In a previous post I computed the probability of the robot long jump attempt scoring 0 under the assumption that the jump threshold was 0.5. In fact this is not the optimal jump threshold. In this post we express the probability of getting a 0 as a function of some variable jump threshold c, and then find the optimal c.
Suppose S is the sample space. Consistent with the previous post, all elements of S are sequences of the form $d_1d_2d_3...[j]d_n$. A sequence s is invalid if $d_1 + d_2 + ... + d_i > 1$ and $d_1 + ... + d_{i-1} < c$ where c is the jump threshold. Now we do the same thing we did in the last post, except this time keeping the jump threshold has a variable c:
$$ P(\text{s is an invalid sequence)} = P(\text{s is an invalid sequence of length 2})\lor P(\text{s is an invalid sequence of length 3})\lor P(\text{s is an invalid sequence of length 4})\lor...\lor P(\text{s is an invalid sequence of length n}) $$
Probability of getting an invalid sequence of length 2 is the probability of the sum of two uniform continuous random variables $D_1 \in U[0,c]$ and $D_2 \in U[0, 1]$ having a sum greater than 1 multiplied by the likelihood of $D_1$ being between 0 and c. The PDF of $D_1$ is $f_{d_1}(x) = 1/c$ with a domain from 0 to c and $f_{d_2}(x) = 1$ with a domain from 0 to 1. Remember that $f_{d_1 + d_2}$ is the convolution of the two PDFs. Convolution produces the following piece-wise function:
$$ \begin{equation*} f_{d_1+d_2}(z)=\begin{cases} z/c \quad &\text{if } \, 0<x<c \\ 1 \quad &\text{if } \, c\le x\le1 \\ 1 - \frac{(z-1)}{c} \quad &\text{if } \, 1<x<1+c \\ \\ \end{cases} \end{equation*} $$
Therefore the probability of getting a sum greater than 1 is $\int_{1}^{1+c}((1-z+c)/c)dz = c/2$. The probability of $D_1$ being between 0 and c is c itself. Therefore the probability of getting an invalid sequence of length 2 is $c^2/2$.
Using the same reasoning as for sequences of length 2, we get
$$ \begin{equation*} f_{d_1+d_2+d_3}(z)=\begin{cases} z^2/c^2 \quad &\text{if } \, 0<x<c \\ 1 \quad &\text{if } \, c\le x\le1 \\ 1 - \frac{(z-1)^2}{c^2} \quad &\text{if } \, 1<x<1+c \\ \\ \end{cases} \end{equation*} $$
Therefore the probability of getting a sum greater than 1 is $2c/3$. The probability of the sum of $d_1+d_2$ being less than c is $c^2/2$. Therefore the probability of getting an invalid sequence of length 3 is $2c/3*c^2/2 = c^3/3$.
$$ \begin{equation*} f_{d_1+d_2+d_3+d_4}(z)=\begin{cases} z^3/c^3 \quad &\text{if } \, 0<x<c \\ 1 \quad &\text{if } \, c\le x\le1 \\ 1 - \frac{(z-1)^3}{c^3} \quad &\text{if } \, 1<x<1+c \\ \\ \end{cases} \end{equation*} $$
The probability of getting an invalid sequence of length 4 is $c^4/8$. Therefore the probability of getting an invalid sequence of length n is:
$$ \frac{c^n}{n*(n-2)!} $$
Therefore we get
$$ P(\text{attempt scores 0}) = \sum_{n=2}^{\infty} \frac{c^n}{n*(n-2)!} = e^c(c-1) + 1
$$