ptmminh / quanttest

Jane Street quant interview/test

Quantitative Research test

October 13, 2016

In Fall 2016, I applied to a quant trading firm (Jane Street). They had me take a HackerRank test for which I was told it was going to be on 'basic math and probability'. To me, these questions require practice so I thought I would post these questions here as sample questions so that future interviewees can practice and be more prepared than I was.

Time: 30 minutes

4 questions

1. We randomly select 4 numbers from the set of the first 16 prime numbers, without replacement. What is the probability their sum is even? Why?
2. You have been chosen to play a game involving a 8-sided die. You get to roll the die once, see the result, and then may choose to either stop or roll again. Your payoff is the sum of your rolls, unless this sum is greater than 8, in which case you "bust" and get nothing. What is your strategy for this game? That is, for each possible outcome of the first roll will you choose stop or to roll again?
3. A bag has three coins in it which are visually indistinguishable, but when flipped, one coin has a 20% chance of coming up heads, another as a 20% chance of coming up heads, and the last has 60% chance of coming up heads. I randomly draw a coin from the bag and flip it, and the result comes up heads. What is the probability that if I reflip this coin, it will come up heads again? Why?
4. We have two urns. You can't tell them apart from the outside, but one has four $1 chips and six $10 chips, and the other has three $1 chips and seven $10 chips. You randomly draw a chip from one of the urns and it happens to be a $1 chip. Without replacing this draw, I offer you a chance to draw and keep a chip from either urn. Should you draw from the same urn or the opposite urn, and what is the expected value of the chip you draw? Why?

Good luck. Feel free to compare your solutions against mine afterwards.

quanttest's Issues

Q4: Expanding on the intuition of the urn probability

I wasn't able to intuitively understand how to mathematically derive this at first glance so I would like to add the explanation for this statement in case anyone faces the same issue:

Denoting $D_1$ as the first draw from the urns, the $P(X=1)$ shown above actually means $P(X=1|D_1=1)$. Using the Bayes's rule,

$P(X=1|D_1=1) = \dfrac{P(D_1|X=1)P(X=1)}{P(D_1|X=1)P(X=1)+P(D_1=1|X=2)P(X=2)}$

Since $P(X=1)=P(X=2)=0.5$, we simplify the equation:

$= \dfrac{P(D_1|X=1)}{P(D_1|X=1)+P(D_1=1|X=2)}=\dfrac{0.4}{0.4+0.3}=\dfrac{4}{7}$

Similarly, with $P(X=2|D_1=1) $, we will get $\dfrac{3}{7}$

Mistake in question 2

There's a mistake in question 2. I think the optimal strategy would be to re-roll if first roll <= 3.

You only calculate probabilities but one should look at the expected value. One should only perform a second roll when its expected value is larger than what we currently have by our first roll.

first roll	Expected value second roll
1	1/8 * (2 + 3 + 4 + 5 + 6 + 7 + 8) = 35/8
2	1/8 * (3 + 4 + 5 + 6 + 7 + 8) = 33/8
3	1/8 * (4 + 5 + 6 + 7 + 8) = 30/8
4	1/8 * (5 + 6 + 7 + 8) = 26/8
5	1/8 * (6 + 7 + 8) = 21/8
6	1/8 * (7 + 8) = 15/8
7	1/8 * (8) = 1
8	0

As soon as the first roll = 4, the expected value that we obtain by making a second roll is lower than the value we already have. As such, optimal strategy is to re-roll when first roll <= 3.

Another extra question that could be asked here is how much you would be willing to bet. For this, we can calculate the expected value of this game. For first roll equal to 1, 2 or 3, we take the expected value of the second roll. For the others, we take the expected value of the first roll.
Giving us: E[X] = 1/8 * (35 / 8 + 33 / 8 + 30 / 8 + 4 + 5 + 6 + 7 + 8) = 5.28125
--> We are willing to bet anything lower than this value.

One could calculate this for an alternative strategy as well. E.g. re-rolling when first roll <= 4 gives us:
E[X] = 1/8 * (35 / 8 + 33 / 8 + 30 / 8 + 26/8 + 5 + 6 + 7 + 8) = 5.1875

Simulation in Python (ugly code!)

>>> import numpy as np
>>> reward_3 = []
>>> reward_4 = []
>>> for _ in range(1_000_000):
...   r1 = np.random.randint(1, 9)
...   if r1 <= 3:
...     strat1 = r1 + np.random.randint(1, 9)
...   else:
...     strat1 = r1
...   if r1 <= 4:
...     strat2 = r1 + np.random.randint(1, 9)
...   else:
...     strat2 = r1
...   if strat1 > 8:
...     strat1 = 0
...   if strat2 > 8:
...     strat2 = 0
...   reward_3.append(strat1)
...   reward_4.append(strat2)
... 
>>> print(np.mean(reward_3))
5.280888
>>> print(np.mean(reward_4))
5.186363

Notice how the average rewards our expected values very well (law of large numbers)

Possible mistake

In the solution of the third exercise, i think that in the part above the first probability of each addendum should be conditional to X=C_i instead that F_i=H.