if u don't know what was picked then it's 25% whether you replace or not. If yes, then it is 1/4 since on the 4th draw you will have 4 clubs out of 16. pick it up. Hence, impossible. I took a Business stats course this year but it was a joke compared to the prob Qs/brainteasers seen on here. this may be the long way but i still believe its correct. "You have a friend who has two children. Am I misreading the question? 11 2 / 36 I'm interested in trading (although the recruitment process for some of the prop shops looks absolutely terrifying/awesome). - 3 queens and Below, we have a couple example questions asked previously during the Jane Street interview process. doesn't matter. But if the cards are not being replaced, then you have to sum up the 4 distinct possibilities. in fact, my guess would be it goes something like this - P(2>1) = 0.5, P(3>2) = 48.5/95 (we know card 1 is one of the cards less than card 2, so the probability of card 3 being greater than card 2 is always skewed by 1 card out of 95? consists of 6 branches, each corresponding to what you roll, each with a prob of 1/6 . Rolling a "3" has a probability of 1/6. Also, you should be multiplying, not adding. Jane Street Trading Probability Interview Questions. the probability in the interviews is nothing that is technically/theoretically advanced, its really the foundation concepts of probability like Bayes theorem but under a great amount of time pressure and in a brainteaser format. ", Competition is a sin. same color? Anyone know how to do it? You have a friend who has two children. 12 1 / 36, The random variable X assumes a value equal to the sum of two dice rolls. You assume that the other 3 of the same cards are somewhere within the 12 remaining cards. by FIASCO. Since X and Y are independent, the P(X>Y | Y=y) = P(X > y) = sum of f(x) from x=y+1 to x=infinity or equivalently, 1 - f(2) - f(3) ... f(y). The problem was missing information, as the expected value depends on the strategy the user uses. What is the expected number of flips to get two heads in a row? at the end of each branch is another node for 2nd roll, each with 6 branches of its own, same as above. = 4.25. Rolling a "6" has a probability of 1/6. It's def not 36, that's just avg # of rolls required to roll 2 6s in general. Anyone have any suggestions for how to learn how to do these? Anyone figure out number 1 yet? how can I solve this card probability question? are you trying to get a qualitative understanding of the greeks? I had an interview with a top BB and not only did they ask me what I would pay for 1 roll, they 2 rolls, then 3 rolls, then they asked what I would pay for 2 rolls should I be allowed to keep the maximum of the two rolls. Or have a quick look into wikipedia: http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem, In reply to balbasur wrote: WSO Free Modeling Series - Now Open Through November 30, 2020, Hardest Probability and Statistics Interview Questions, Caring for terminally ill parent while in IB, 1st Year Analyst in Investment Banking - Mergers and Acquisitions">, Prospective Monkey in Investment Banking - Mergers and Acquisitions">, 1st Year Analyst in Investment Banking - Industry/Coverage">, Q&A - I'm a Non-target who broke into MM IB with no prior experience, http://pratikpoddarcse.blogspot.com/2009/10/lets-say-keep-tossing-fair-…, Investment Banking Interview Case Samples, Investment Banking Interview Questions and Answers, Financial Modeling Training Self Study Courses, Certified Sales & Trading Professional - Director, Investment Bank Interview - Toughest Questions, Hedge Fund Interview Sample Pitches - Long/Short, Certified Sales & Trading Professional - 1st Year Analyst, Investment Banking Interview Brainteasers, Modeling Training - Special for WSO Members. If you struggle with bayes approach (probability tree) your going to hate game theory where you then have to use backwards deduction. Sales and Trading: Interview Guide written by Hedge Fund Trader, http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem, Certified Prop Trading Professional - Trader, Certified Equity Research Professional - Vice President. For each draw {A , B , C , D} , isn't there a 1/4! Conditional expected value: (2/6)2 + (1/6)(3+4+5+6) = 3.6667, First roll outcome: 1 Not a textbook, but covers just about every quant question you'll face in interviews: http://www.amazon.com/Heard-Street-Quantitative-Questions-Interviews/dp…. Same with the first, second, etc. There. f'(0) (1-k) = a Game theory was so long ago.... Well according to Murphy's law, there is a 100% chance that you will get a probability question, given how worried you seem. Intuitively, I think it would be: leaving out the first queen, you now have 51 cards to choose from to form the remaining 12 cards in your hand. For the second question, you'll eventually win, i.e. First, it is ridiculed. Whether you are drawing from 97, 500, or 1,000,0000 cards, the probability of drawing 4 cards in ascending order is the same. the above 2 questions are essentially the same. For a discrete random variable, E(X) is calculated as. Then count how many ways that can happen, which is (48 C 9). In reply to Anyone figure out number 1 by Anal Analyst. Not sure about the probability, because you could make an argument about the .600 might not be a good predictor if it is not many games, and it probably depends on what team they are playing. Doesn't matter when A gets his heads as long as B gets it first. I think I got the answers right? Prove that no three positive integers a, b, and c can satisfy the equations (a^n)+(b^n) = (c^n) for any integer value of n greater than two. take derivative of top and bottom (L'Hopital's rule) and get In reply to probability question by sdumb. That's the probabilty that no goals are scored in any game. Why is it not 36? then multiply that by 1/6 twice, i.e.e 1/36. Im thinking its symetric so 1/2^3 either that or find expected first pick then prob of getting higher than that for the second card and so on. There are 6 possible outcomes for the 1st roll (1, 2, 3, 4, 5, 6), each with a probability of 1/6. 2 ways to make X=4 If you roll over 3.5, then the score you got must have been either 4, 5, or 6. Disregard this. Ok obviously for (a) the answer is $3.50 because you just mult. know dice/coinflip EV etc.. To show the P(X>Y) for all Y, we must calculate the infinite sum: Luckily I noticed the second time it happened, but watch out for that. I'm looking deeper into options and realize that I need a better base of prob knowledge to really go any further. Does anybody have any website recommendations to refresh my probability and stat. To get my head back in the game? Instead, the interview questions mirrored the ones used by quantitative trading firm today! Solutions To Probability Interview Questions . In reply to Probability Problem has me stumped by testman. You have a friend who has two children. Neither of those companies will ask you about concepts you haven't seen before. The other possibility (which is equally likely) is that you get 1, 2, or 3. Are you expected to do this all in your head where you think outloud and explain HOW you would arrive at the answer? Such thing converges since it is a geometric series. since each xi is bounded by 0 and 1, the latter probability is gonna be a finite (1 to n) sum of 2exp (-2d^2*n). It doesn't matter if it's the fourth or fourteenth card, if you don't know anything about the other cards being drawn.

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