$X = x_1, x_2, x_3$
$Y = y_1, y_2, y_3$
...
For any person $X$ in this list:
- $x_1$ is the time he takes to do the work,
- $ \dfrac{x_2}{x_3}$ is the probability he gets the work done.
Consider this example, only $2$ persons:
$X = 2, 1, 4$ and $Y = 5, 2, 3$
- $X$ takes $2$ hrs and $\dfrac{1}{4}$ chances he does the work
- $Y$ takes $5$ hrs and $\dfrac{2}{3}$ chances he does the work
If I choose $X$,
- $X$ does the work, time taken = $2$
- $X$ does not do the work, time taken = $2 + 5$
I will have to resort to $Y$ to do the work after having wasted time with $X$, hence $Y$'s time has to be added.
Total time taken = $2\times\dfrac{1}{4} + (2+5)\times\dfrac{3}{4}$
Similarly, if I choose $Y$ first,
Total time taken = $5\times\dfrac{2}{3} + (2+5)\times\dfrac{1}{3}$
If there were more persons to choose from, I will have to add their time too.
So choosing $Y$ from $X, Y, Z$
Total time taken = $5\times\dfrac{2}{3} + (2+5+Z$'s time$)\times\dfrac{1}{3}$
How can I generalize a formula for calculating the total time taken, if I choose a person?
I came up with this:
Time taken if I choose a person $X$ from a list of $P$ persons to do the work:
$x_1 \times \dfrac{x_2}{x_3}\ + T\times\dfrac{x_3-x_2}{x_3}$
where,
- T is sum of individual time of all persons, a constant,
- $x_2 < x_3$, $x_1 > 0$
- All numbers are positive integers.
Is this correct?
For what values of $x_1, x_2, x_3$ can it fail?
Suppose there is a total of $n$ people, and you have already specify the working order from person $1, 2, \ldots$, up to $n$. Let
Assuming each person decision is independent, then the probability that the work is finally done by the $i$-th person is $$ \left[\prod_{j=1}^{i-1}(1 - p_j)\right]p_i, i = 2, 3, \ldots, n$$ The total working time if the work is finally done by the $i$-th person is $$ \sum_{j=1}^i t_j$$
As a result, the expected total working time is given by $$ p_1t_1 + \sum_{i=2}^n \left[\prod_{j=1}^{i-1}(1 - p_j)\right]p_i\sum_{j=1}^i t_j$$