The Expected Value of Rolling Once and Twice ,, WSO Discount - Financial Modeling Courses, Certified Sales & Trading Professional - Assistant, Certified Equity Research Professional - 1st Year Associate, Certified Hedge Fund Professional - Portfolio Manager, Certified Hedge Fund Professional - 1st Year Associate,[/quote,,, Question with Card Probability brainteaser, Certified Consulting Professional - 3rd+ Year Analyst, Certified Hedge Fund Professional - 1st Year Analyst. roll a die, and you get paid what the dice shows. This is due to each mutation either closing the gap by 3 or keeping them equal. you'll eventually roll a 6 if you are allowed to keep rolling indefinitely. certainly challenging (especially with the 15 minute time frame). yea you are right the red herring threw me off, PROBABILITY brainteasers (Originally Posted: 10/07/2010). If you want a textbook, here's an old, but well written probability text:…, In reply to Book to learn how to answer tough probability questions? This covers all cases because this defines when X>Y, and it becomes an easy problem. Anyone figure by aparnaf2f106, In reply to For the 1st question, I get 0 by Anonymous Monkey (not verified). The easiest/quickest idea to pull out of this is that the expectation of a geometric random variable is just 1/p, where p is the probability that your event will happen in a given roll. That leaves you 9 spots to fill, and 48 choices to fill them with (since the 4 of the same card are already there). The probability that the couple has "a boy" is also 3/4. The eldest is a boy, what are the odds that your friend has a girl?". In reply to Probability review? 2 - represent number of reds blues and yellows with 'a', 'b' & 'c' where a>b>c. Probability Question (Originally Posted: 01/25/2008), a player has a 30% probability to score in a game. (Hoeffding's inequality) I'm looking for something textbook like, I already have the Crack book. It was only solved in 1994 by Andrew Wiles and Richard Taylor. ), P(4>3) = 49/94 (repeat logic from above but skewed by 2 cards out of 94). I had these two questions in one of the interviews: a) you have a 10x10 in cube of ice suspended in the air, it is made up of 1x1 in smaller cubes, when the ice starts to melt, the outer layer of cubes falls away, how many 1x1 cubes are left still together? I know I've already posted in your other thread but.. 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 3.5, EX. 5 4 / 36 (you aren't told what you drew in cards 1 2 or 3). the only scenario in which he loses more all games is FFF so that leave 5 success. Conditional expected value: 6, First roll outcome: 5 Anyone know how to do it? Y is a little bit more dfficult. So if x is the true probability of heads and x_n_trials is the empirical probability of heads after n flips of the coin, we want to show: P( max | x - x_n_trials | > d ) = O(1/d^2 * log( 1/(g*d) ) ). This would be zero. If you are looking for something about the history of risk and probability, this was an interesting read. You're given a loop of copper wire. yesman has it spot on. got it though after about 5 minutes of hard work and got the job. In reply to Prove that no three positive by blastoise, In reply to balbasur wrote: From here we can calculate the EV of the game 1/24*10 - 23/24 = -13/24. On an island there are snakes of 3 different colors. same color? 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. The answer is 1/4. That is, the expected hitting time to "6-6" = expected hitting time to "6" + expected hitting time to "6-6" from "6". solving for a we get 1/36a = 7/6 --> a = 42. Out of all the different combinations you can make, I assume 1/4th will end in clubs 1/4th will end in diamonds and 1/4th will end in spades and 1/4 will end in hearts. Rolling a "2" has a probability of 1/6. Treating n as continuous will give us a maximum at n = 3.5. Guess I'll just look over some of my old notes, good to know it's not going to be too bad. You're instructed to insert the wire loop into a machine that will make three random cuts (independently, uniformly distributed) along the wire. Don’t look at the solutions until you try each question! i prefer the former, but others can make the case for 60%. I had one of the hardest interview probability questions and it was just expectation however it was near impossible to structure the answer.