# Quantitative Researcher Interview Questions

Quantitative researcher interview questions shared by candidates

## Top Interview Questions

3) Poker. 26 red, 26 black. Take one every time, you can choose to guess whether it’s red. You have only one chance. If you are right, you get 1 dollar. What’s the strategy? And what’s the expected earn? expected earn is 25 cents. 1/2*1/2*1, prob of choosing to guess is 1/2, prob of guessing right is 1/2, and the pay is $1 I would start picking cards without making a decision to reduce the sample size. This is risky because I could just as easily reduce my chances of selecting red by taking more red cards to start, as I could increase my chances of selecting red by picking more black cards first. But I like my chances with 52 cards, that at some point, I will at least get back to 50% if I start off by picking red. Ultimately, I can keep picking cards until there is only 1 red left. But I obviously wouldn't want to find myself in that situation so I would do my best to avoid it, by making a decision earlier rather than later. Best case scenario, I pick more blacks out of the deck right off the bat. My strategy would be to first pick 3 cards without making a decision. If I start off by selecting more than 1 red, and thus the probability of guessing red correctly is below 50%, then I will look to make a decision once I get back to the 50% mark. (The risk here is that I never get back to 50%) However, if I pick more than 1 black card, then I will continue to pick cards without making a choice until I reach 51% - ultimately hoping that I get down to a much smaller sample size, and variance is reduced, while odds are in my favor that I choose correctly. The expected return, in my opinion, all depends on "when" you decide to guess. If you decide to guess when there is a 50% chance of selecting correctly, then your expected return is 50 cents (50% correct wins you $1 ; 50% incorrect wins $0 --- 0.5 + 0 = .5) If you decide to guess when there is a 51% chance of selecting red correctly, then the expected return adjusts to (0.51* $1) + (0.49 * $0) = 51 cents. So, in other words, your expected return would be a direct function of the percentage probability of selecting correctly. i.e. 50% = 50 cents, 51% = 51 cents, 75% equals 75 cents. Thoughts? There is symmetry between red and black. Each time you pull a card it is equally likely to be red or black (assuming you haven't looked at the previous cards you pulled). Thus no matter when you guess you odds are 50% and the expected return should be 50 cents. Show More Responses |

2) A. 10 ropes, each one has one red end and one blue end. Each time, take out a red and a blue end, make them together. Repeat 10 times. The expectation of the number of loops. B. 10 ropes, no color. All the other remains the same. |

A tosses n+1 coins. B tosses n coins. B wins if he has at least as many heads as A. What is the probability that B wins? |

1) Tow coins, P(head)=1/3, P(tail)=2/3, design a way to get the effect of fair coin |

If X, Y and Z are three random variables such that X and Y have a correlation of 0.9, and Y and Z have correlation of 0.8, what are the minimum and maximum correlation that X and Z can have? |

5 babies in a room, 2 boys and 3 girls. one baby with unknown sex is added. Randomly choose one baby and the result is a boy. What's the prob that the added baby is a boy? |

You play a game with someone, the rule is to take turns to put a quarter on a round table ( can be any size, but must be of symmetrical shape). You are trying to cover the table up with quarters, you lose when there's no place for you to put down the quarter. Should you be the first person to play ? |

How to generate a gaussian distribution with rand() |

Most difficult question i received was what were the 5 steps to a conersation? |

Suppose there is a series of 100 light bulbs labeled 1 through 100 that all start in the off position. Person 1 walks in the room and turns on every light bulb. Person 2 walks in the room and flips the switch on every 2nd light bulb (i.e. he turns off all even numbered light bulbs). Person 3 walks in the room and flips the switch on every 3rd light bulb (turning some on and some off). This process continues until person 100 goes through and flips the switch on the 100th light bulb. Which light bulbs will be lit up at the end? |

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