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24 Feb 2012
 3) Imagine you are out playing golf with 2 friends. You need to decide randomly, and fairly, the player to go first. You have a fair coin with you, which you can flip. How do you use the coin to decide who goes first, with an emphasis on minimizing the number of flips you need to make?4 AnswersYou group your friends as if they're one person, so if you lose, you can make another flip for just the both of them.That doesn't seem fair as you can win on the first flip but it would take one of your friends two flips to win. So chances of winning: You: 50%, Friend1: 25%, Friend 2: 25% The only fair way that I can think of is that you do it like this: 1. You vs Friend 1 2. Winner of #1 vs Friend2 If Winner of #1, wins this then they go first, otherwise 3. Loser of #1 vs Friend2 If Friend2 wins flip #2 and #3 then they go first If everyone wins 1 flip each then you start again. This will happen 25% of the time. When you factor this in the above method finds a winner after 4 flips (3+3/4+3/16+.....) whilst giving everyone an equally fair opportunity to win.I should clarify: That method finds a winner after 2 flips 50% of the time, after 3 flips 25% and ties after three flips the other 25% of the time. On average it will take 4 flips to win although this will never happen as, if it is tied after 3 flips, then it will take at least 2 more to determine a winner.Show more responsesFlip the coin twice and record the result of both tosses. We'll let the sequence of flips - HH - represent you winning the coin toss. Let the sequence - TH - represent Friend 1 winning and let the sequence - TT - represent Friend 2 winning. Notice that everybody has an equal chance of winning. If the coin tosses result in the sequence HT, repeat the process and flip the coin twice again. This will take an average of 8/3 coin tosses. To find this - let x be the expected number of coin tosses this algorithm will take. Then x = 3/4 * 2 + 1/4 * (x + 2). Solving this yields x = 8/3.

24 Feb 2012
 2) Given a 100 floor building, and 2 identical eggs (with an unknown capacity to survive a fall from any floor up to floor "x" of the building), how would you minimize the number of drops you need to make in order to determine the highest floor the eggs can be safely dropped from?3 Answersyou should go by the window, and drop both eggs simultaneously, with one hand outside the window and one hand within the building.Drop the first egg from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100... (i.e. move up 14 then 13, then 12 floors, etc) until it breaks (or doesn't at 100). The second is used to search floor by floor until it breaks!Let's assume we can minimize the number of egg drops N with the following strategy. 1. Start from ground floor 2. Go up x floors. 3. Drop first egg. 4. If first egg does not break, continue from step 2 5. If first egg breaks, go back to the last floor where the first egg did not break, and drop the second egg floor by floor to determine the highest floor. By applying this strategy, we can guarantee for max number of drops N <= 100/x + x where the first term is the number of drops of first egg and likewise for the second egg. We can now use basic calculus to minimize N in terms of x, and ensuring that our resulting x is a discrete number of floors. Setting (dN/dx) equal to zero, we get an exact solution x = 10 floors for our strategy. This would give a guaranteed N <= 20 for the number of drops based on the formula we chose. Although, considering the top floor cases where the last drop is not necessary, the max number of drops is actually N = 18. (Accounting for constant terms in N(x) formula would not change the calculation of x = 10)

13 Sep 2012
 There’s a 60% chance it will rain on Saturday AND Sunday. What is the probability that there will be rain on the weekend.3 AnswersTake the square root of 0.6 to give the individual probability on Sat or Sun = 0.7746. The chance of it being dry all weekend is therefore one minus this number squared = 0.051, so the chance of rain on the weekend is 95%.Your answer assumes that the probability of it raining on sat or sunday is identical*. Should be stated in question. Ie if it is more likely to rain on sunday than saturday than it changes probabilities....such as probability is 0.9 it rains on saturday and 0.6667 on sunday.Let the probability it rains on Saturday be p and the probability that it rains on Sunday be q Probability it rains on Sat AND Sun = p*q = 0.6 Also, we know that there are 4 scenarios Rain on Sat, No Rain on Sun Rain on Sat, Rain on Sun No Rain on Sat, Rain on Sun No Rain on Sat, No Rain on Sun The sum of probabilities of these 4 cases = 1 Therefore p*(1-q) + p*q + (1-p)*q + (1-p)*(1-q) = 1 p - p*q + p*q + q - p*q + 1 - p - q + p*q = 1

### Spring Insight Sales, Trading and Research at J.P. Morgan was asked...

20 Nov 2014
 What's the name of JPMorgan's CEO? 2 AnswersI had read his name 10 times before the interview, but I forgot it when I actually needed it.Dimon

31 Aug 2015
 With one die, suppose in a round, you earn the amount of dollars equal to the value of the upwards face of the die. Now also suppose after your first roll, you are given the opportunity to cancel your first and roll again, taking that value as the final value. What should your strategy be?2 AnswersExpected Value of next roll is 3.5. If current die less than or equal to 3, roll again. Else, take the current payoff.expected value of one roll 3.5 expected value to role again: 0.5*3.5+ 0.5 * 5 = 4.25

21 Jul 2018
 How would you feel with underperforming team2 AnswersHow would you handle putting together a report of recommendations for another team at the last minuteProactive

4 Jan 2011
 Do u prefer to be a leader or a follower? 1 AnswerI prefer to be in a position where I am trusted to make good decisions and to follow through on expectations. I understand that there will be situations where I am the most knowledgeable or best equipped to address the situation. As an example, I am responsible for reviewing our RFPs. This process requires that I know the RFP inside and out, so I will lead these discussion as well as assign others tasks and deadlines. But, when it comes to partnering or deciding how exactly to respond, our executive leadership should be directing that effort. In that situation, it's my job to ensure that our response aligns with the comapny's strategic goals. It's important to lead confidently, but always be open to learning from others.

7 Jan 2013
 You are going to see a sports tournament between two teams. Best of 7. Can only have win/loss outcomes. Devise the optimum betting strategy such that at the end of the tournament you are either up 100 or down 100. Was initially confused about starting money but he told me that you can bet any amount. 2 AnswersRight, so you were told you can bet any amount of money........surely you can just bet 100 on the first game then you'll either be up or down by 100 (as required) whatever the outcome, then for the rest of the games just bet 0 and nothing will change......It is Best of 7, meaning you should win 100 dollar if your home team won the tournament and lose 100 dollars otherwise. Just think through all possible cases.... BACKWARDS, like in pirates and golds problem. Step 1: How much you should bet on last game? Well if you team has won 4, 5 or 6 games so far, you are already up 100 dollars and therefore bet 0 dollars. If you team has won 0, 1 or 2 games you are already down 100 dollars and bet 0 dollars as well. So the only case left is you team has tied with score 3 : 3. If now you are up X dollars from initial amount and bet Y dollars, you will be up X+Y if your team won or up X - Y if your team lose. now X+Y=100 and X-Y = -100 by condition; meaning X=0 and Y=100. Step 2: apply the same reasoning. Carry this one until you reach step 7, which should tell you how much to bet in game 1. I am too lazy to carry it out though.