roll again if the expected value of rolling is higher than the current sum, and to stop otherwise.
10. After calculating, we find that the optimal strategy is:
    • Roll if s < 8, 18, 28, 38…when it closer to 10, 20, 30, 40 …
    • Stop if s ≥ 8, 18, 28, 38…when it closer to 10, 20, 30, 40…

For games with values other than 10, the same principle applies, but the specific thresholds will change. For example:

• If the losing condition is multiples of 5, the optimal strategy would be to stop at 4.
• If the losing condition is multiples of 15, the optimal strategy would be to stop at 12 or higher.

Question

Divide a pack of cards into two piles. What is the expected value of the lowest card value on the left pile.

Answer

First, let’s clarify some assumptions:

• We’re using a standard 52-card deck.
• We’re considering Ace as the lowest card (value 1), and King as the highest (value 13).
• The division into two piles is random.

The key insight here is that the lowest card on the left pile will be the lowest card in the entire deck, unless that lowest card happens to be in the right pile.

Let’s consider the probability of the lowest card being in the left pile:

• There’s only one lowest card in the deck.
• The probability of this card being in the left pile is 1/2.

If the lowest card is not in the left pile (probability 1/2), then the lowest card in the left pile will be the second lowest card in the deck, unless that’s also in the right pile. This continues for the third lowest, fourth lowest, and so on.

E(lowest on left) = 1 * (1/2) + 2 * (1/4) + 3 * (1/8) + 4 * (1/16) + … = 2

Question

What is the expected number of rolls of a 6 sided die to get each face at least once?

Answer

Let E(k) be the expected number of additional rolls needed to collect a new face when we already have k different faces.

When we have k faces, the probability of getting a new face on a single roll is (6-k)/6.

The expected number of rolls to get a new face is the reciprocal of this probability:
E(k) = 6/(6-k)

The total expected number of rolls is the sum of E(k) for k from 0 to 5:

E(total) = 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1
= 1 + 1.2 + 1.5 + 2 + 3 + 6
= 14.7

Therefore, the expected number of rolls to get each face at least once is approximately 14.7.

Question

When you roll a six sided die, you receive $1 for every pip on the side rolled (i.e. rolling “2” earns you $2, “3” earns you $3 and so on). Assuming you must pay an agreed amount each time the dice is rolled, how would you make that market?

Answer

• Expected Payout: The expected payout of a single roll is (1+2+3+4+5+6)/6 = 3.5.
• Profit Margin: You need to add a profit margin to the expected payout. Let’s say you want a 10% profit margin. This would add 0.35 to the expected payout.
• Bid/Ask Spread: You’ll set your bid to 3.15 and ask to 3.85

Question

There are 100 prisoners and a room with a lightbulb. Prisoners are called in at random to turn the bulb on or off. Upon exiting, the room, they must guess whether they are the last of the 100 to enter the room. If they are correct, they go free, if not, everyone fails. Prisoners may communicate beforehand, but once the game begins, they are separated and have no idea who has or hasn’t entered the room yet. What strategy would ensure the prisoners would be successfully released?

Answer

Designate a Counter: Before the game begins, the prisoners agree that one specific prisoner (let’s call them the “Counter”) will be responsible for counting how many unique prisoners have entered the room.

1. Lightbulb Rules:
   • The lightbulb starts in the OFF position.
   • The Counter will turn the light ON whenever they enter the room and find it OFF.
   • All other prisoners will turn the light OFF the first time they enter the room and find it ON. After that, they will leave the light as they found it (either ON or OFF).
2. Counting Mechanism:
   • Each time the Counter enters the room and finds the light ON, they will turn it OFF and increment their count by one.
   • The Counter will keep track of how many times they have turned the light ON.
3. Winning Condition:
   • Once the Counter has turned the light ON a total of 99 times, they can confidently declare that all other prisoners have entered the room at least once. This is because each of the other 99 prisoners has turned the light OFF exactly once when they found it ON.
   • The Counter can then guess that they are the last prisoner to enter the room.

Why This Works

• Unique Identification: By using the lightbulb as a signaling mechanism, the prisoners can uniquely identify when each of the other prisoners has entered the room.
• No Communication After Start: Since the prisoners cannot communicate after the game starts, this strategy allows them to rely on the lightbulb’s state to convey information about who has entered the room.

Question

I’m dealing a deck of cards. You can stop it at any time and if the next card is red, you win. What is the optimal strategy for winning?

First, let’s consider the probability of winning at any given point:

• In a standard 52-card deck, there are 26 red cards (13 hearts and 13 diamonds) and 26 black cards.
• As cards are dealt, this ratio will change.

1. The key insight is that the optimal decision at any point depends on the proportion of red cards remaining in the deck.
2. Let’s define:
   r = number of red cards remaining
   b = number of black cards remaining
   n = total cards remaining (r + b)
3. The probability of the next card being red is r/n.
4. If we don’t stop, we have two possibilities for the next round:
   • The next card is black (probability b/n), and we’re in a similar situation with one less black card.
   • The next card is red (probability r/n), and we’ve lost the opportunity to win at this point.
5. The optimal strategy is to stop when the probability of winning now (r/n) is greater than the expected probability of winning if we continue.
6. After some mathematical analysis (which involves solving a recurrence relation), the optimal strategy is stop when r > b In other words, stop as soon as there are more red cards remaining than black cards.
7. At the start of the game, there are equal numbers of red and black cards, so we should not stop immediately.
8. As cards are dealt, we need to keep track of the count of red and black cards that have been revealed.
9. The moment the count of revealed black cards exceeds the count of revealed red cards, we should stop, as this indicates there are now more red cards remaining than black cards.

This strategy gives the highest probability of winning the game. The exact probability of winning with this optimal strategy is approximately 0.7545 (about 75.45%).

Question

What is the probability that two people in a room full of 12 people share the same birthday?

Answer

There are 365 possible birthdays (ignoring February 29th for simplicity). The probability of the first person having a unique birthday is 365/365 = 1. The probability of the second person having a different birthday is 364/365, since one birthday is already taken. The probability of the third person having a different birthday is 363/365, and so on.

The probability of all 15 people having different birthdays is:

(365/365) × (364/365) × (363/365) × … × ((365-11)/365) ≈ 0.8330

So, the probability of at least two people sharing the same birthday is:

1 – 0.8330 ≈ 0.1670

Therefore, the probability that two people in a room of 15 share the same birthday is approximately 50.73%.

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