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HHT is more likely to appear first than HTT. The probability of HHT appearing first is 2/3 and thus the probability of HTT appearing first is 1/3. Indeed, both sequences need H first. Once H appeared, probability of HHT is 1/2 (b/c all you need is one H), and probability of HTT is 1/4 (b/c you need TT). Thus HHT is twice is likely to appear first. So, if the probability that HTT appears first is x, then the probability that HHT appears first is 2x. Since these are disjoint and together exhaust the whole probability space, x+2x=1. Therefore x=1/3. Moins

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There is symmetry between red and black. Each time you pull a card it is equally likely to be red or black (assuming you haven't looked at the previous cards you pulled). Thus no matter when you guess you odds are 50% and the expected return should be 50 cents. Moins

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Ok grazie,io ho fatto il colloquio orale tecnico giovedì scorso e non ho ancora ricevuto l'esito. Moins

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Simply saying "binary search tree" does not solve this problem in its entirety because of the five week time limit. If you just naively used a binary search, you'd take four weeks for the first iteration alone (the problem says the poison's effects take a month to surface, so that definitely wouldn't work. 10 prisoners lined up. Number the bottles from 0 - 999. Now, divide the bottles into blocks of powers of 2. Prisoner 9 drinks from 0 - 511. Prisoner 8 drinks from 0 to 255 and 512 to 999. Prisoner 7 drinks from 0 to 127, 256 to 383, 512 to 639, 768 to 896 Prisoner 6 drinks from 0 to 63, 128 to 191, 256 to 319, 384 to 447, 512 to 575, etc. Prisoner 1 drinks from every second bottle A month later, line up the Prisoners and treat the dead ones as 1s and the living ones as 0s - you'll have the answer in binary and find the dreaded bottle! Moins

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This is all very interesting and I'm sure has some application...but to trading? I don't think so. I own a seat on the futures exchange and was one of the largest derivatives traders on the floor. Math skills and reasoning are important but not to this level. I would associate day trading/scalping more to race car driving i.e. getting a feel for what's going on during the day, the speed at which the market is moving and the tempo for the up and down moves. If I were the interviewer at one of these firms, I throw a baseball at your head and see if you were quick enough and smart enough to duck. Then if you picked it up and threw it at my head I'd know that you had the balls to trade. I know guys who can answer these questions, work at major banks, have a team of quants working for them and call me up to borrow money from me because they're not making money. At the end of the day, if you want to be a trader then...be a trader. If you want to be a mathematician then be a mathematician. It's cool to multiply a string of numbers in your head, I can do this also, but never in my trading career did I make money because in an instant I could multiply 87*34 or answer Mensa questions which...realistically the above answer is: it depends on the market as the market will dictate the price. You may want to charge $87 to play that game but you'd have to be an idiot to play it. In trading terms this means that when AAPL is trading at $700 everyone would love to buy it at $400. Now that it's trading at $400 everyone is afraid that it's going to $0. Hope this helps. No offense to the math guys on this page, just want to set the trading record straight. Moins

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chainsaw is right - the question is subtle. Outcome space is (b1,b2), (b1,g2), (g1,b2) and (g1,g2). Now - if you HAVE SEEN the girl (specific girl let's say girl 1) that removes outcomes (b1,b2)&(b1,g2) leaving us with p=1/2 of choice between outcomes (g1,b2) and (g1,g2). If you KNOW (was told) that one is a girl, but do not know which one, than only (b1,b2) is removed from outcome space and the chance is 2/3. Moins

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Let me show you a mathematical approach. Common sense dictates that the minute hand moves at a faster rate of 5.5 degrees a minute (because the hour hand moves 0.5 degrees a min and the minute hand moves 6 degrees a minute). We start at 12 midnight. The hands are together. For subsequent 90 degree angles to occur, the minute hand must "overtake" the hour hand by 90 degrees, then 270 degrees, then 360 + 90 degrees, then 360 + 270 degrees, then 360 + 360 +90 degrees.. and so on. This can be re-expressed as: (1)90, 3(90), 5(90), 7(90), 9(90), 11(90)... n(90). The number of minutes this takes to happen can be expressed as (1)90/5.5, 3(90)/5.5, 5(90)/5.5, 7(90)/5.5, 9(90)/5.5, 11(90)/5.5... n(90)/5.5. In one day, there are 24 hr * 60 mins = 1440mins To find the maximum value of n, n(90)/5.5 = 1440 n = 88 but as seen from above, n must be an odd number (by pattern recognition and logic) hence n must be the next smallest odd number (87) counting 1,3,5,7,9,11......87, we see that the number of terms = (87-1)/2 +1 = 44. In other words, the minute hand "overtakes" the hour hand on 44 occasions in 24 hours in order to give a 90 degree angle. Therefore the answer to your question is 44. Moins

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The correct answer is 3/4, as this problem is equivalent to the famous 3-points-on-semicircle problem. Why? If one of the pieces has length greater than 1/2 the circumference, then the three points of cutting must lie in the same semicircle. On other hand, if the three points of cutting lie on the same semicircle, then the longest piece must be at least 1/2 of the circumference. Moins

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