Quantitative Researcher Interview
You throw 1000 darts. Each one has a 50% chance to score. For the first 500 darts each is worth 1 point, for the second 500 darts each is worth 3 points. If you score 1500 points. Most likely how many 3point darts have you scored.
The number of ways of picking k 3point darts to hit is 500Ck, and then you need 1500-3k 1 point darts to hit. The number of ways fulfilling the required conditions is then (500 Choose k)*(500 Choose 1500-3k). You can clearly see that k needs to be bigger than 333, and making a plot of the function shows that k=398 gives the maximum probability.
TK answer is right, but I think they probably want you to approximate with Gaussians. Ignoring a lot of constants, and the standard deviations since they don't matter, the distribution for x 1pt hits and y 3pt hits is: p(x,y) ~ exp(-(x-250)^2-(y-250)^2) log likelihood: l(x,y) ~ -(x-250)^2-(y-250)^2 use x+3y=1500 l(y) ~ -(1250 -3y)^2 - (y-250)^2 l'(y) = 6(1250-3y) -2(y-250) Find maximum: 6*1250 - 18y = 2y -500 y = 8000/20 = 400 This way you get 400 3pt hits and 300 1pt hits, which is pretty close to the real answer 398/306, and doesn't require evaluating massive binomial coefficients over the phone.
average score per hit: (1+3)/2=2, then expected hits: 1500/2=750, each gets 750/2=375 hits
This solution is for calculating the expected number of 3 point darts scored. If `most likely' means highest probability then the previous solution is the correct one. Let X and Y be number of successes among the two types of darts. X and Y are independent binomially distributed random variables. We need to find E[ Y | X+3Y] Since X and Y are identically distributed, E[ X | X + 3Y] = E[ Y | X + 3Y] Using the linearity of expectation on E[ X + 3Y | X + 3Y ] = X+3Y, we obtain E[ X | X + 3Y ] = E[ Y | X + 3Y ] = (X+3Y)/4 = 1500/4 = 375
The end of the last line of my previous post dropped some characters. It is 1.5 (.5) 500 = 375 hits of 3-point darts.
Jaime Cuevas Dermody on
The 375 answer is correct but the computation involved more structure than necessary. Here is the same logic in a simpler context. Given a 50% chance of scoring 1-point darts and the same for 3-point darts, the expected (mean) value of total points is: half the 500 1-point darts score plus half the 500 3-points darts score. The expected total points is: 1/2 (500) 1 point + 1/2 (500) 3 points = 1000 points. If 1500 point were actually scored in total, then the equality of distribution among the 1-point and the 3-point scoring implies you scored 1500/1000 = 1.5 times the means. That implies 1.5 (.5) 500 (1 point) + 1.5 (.5) 500 (3 points) = 1500 points. So you scored 1.5 (.5) 375 hits of 3-point darts.
Jaime Cuevas Dermody on
Answer is 375 each: We know that P(x) = P(y); The chances of hitting x is equal to that of y is given. We also know that x+3y = 1500 points. After algebra we get: 1500 = 4y, divide it out and y = 375, making x also = 375