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The way I approached it was as follows: Starting from the initial state, one player has to win 2 points in a row to win. If that does not happen, we're back to the initial state. Therefore, P(A) = 0.49, P(B) = 0.09, P(repeating) = 1 - P(A) - P(B) = 0.42 I enjoy using the intuition that A is 49/9 more likely to win and then find the probabilities by dividing both by 0.42 (making 49 and 9 add up to 100 while preserving their proportions). Grats on the JS offer! Less
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Shouldn't it be 0.7?
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84.5 % (sum of 0.49x(0.42)^n)
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5/36
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5/36 is right.... Combinations that add up to six: 1 and 5, 2 and 4, 3 and 3, 4 and 2, 5 and 1. So we have 5 successes. There are 36 individual outcomes because there are 6 sides on each die and 6*6 = 36 So we have 5/36. Less
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1/6
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12
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Actually, +/- 12
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Hey, could you please tell me more about the final round interview at their office? Less
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Chance of 1 Heads = .5 Chance of 7 heads in a row = (.5)^7 to do this in your head think of fractions... (1/2)^7 = (1/2*1/2=1/4, * 1/2=1/8, * 1/2 = 1/16....if you do it seven times you get 1/128. so you have a better chance to pick out the penny with two heads (1/100) than to get 7 heads in a row (1/128) Less
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99 times out of 100 you take the normal penny. prob that normal penny give 7 heads in a row is 1/128. so, prob that he has special penny is 1/100 / (1/100 + 0.99*1/128) = 56.2% Less
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~56.4% that you have the 2-headed penny.
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You’re making a profit. Selling at a higher price than what you bought
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The simple answer is that if you are short on a contango curve then you are losing money, for the curve is sloped positively. Less
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1+2+3+4+5+6+7+8+9+10=55 55/10 = 5.5 = Average Expected Winning Price it higher then that number to be profitable Less
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Whatever will maximize the value from your customers.
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Don't say you weren't interested in your subjects, because then "That means you're not hardworking!" After keeping the CG criteria 6, and clearing their round 1, they were not satisfied with your answer. If CG was such a problem, should've increased the minimum criteria Less