By solving the problem through manipulating the pirates desire to be the only one left alive, there is an underlying assumption that only one pirate is necessary to sail a large vessal at sea, which is not the case, and even pirates know that. If a person is working in a commission environment, most of the time a salesperson wants and needs to be number one, and is expected to be so. Otherwise the 3rd and 5th will probably not receive any gold. Now you may think this is good, but the odds of one of the 4 of you being alive and being the voter is 1 in 4 or 25% (see mister interviewer, I can to simple math). As Interview Candidate said, this is clearly a question designed to explore your problem solving skills rather than to elicit a specific unitary answer. Also - as some others have pointed out, this probably isn't the right question. Thus, a generic technocratic solution (considering that I am applying for a job with a technology company) to cater to all parameters is sought here. That however is a biased estimator. Let's say I'm the captain, and my pirates are 'greedy'. The time value of money is represented by the risk-free (rf) rate in the formula and compensates the investors for placing money in any investment over a period of time. Me: I'd be number 15, because I probably helped 1-14 that's why they like me so much. As stated the answer is: 25 gold pieces should go to any three pirates, with the leader keeping the remaining 25 for himself. We pose our standard question: "If the captain loses the vote, who gets shafted?" Let's say pirates are in nature greedy. The leader can keep 98 of the gold coins and give 1 to the 2nd ranked and 1 to the 4th ranked pirates. Therefore: I keep 22 Gold, and I give pirates 1,2,3 each 26 GOLD. Alternately, one could use a boolean array of length n with all values set to false and then for each value, set array[val - 1] to true. Often the problem or inquiry is based on the real … They are generally more interested in sales results, period. BUT, if it only takes 2 Votes (i.e, my vote counts), then I keep 34 Gold, and give pirate 2 and 3 33 Gold each. This ensures that only milk is poured and kept in the container. A return over six months, for example, cannot be compared to a return over 12 months. However, nothing says you need to play this game. It doesn't matter what the other two pirates who were left out think because he has the two votes he needs. Option #1: I give myself and 4 pirates 20 gold each. Discuss the runtime of the algorithm and how you can be sure there won't be any infinite loops. If one of the pieces is bigger than the 0.5, then the triangle could not capture the center so the answer is 1/4. Pirate 1 only needs two other supporters. In no average case are they better off than taking your 26-33 piece offer and the certainty of life. I think it should be (N+1)/N * max(X1,..XN). But ask the one on the floor to pull the other down with him, he can do it easily. it's a machine question, not a human question. Yargh, but then me and me accomplice, the man without any gold, we shivved all the rest and threw the bodies overboard, each claiming 50 gold, yar! Consider this, being #1 is definitely what anyone wants (mainly because the sense of reward for being n.1 is attractive $$$), but deciding on being hated or loved hugely depends on the social needs of one, according to Maslow's theory these needs (the need for love and appreciation) must be achieved to gain real fulfillment in life. All buccaneers vote on the proposal, and if half the crew or more go "Aye", the loot is divided as proposed, as no buccaneer would be willing to take on the captain without superior force on their side. I think I get it. I would prefer #15. I'd analyse the tail of cumulative density function. Subtracting from 5050 is an elegant solution, but not obvious as to why it works. 1/pi * integral from 0 to pi (x/pi). I was hated in the position I was in by most because I followed the rules and reminded (quite regularly) the others in the group to do the same. If that includes my boss and people above me, then #15. Apologies, but both of the above answers are incorrect. So is the question a conundrum or a fallacy? I came up a solution: if we know the distribution of actual measurement and the value of actual measurement, then the expected probability of getting wrong measurement should equal to the probability of actual measurement greater than length of ruler.