Probability that a selection from a random set from a multivariable equation $z=f(x,y)$ is the maximum, given $y$.

Question

I was watching a Veritasium video about the effect of luck and skill on success, and to test, he did the following:

He randomly generated values for luck and skill (say from 0 to 1). He then added them, such that the sum $z=0.95S+0.05L$. He did this for a large number of samples, and then found that the top 11 samples had an average luck score of 0.947.

So I wanted to investigate this analytically in the following manner:

1. Have an equation $z=0.95x+0.05y$ (x,y are replacements for S and L)
2. For a sample $S$ of n values drawn from this equation $z$, such that $x\in[0,1], y\in [0,1]$, what is the probability that the number, $s$, chosen from this sample is the maximum value of this sample, given that $y=k$. That is, find $Pr(s=S_{max}|y=k)$

• Note: when drawing values from the equation, the choices of $x$ and $y$ follow a uniform distribution from $[0,1]$..

3. I then want a theoretical graph where the y-axis is this probability, and the x-axis is $k$, (k ranging from 0 to 1).

Perhaps this graph/equation is given by some equation.

How do I find this graph (Analytically not numerically).

EDIT : The sample size n is not a parameter in this instance, it is just for the sake of visualisation. As we are looking at expected probabilities, $n$ does not make a difference

Answer 1

We have $f_{0.95X}(x) = 1/0.95 \cdot 1_{[0,0.95]}(x)$ and $f_{0.05Y}(x)=20 \cdot 1_{[0,0.05]}(x)$. Assuming $X$ and $Y$ are independent, the pdf of $Z$ is:

$$f_Z(t) = \int_0^t f_{0.95X}(y)f_{0.05Y}(t-y) dy = \begin{cases} \frac{20}{0.95}t & \text{if } 0 \leq t \leq 0.05 \\ \frac{1}{0.95} & \text{if } 0.05 \leq t \leq 0.95 \\ \frac{20}{0.95}(1-t) & \text{if } 0.95 \leq t \leq 1 \\ 0 & \text{otherwise.} \end{cases}$$

Integrate this function to get the cdf: $$F_Z(t) = \begin{cases} 0 & \text{if } t < 0 \\ \frac{10}{0.95}t^2 & \text{if } 0 \leq t \leq 0.05 \\ \frac{1}{40 \cdot 0.95} + \frac{t-0.05}{0.95} & \text{if } 0.05 \leq t \leq 0.95 \\ \frac{400}{19}(t-\frac{1}{2} t^2) -\frac{181}{19} & \text{if } 0.95 \leq t \leq 1 \\ 1 & \text{otherwise.} \end{cases}$$

The probability that you ask for is $g(k) = P(0.05k+0.95X \geq \max\{Z_1,Z_2,\ldots,Z_{s-1}\})$. where $X$ and $Z_i$ are iid. You can compute this probability as: $ \int_0^1P(0.05k+0.95x \geq \max\{Z_1,Z_2,\ldots,Z_{s-1}\})f_x(x) dx$, which is $ \int_0^1 F_Z(0.05k+0.95x)^{s-1} f_x(x) dx.$ I will simplify this integral if/when I have more time.