## Credit Suisse Interview Question for Analysts

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Country: India
Interview Type: In-Person

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If I understand correctly, the expected pay is just the expected distance of the maximum of the 2 throws. To find the probability distribution of the maximum, it's easier to start from the CumulativeDistributionFunction, which is just a product of 2 identical probabilities, because the shots are independent.
I understand that the the distribution is uniform over the circle, and NOT over the distance from origin alone, so the CDF is simply :

P(max{r1,r2} < R) = P(r1 < R) * P(r2 < R) = [P(r1 < R)]^2 = [Area(r-Circle)/Area(unit-circle)]^2 = [r^2]^2 = r^4

so p(r) = ∂rP(r) = 4r^3

and <pay> = <max> = E[r] = ∫r∙4r^3dr = 4/5 = 0.8

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Let the distances from center at which the darts lend be random variables X_1 and X_2. They are independent and uniformly distributed over [0,1].
F(r)=P[max{X_1, X_2}<r]= P[X_1<r]*P[X_2<r]
= r*r
= r^2
Expected Value= integration of survival function(1-F(r))
E=integral 0 to 1 (1-r^2)dr
E=2/3

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(Assume all throws fall inside a circle with radius r = 1)

With uniform probability, all darts can fall anywhere on the area of the unit circle with the same probability. I’d like to think of the problem as a monte-carlo integration problem. After 1000s of simulations (dart throws), if we split the area of the circle in two equal half, with an inner circle of radius e (where intuitively r > e > 0.5r), e would be the expected, average payoff.

A = PI * r^2
0.5 A = PI * e^2
r = 1
e = sqrt(0.5) = 0.707107

Comment hidden because of low score. Click to expand.
0
of 0 vote

(Assume all throws fall inside a circle with radius r = 1.)

With uniform probability, all darts can fall anywhere on the area of the unit circle with the same probability. I’d like to think of the problem as a monte-carlo integration problem. After 1000s of simulations (dart throws), if we split the area of the circle in two equal half, with an inner circle of radius e (where intuitively r > e > 0.5r), e would be the expected, average payoff.

A = PI * r^2
0.5 A = PI * e^2
r = 1
e = sqrt(0.5) = 0.707107

Comment hidden because of low score. Click to expand.
0
of 0 vote

(Assume all throws fall inside a circle with radius r = 1)

With uniform probability, all darts can fall anywhere on the area of the unit circle with the same probability. Think of the problem as a monte-carlo integration problem, where after 1000s of simulations (dart throws). We can split the area of the circle in two equal halfs using a concentric inner circle of radius e (where intuitively r > e > 0.5r). In this case, e would be the expected average payoff.

A = PI * r^2
0.5 A = PI * e^2
r = 1
e = sqrt(0.5) = 0.707107

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