Research associate Interview Questions

Did you get the final round ?

No yet

Did you hear back from them yet? I still haven’t heard anything.

Had a shot at this, Basically, the trick is to work backwards. Firstly, consider a series with only 1 game, build to 3 games, then to 5 and finally to 7 1 Game Series If you start with no money, you simply bet $100 on A and you obtain the desired result. If you start with some money (X, negative or positive), you can never obtain desired result: Bet 0: X, X (need 100, -100) Bet X: 2X, 0 Bet 2X: 3X, -X Important result (1) 3 game series Starting with no money, make a bet of X on A. Due to symmetry, we will consider only the situation if A wins (B is skew symmetric, i.e. identical bit negative). A:B = 1:0 If A wins again, they have won the series hence they need to win a 100 total. However, if B wins, we are back to a 1 Game Series, and need to enter this stage with no money too. Hence, the bet is identical to the first bet, i.e X. If A wins, you have 2X = 100, X=50 and if A loses you have a 1 game series, following above strategy. Summary, bet $50 on A on game 1 and $50 on A on game 2. If they win both, you're up a $100, they lose both, you're down $100 and they draw, you have zero and it's a 1 game series. If you start with some money for a game 3 series, symmetry no longer holds, as after stage 1, you don't have skew symmetry. Say, you have X if A wins and -Y if B wins, in the event of a tie, you need to have 0 for a 1 game series to lead to desired result. However, if A wins game 2, you have 2X and if B wins you have -2Y, which means for 2X=100=2Y, we can't start with money. This argument holds for 5 and 7. 5 game series: Similarly as before, bet X on A and then X again on A. 1:1 Now a 3 game series with 0 start, recurse 2:0 You have 2X (or -2X in the converse) Now if A wins we need to bet the amount such that the end total is $100. If A loses, we have a 2:1, or a 1:0 lead in a 3 game series. Assume you bet Y 3:0 You have 2X+Y 2:1 You have 2X-Y Following the 2:1 result, which is identical to a 1:0 lead in a 3 game series, we look back and note that 2X-Y = 50 for the same stage in the 3 game series. And 2X+Y=100 for the winning total to be correct: 2X-Y = 50 2X+Y = 100 4X = 150 X = 75/2 Y=25 Summary: Bet 75/2 on A first 2 games. If it's a tie, follow 3 game series. If you win both (or lose both), bet 25 on the next round. If you win, you have a 100, if you lose you're at the 1:0 stage of the 3 game with 50 which can give desired outcome. 7 Game Series: Again as before, bet twice on A with X 1:1 - follow 5 game series 2:0: Bet Y on A (You have 2X in hand) If Lose the game: is a 2:1 result, or a 1:0 lead in a 5 game series You have 2X-Y in hand, and at the same stage in a 5 game series, you need 75/2 3:0: Now bet Z on A (2X+Y in hand) If you win, 2X+Y+Z = 100 You lose, you have 3:1 which is identical to the 2:0 lead in the 5 game series. From above, you need to have 75 at this stage (2X+Y-Z=75) Solving, using the last two eqn alone 2Z = 25 Z = 25/2 Everything else 2X+Y = 175/2 2X-Y = 75/2 4X = 250/2 X = 125/4 Y = 25 In summary, bet 125/4 on A for the first 2 games. If tie, follow 5 game series. Otherwise you've won both (or lost both). Then bet 25 on the next game. If you lose [2:1], follow 5 game series with 1:0 lead. Otherwise, bet 25/2 on the final game. If you win, you have the 100, otherwise [3:1] follow the 5 game series with a 2:0 lead.

I got the same answer as Varun. But I think it's easier to do in terms of a tree. Start with (0,0) and then two possible nodes (0,1) and (1,0). If either team reaches a 4, it is a terminal state, with value -100 or +100. You'll end up with a tree that widens and then narrows to (3,4) and (4,3). Now use backward induction. Label each node with Value|Bet.

You immediately bet 100$ that A will win the series and - that's it.

P: The 5th fish has been caught, P=P1+P2+P2, where, P1: The 5th fish been caught as the first one P2: ... second one... P3: ....third one (or the last one). P1 = 0.5^5 P2 = 4*0.5^5 P3 = 6*0.5^5 The fifth fish will not be caught Pc = 1-P = 1-P1-P2-P3

I got a different answer :/ There's 2 ways the fish can survive: either the bear leaves before the 5th fish, or the bear fails to catch it. The probability the bear leaves is the probability the bear gets 3 fish in 4 attempts, or 4C3/16 = 1/4 The probability the bear tries * fails to catch the fish is 3/4 (the probability of trying) * 1/2 (the probability of failing) = 3/8. The total probability is thus 5/8? Not sure what I did differently

I had read his name 10 times before the interview, but I forgot it when I actually needed it.

Dimon

Masters in computer science

With 20 years experience as a lecturer in university

Yes

Yes

straight forward O(n) through memorization.

This is similar to the maximum sub-array algorithm questions. Can look it up.